what are the solutions to the quadratic equation (5y+6)2=24

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Page No 34:

Question 1:

Write any ii quadratic equations.

Answer:

2 quadratic equations are
${\mathrm{10}}^2+\mathrm{ten}\mathrm{ten}-200=0$ and$x^2+5x-6=0$.

Page No 34:

Question 2:

Determine which of the post-obit are quadratic equations.
(1) ten ⁱⁱ + 5ten – two = 0
(ii) y ⁱⁱ = 5y – x
(three) $y^2+\frac{1}{y}=2$
(4) $x+\frac{1}{\mathrm{ten}}=-\mathrm{ii}$
(5) (m + 2) (1000 – 5) = 0
(six) m ⁱⁱⁱ + 3m ^two – 2 = iiim ³

Reply:

(1)x ² + 5x– 2 = 0
Just one variable x.
Maximum index = two
So, it is a quadratic equation.

(2)y ²= 5y – 10
Merely one variable y.
Maximum index = 2
So, it is a quadratic equation.

(3)

$y^{\mathrm{two}}+\frac{\mathrm{one}}{y}=2$
$\Rightarrow y^3+1=2y$

Only one variable y.
Maximum alphabetize = 3
And then, information technology is not a quadratic equation.

(4)

$\mathrm{ten}+\frac{1}{\mathrm{10}}=-2$
$\Rightarrow {\mathrm{ten}}^2+\mathrm{i}=-2\mathrm{10}$

Only i variablex.
Maximum alphabetize = ii
And then, information technology is a quadratic equation.

(5) (m + two) (m– five) = 0

$\Rightarrow k^{\mathrm{two}}-\mathrm{three}\mathrm{grand}-10=0$

Just one variable thou.
Maximum alphabetize = ii
So, it is a quadratic equation.

(6)grand ³+ 3thou ² – 2 = threegrand ³
Only one variable 1000.
Maximum index = 3
So, it is not a quadratic equation.

Folio No 34:

Question 3:

Write the following equations in the form ax ² + bx + c= 0, and then write the values of a, b, c for each equation.
(i) 2y = 10 – y ²
(2) (ten – i)ⁱⁱ = twox + 3
(3) x ² + 5x = –(3 – ten)
(four) threechiliad ² = 2m ² – nine
(5) P(3+sixp) = –v
(half dozen) 10 ² – 9 = 13

Respond:

(1) iiy = 10 –y ^{2
$$\begin{gathered} \Rightarrow y^{\mathrm{two}}+\mathrm{ii}y=\mathrm{ten} \\ \Rightarrow y^2+2y-10=0 \end{gathered}$$}
And so,it is of the formax ² +bx +c= 0 where a = one, b = 2 and c = −10.

(two) (x – 1)² = twox + 3

$$\begin{gathered} \Rightarrow \left(x^{\mathrm{two}}-2\mathrm{ten}+\mathrm{one}\right)=2\mathrm{10}+3 \\ \Rightarrow x^2-4x-\mathrm{ii}=0 \end{gathered}$$

So,it is of the formax ² +bx +c= 0 wherea = one,b = −4 andc = −2.

(3)x ² + 5x = –(3 – 10)

$$\begin{gathered} \Rightarrow x^2+5\mathrm{ten}+3-\mathrm{10}=0 \\ \Rightarrow x^2+4\mathrm{ten}+3=0 \end{gathered}$$

Then,information technology is of the formax ^two +bx +c= 0 wherea = 1,b = 4 andc = 3.

(4) threem ² = 21000 ² – 9

$$\begin{gathered} \Rightarrow 3m^2-2m^2+9=0 \\ \Rightarrow m^2+0m+\mathrm{nine}=0 \end{gathered}$$

Then,information technology is of the formax ^two +bx +c= 0 wherea = 1,b = 0 andc = 9.

(5) p(3+6p) = –5

$$\begin{gathered} \Rightarrow 3p+\mathrm{half\; dozen}{p}^{\mathrm{two}}+\mathrm{v}=0 \\ \Rightarrow 6p^2+\mathrm{three}p+5=0 \end{gathered}$$

So,information technology is of the formax ⁱⁱ +bx +c= 0 wherea = 6,b = 3 andc = v.

(6)ten ² – 9 = 13

$$\begin{gathered} \Rightarrow {\mathrm{ten}}^2-9-\mathrm{thirteen}=0 \\ \Rightarrow {\mathrm{10}}^{\mathrm{two}}-22=0 \\ \Rightarrow x^{\mathrm{two}}+0x-22=0 \end{gathered}$$

So,it is of the gradeax ⁱⁱ +bx +c= 0 wherea = 1,b = 0 andc =  −22.

Folio No 34:

Question 4:

Determine whether the values given confronting each of the quadratic equation are the roots of the equation.
(1) x ^two + 4x – 5 = 0 , x = 1, –1
(2) 2g ² – fiveg = 0, $m=\mathrm{ii},\frac{\mathrm{five}}{\mathrm{two}}$

Answer:

(1) x ² + 4x – v = 0 ,ten = one, –1

$$\begin{gathered} Forx=\mathrm{i} \\ {\left(1\right)}^2+\mathrm{four}\left(\mathrm{ane}\right)-\mathrm{five}=0 \\ \Rightarrow 1+4-5=0 \\ \Rightarrow 5-5=0 \end{gathered}$$

So,x = 1 is a solution of the given equation.
Forx = –one

$$\begin{gathered} {\left(-\mathrm{i}\right)}^{\mathrm{ii}}+4\left(-\mathrm{i}\right)-\mathrm{v}=0 \\ \Rightarrow 1-4-\mathrm{five}=0 \\ \Rightarrow -\mathrm{iii}-\mathrm{v}=0 \\ \Rightarrow -8\ne 0 \end{gathered}$$

So,x = –1 is non a solution of the given equation.
Thus, justx = i is root of the given equation.

(ii) 2m ⁱⁱ – 5m = 0,

$m=\mathrm{two},\frac{5}{2}$
$$\begin{gathered} Whenk=2 \\ 2{\left(2\right)}^2-5\left(\mathrm{two}\right)=0 \\ \Rightarrow \mathrm{viii}-\mathrm{ten}=0 \\ \Rightarrow -2\ne 0 \end{gathered}$$

So, m= 2 is not a solution of the given equation.

$$\begin{gathered} \mathrm{When}k=\frac{5}{2} \\ \mathrm{ii}{\left(\frac{5}{\mathrm{ii}}\right)}^{\mathrm{ii}}-5\left(\frac{5}{2}\right)=0 \\ \Rightarrow \frac{25}{\mathrm{two}}-\frac{25}{2}=0 \end{gathered}$$

So,

$m=\frac{\mathrm{v}}{2}$

 is a solution of the given quadratic equation.
Thus, simply

$g=\frac{5}{\mathrm{ii}}$

is a root of the given quadratic equation.

Folio No 34:

Question five:

Find chiliad if x = iii is a root of equation kx ² – xx + 3 = 0

Answer:

Givenx  = iii is a root of equation kx ⁱⁱ  – x x  + 3 = 0
And then, ten  = 3 must satisfy the given quadration equation.
$$\begin{gathered} \mathrm{chiliad}{\left(3\right)}^2-10\left(3\right)+3=0 \\ \Rightarrow \mathrm{nine}\mathrm{grand}-\mathrm{xxx}+\mathrm{three}=0 \\ \Rightarrow \mathrm{nine}k-27=0 \\ \Rightarrow 9\mathrm{thousand}=27 \\ \Rightarrow k=\frac{27}{\mathrm{nine}}=3 \end{gathered}$$
Thus, k= 3.

Page No 34:

Question half-dozen:

One of the roots of equation vone thousand ² + 2m + grand = 0 is $\frac{-7}{5}$ . Complete the following activeness to detect the value of 'thou'.

Answer:

$\boxed{\frac{-\mathrm{vii}}{5}}$ is a root of the quadratic equation 5m ² + iithou +g= 0.
∴ Put k =$\frac{-\mathrm{seven}}{5}$ in the equation.
$$\begin{gathered} \mathrm{v}\times {\boxed{\frac{-7}{5}}}^2+2\times \boxed{\frac{-7}{5}}+\mathrm{yard}=0 \\ \boxed{\frac{49}{5}}+\boxed{\frac{-\mathrm{fourteen}}{\mathrm{v}}}+\mathrm{grand}=0 \\ \boxed{\mathrm{seven}}+k=0 \\ k=\boxed{-\mathrm{vii}} \end{gathered}$$

Page No 36:

Question 1:

Solve the following quadratic equations by factorisation.
(1) x ² – 15x + 54 = 0
(2) x ² + x – 20 = 0
(three) 2y ⁱⁱ + 27y + 13 = 0
(iv) fivem ² = 22yard + 15
(5) iiten ² – iix + $\frac{\mathrm{ane}}{\mathrm{two}}$ = 0
(6) $\mathrm{vi}x-\frac{\mathrm{ii}}{x}=\mathrm{ane}$
(seven) $\sqrt{\mathrm{two}}{\mathrm{10}}^{\mathrm{ii}}+7\mathrm{10}+5\sqrt{2}=0$
to solve this quadratic equation by factorisation, complete the following activity.
(8)$3x^2-2\sqrt{6}x+2=0$
(9)$\mathrm{ii}m\left(m-24\right)=\mathrm{fifty}$
(10)$25m^2=9$
(11)$7k^2=21m$
(12)${\mathrm{yard}}^2-11=0$

Answer:

(ane)x ² – 15x + 54 = 0

$$\begin{gathered} \Rightarrow x^{\mathrm{two}}-9x-6\mathrm{10}+54=0 \\ \Rightarrow x\left(x-\mathrm{nine}\right)-6\left(x-\mathrm{ix}\right)=0 \\ \Rightarrow \left(x-6\right)\left(x-9\right)=0 \\ \Rightarrow \left(\mathrm{ten}-\mathrm{half\; dozen}\right)=0\mathrm{or}\left(x-9\right)=0 \\ \Rightarrow \mathrm{10}=6\mathrm{or}x=9 \end{gathered}$$

So, 6 and 9 are the roots of the given quadratic equation.

(2)10 ^two +x – twenty = 0

$$\begin{gathered} \Rightarrow x^{\mathrm{ii}}+5x-4x-\mathrm{xx}=0 \\ \Rightarrow x\left(x+5\right)-4\left(x+5\right)=0 \\ \Rightarrow \left(x-4\right)\left(x+5\right)=0 \\ \Rightarrow \left(x-\mathrm{four}\right)=0\mathrm{or}\left(x+\mathrm{five}\right)=0 \\ \Rightarrow x=\mathrm{four}\mathrm{or}\mathrm{ten}=-5 \end{gathered}$$

So, iv and

$-$

5 are the roots of the given quadratic equation.

(three) twoy ² + 27y + 13 = 0

$$\begin{gathered} \mathrm{ii}{y}^2+26y+y+13=0 \\ \Rightarrow 2y\left(y+13\right)+\left(y+13\right)=0 \\ \Rightarrow \left(\mathrm{two}y+1\right)\left(y+13\right)=0 \\ \Rightarrow \left(\mathrm{two}y+\mathrm{ane}\right)=0\mathrm{or}\left(y+\mathrm{xiii}\right)=0 \\ \Rightarrow y=\frac{-1}{\mathrm{two}}\mathrm{or}y=-13 \end{gathered}$$

And so,

$\frac{-\mathrm{one}}{\mathrm{ii}}\mathrm{and}-13$

 are the roots of the given quadratic equation.

(four) 5g ² = 22m + 15

$$\begin{gathered} \Rightarrow 5m^2-22\mathrm{thou}-\mathrm{fifteen}=0 \\ \Rightarrow \mathrm{v}{\mathrm{thou}}^2-25k+3\mathrm{1000}-\mathrm{fifteen}=0 \\ \Rightarrow \mathrm{v}m\left(\mathrm{chiliad}-\mathrm{five}\right)+3\left(m-5\right)=0 \\ \Rightarrow \left(5m+3\right)\left(m-5\right)=0 \\ \Rightarrow \left(5k+3\right)=0\mathrm{or}\left(g-\mathrm{five}\right)=0 \\ \Rightarrow \mathrm{chiliad}=\frac{-3}{5}\mathrm{or}m=5 \end{gathered}$$

So,

$\frac{-3}{5}\mathrm{and}5$

 are the roots of the given quadratic equation.

(5) 2x ² – 2x +

$\frac{\mathrm{i}}{2}$

 = 0

$$\begin{gathered} \Rightarrow 4x^2-4\mathrm{10}+1=0 \\ \Rightarrow {\left(2x\right)}^2-2\times 1\times 2\mathrm{10}+\mathrm{one}=0 \\ U\sin gth\mathrm{east}id\mathrm{east}\mathrm{due\; north}tity, \\ {\left(a-b\right)}^2=a^{\mathrm{two}}+b^2-\mathrm{ii}ab \\ {\left(2\mathrm{10}\right)}^2-2\times 1\times \mathrm{two}x+\mathrm{one}=0 \\ \Rightarrow {\left(\mathrm{ii}x-1\right)}^2=0 \\ \Rightarrow x=\frac{1}{2},\frac{\mathrm{one}}{2} \end{gathered}$$

And then,

$\frac{1}{2}\mathrm{and}\frac{1}{2}$

 are the roots of the given quadratic equation.

(6)

$6\mathrm{10}-\frac{2}{x}=\mathrm{one}$
$$\begin{gathered} \Rightarrow \mathrm{half\; dozen}{x}^2-2=1x \\ \Rightarrow 6x^2-1x-2=0 \\ \Rightarrow 6x^{\mathrm{ii}}-4\mathrm{ten}+3x-\mathrm{ii}=0 \\ \Rightarrow 2\mathrm{10}\left(3\mathrm{ten}-\mathrm{ii}\right)+1\left(\mathrm{iii}\mathrm{ten}-\mathrm{two}\right)=0 \\ \Rightarrow \left(3x-\mathrm{two}\right)\left(\mathrm{two}\mathrm{10}+\mathrm{ane}\right)=0 \\ \Rightarrow \left(3x-2\right)=0\mathrm{or}\left(\mathrm{ii}x+1\right)=0 \\ \Rightarrow \mathrm{ten}=\frac{\mathrm{two}}{3}\mathrm{or}\mathrm{10}=\frac{-\mathrm{ane}}{\mathrm{ii}} \end{gathered}$$
$\frac{2}{3}\mathrm{and}\frac{-1}{\mathrm{two}}$

 are the roots of the given quadratic equation.

(7)

$\sqrt{2}{\mathrm{ten}}^{\mathrm{two}}+\mathrm{seven}x+\mathrm{v}\sqrt{2}=0$
$$\begin{gathered} \sqrt{2}{x}^2+\boxed{5\mathrm{ten}}+\boxed{\mathrm{ii}x}+5\sqrt{2}=0 \\ x\left(\sqrt{2}\mathrm{ten}+5\right)+\sqrt{2}\left(\sqrt{2}x+5\right)=0 \\ \left(\sqrt{\mathrm{two}}x+5\right)\left(x+\sqrt{2}\right)=0 \\ \left(\sqrt{2}\mathrm{10}+5\right)=0\mathrm{or}\left(x+\sqrt{\mathrm{ii}}\right)=0 \\ \therefore x=\boxed{\frac{-\mathrm{v}}{\sqrt{\mathrm{ii}}}}\mathrm{or}\mathrm{ten}=-\sqrt{\mathrm{ii}} \\ \therefore \boxed{\frac{-5}{\sqrt{2}}}\mathrm{and}-\sqrt{\mathrm{ii}}\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}. \end{gathered}$$

(eight)

$\mathrm{iii}{\mathrm{ten}}^2-2\sqrt{\mathrm{vi}}\mathrm{10}+2=0$
$$\begin{gathered} \Rightarrow 3{\mathrm{10}}^{\mathrm{two}}-\sqrt{\mathrm{half-dozen}}x-\sqrt{6}\mathrm{ten}+2=0 \\ \Rightarrow \sqrt{3}\mathrm{10}\left(\sqrt{3}x-\sqrt{2}\right)-\sqrt{\mathrm{two}}\left(\sqrt{3}x-\sqrt{\mathrm{two}}\right)=0 \\ \Rightarrow \left(\sqrt{3}x-\sqrt{\mathrm{two}}\right)\left(\sqrt{\mathrm{three}}\mathrm{ten}-\sqrt{\mathrm{two}}\right)=0 \\ \Rightarrow x=\frac{\sqrt{2}}{\sqrt{\mathrm{iii}}},\frac{\sqrt{2}}{\sqrt{3}} \end{gathered}$$

(9)

$2m\left(k-24\right)=\mathrm{fifty}$
$$\begin{gathered} \Rightarrow m\left(m-24\right)=25 \\ \Rightarrow {\mathrm{one\; thousand}}^2-24g=25 \\ \Rightarrow m^2-24\mathrm{chiliad}-25=0 \\ \Rightarrow {\mathrm{chiliad}}^{\mathrm{ii}}-25m+\mathrm{one\; thousand}-25=0 \\ \Rightarrow m\left(m-25\right)+1\left(\mathrm{thou}-25\right)=0 \\ \Rightarrow \left(m+1\right)\left(m-25\right)=0 \\ \Rightarrow \mathrm{thou}=-1,25 \end{gathered}$$

(ten)

$25m^2=\mathrm{ix}$
$$\begin{gathered} \Rightarrow 25g^{\mathrm{two}}-9=0 \\ \Rightarrow {\left(5\mathrm{chiliad}\right)}^{\mathrm{two}}-3^2=0 \\ \Rightarrow \left(\mathrm{five}\mathrm{chiliad}-3\right)\left(5\mathrm{grand}+3\right)=0\left(\because \left(x-a\right)\left(x+a\right)=x^2-a^2\right) \\ \Rightarrow \left(5\mathrm{one\; thousand}-3\right)=0or\left(\mathrm{v}g+\mathrm{iii}\right)=0 \\ \Rightarrow k=\frac{\mathrm{three}}{5},\frac{-3}{5} \end{gathered}$$

(11)

$7m^{\mathrm{ii}}=21\mathrm{chiliad}$
$$\begin{gathered} \Rightarrow 7m^2-21m=0 \\ \Rightarrow 7m\left(m-3\right)=0 \\ \Rightarrow 7k=0\mathrm{or}m-3=0 \\ \Rightarrow \mathrm{thou}=0\mathrm{or}m=\mathrm{three} \end{gathered}$$

(12)

${\mathrm{chiliad}}^2-11=0$
$$\begin{gathered} \Rightarrow {\mathrm{chiliad}}^2-{\left(\sqrt{11}\right)}^{\mathrm{ii}}=0 \\ \Rightarrow \left(g-\sqrt{11}\right)\left(\mathrm{grand}+\sqrt{11}\right)=0 \\ \Rightarrow m=\sqrt{11},-\sqrt{11} \end{gathered}$$

Page No 39:

Question 1:

Solve the following quadratic equations by completing the foursquare method.
(ane) 10 ² + 10 – xx = 0
(2) x ² + twox – 5 = 0
(3) k ² – 5m = –3
(4) 9y ² – 12y + 2 = 0
(5) twoy ² + 9y +ten = 0
(6) 510 ⁱⁱ = 4x + 7

Answer:

(1)x ^two +x – 20 = 0

$$\begin{gathered} {\mathrm{ten}}^2+\mathrm{ten}+{\left(\frac{1}{2}\right)}^2-{\left(\frac{\mathrm{i}}{\mathrm{ii}}\right)}^2-\mathrm{twenty}=0 \\ \Rightarrow {\left(x+\frac{\mathrm{one}}{2}\right)}^2={\left(\frac{1}{\mathrm{two}}\right)}^{\mathrm{ii}}+\mathrm{xx} \\ \Rightarrow {\left(x+\frac{\mathrm{ane}}{2}\right)}^{\mathrm{ii}}=\frac{1}{4}+20 \\ \Rightarrow {\left(x+\frac{\mathrm{i}}{2}\right)}^2=\frac{81}{\mathrm{iv}} \\ \Rightarrow x+\frac{\mathrm{i}}{2}=\frac{\mathrm{nine}}{\mathrm{two}}\mathrm{or}x+\frac{\mathrm{i}}{2}=-\frac{\mathrm{ix}}{2} \\ \Rightarrow x=4\mathrm{or}x=-\mathrm{v} \end{gathered}$$

(2)ten ² + iix – v = 0

$$\begin{gathered} \Rightarrow {\mathrm{ten}}^2+2x+{\left(\frac{2}{2}\right)}^2-{\left(\frac{2}{\mathrm{ii}}\right)}^{\mathrm{two}}-5=0 \\ \Rightarrow \left({\mathrm{10}}^{\mathrm{two}}+\mathrm{ii}x+\mathrm{i}\right)-1-\mathrm{five}=0 \\ \Rightarrow {\left(x+\mathrm{one}\right)}^2-6=0 \\ \Rightarrow {\left(\mathrm{ten}+1\right)}^{\mathrm{two}}=6 \\ \Rightarrow {\left(x+\mathrm{one}\right)}^{\mathrm{two}}={\left(\sqrt{\mathrm{half\; dozen}}\right)}^{\mathrm{two}} \\ \Rightarrow x+1=\sqrt{6}\mathrm{or}x+\mathrm{one}=-\sqrt{\mathrm{half-dozen}} \\ \Rightarrow x=\sqrt{6}-1\mathrm{or}\mathrm{ten}=-\sqrt{6}-\mathrm{ane} \end{gathered}$$

(3)g ^two – fivechiliad = –3

$$\begin{gathered} \Rightarrow m^2-5\mathrm{yard}+{\left(\frac{-\mathrm{v}}{\mathrm{two}}\right)}^{\mathrm{ii}}-{\left(\frac{-5}{2}\right)}^2=-\mathrm{three} \\ \Rightarrow \left({\mathrm{1000}}^2-5\mathrm{one\; thousand}+\frac{25}{4}\right)-\frac{25}{4}=-3 \\ \Rightarrow {\left(m-\frac{\mathrm{v}}{2}\right)}^2=-3+\frac{25}{\mathrm{four}} \\ \Rightarrow {\left(m-\frac{5}{\mathrm{two}}\right)}^2=\frac{\mathrm{xiii}}{4} \\ \Rightarrow {\left(m-\frac{5}{\mathrm{two}}\right)}^2={\left(\frac{\sqrt{\mathrm{xiii}}}{\mathrm{two}}\right)}^{\mathrm{two}} \\ \Rightarrow g-\frac{5}{2}=\frac{\sqrt{13}}{2}\mathrm{or}m-\frac{5}{2}=-\frac{\sqrt{13}}{\mathrm{two}} \\ \Rightarrow \mathrm{thousand}=\frac{\sqrt{\mathrm{xiii}}}{\mathrm{ii}}+\frac{5}{2}\mathrm{or}m=-\frac{\sqrt{\mathrm{thirteen}}}{2}+\frac{5}{2} \\ \Rightarrow m=\frac{\sqrt{13}+5}{2}\mathrm{or}m=\frac{5-\sqrt{13}}{\mathrm{ii}} \end{gathered}$$

(4) ixy ^two – 12y + 2 = 0

$$\begin{gathered} \Rightarrow y^{\mathrm{two}}-\frac{12}{9}y+\frac{\mathrm{two}}{9}=0 \\ \Rightarrow y^2-\frac{4}{\mathrm{iii}}y+\frac{2}{9}=0 \\ \Rightarrow y^2-\frac{4}{3}y+{\left(\frac{{\displaystyle \frac{4}{\mathrm{iii}}}}{\mathrm{ii}}\right)}^{\mathrm{two}}-{\left(\frac{{\displaystyle \frac{\mathrm{iv}}{\mathrm{three}}}}{\mathrm{ii}}\right)}^2+\frac{2}{\mathrm{nine}}=0 \\ \Rightarrow y^{\mathrm{ii}}-\frac{\mathrm{four}}{\mathrm{iii}}y+{\left(\frac{2}{3}\right)}^2-{\left(\frac{2}{3}\right)}^{\mathrm{two}}+\frac{\mathrm{ii}}{9}=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left[y^2-\frac{4}{3}y+\left(\frac{\mathrm{four}}{9}\right)\right]-\left(\frac{4}{9}\right)+\frac{2}{\mathrm{ix}}=0 \\ \Rightarrow {\left(y-\frac{2}{3}\right)}^{\mathrm{two}}-\frac{2}{\mathrm{ix}}=0 \\ \Rightarrow {\left(y-\frac{2}{3}\right)}^2=\frac{2}{\mathrm{nine}} \\ \Rightarrow {\left(y-\frac{2}{3}\right)}^2={\left(\frac{\sqrt{2}}{3}\right)}^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow y-\frac{\mathrm{ii}}{3}=\frac{\sqrt{\mathrm{two}}}{3}\mathrm{or}y-\frac{2}{3}=-\frac{\sqrt{2}}{3} \\ \Rightarrow y=\frac{\sqrt{2}}{3}+\frac{2}{\mathrm{iii}}\mathrm{or}y=-\frac{\sqrt{2}}{3}+\frac{2}{\mathrm{iii}} \\ \Rightarrow y=\frac{\sqrt{\mathrm{two}}+2}{\mathrm{three}}\mathrm{or}y=\frac{-\sqrt{\mathrm{two}}+\mathrm{two}}{3} \end{gathered}$$

(5) iiy ² + niney +x = 0

$$\begin{gathered} \Rightarrow y^2+\frac{9}{2}y+\mathrm{v}=0 \\ \Rightarrow y^{\mathrm{two}}+\frac{9}{\mathrm{ii}}y+{\left(\frac{{\displaystyle \frac{9}{2}}}{2}\right)}^2-{\left(\frac{{\displaystyle \frac{\mathrm{nine}}{2}}}{\mathrm{ii}}\right)}^2+5=0 \\ \Rightarrow \left[y^{\mathrm{ii}}+\frac{9}{2}y+{\left(\frac{9}{4}\right)}^2\right]-{\left(\frac{9}{\mathrm{iv}}\right)}^{\mathrm{two}}+5=0 \\ \Rightarrow {\left(y+\frac{\mathrm{ix}}{4}\right)}^{\mathrm{two}}=\frac{81}{16}-\mathrm{five} \\ \Rightarrow {\left(y+\frac{9}{4}\right)}^2=\frac{1}{\mathrm{xvi}} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(y+\frac{\mathrm{nine}}{\mathrm{four}}\right)}^2={\left(\frac{1}{\mathrm{iv}}\right)}^{\mathrm{ii}} \\ \Rightarrow y+\frac{9}{4}=\frac{\mathrm{one}}{4}\mathrm{or}y+\frac{9}{4}=-\frac{1}{4} \\ \Rightarrow y=\frac{1}{4}-\frac{\mathrm{nine}}{4}=\frac{-8}{4}=-\mathrm{two}\mathrm{or}y=-\frac{\mathrm{ane}}{4}-\frac{9}{\mathrm{four}}=\frac{-10}{4}=-\frac{5}{\mathrm{two}} \\ \Rightarrow y=-2,-\frac{5}{\mathrm{two}} \end{gathered}$$

(6) vx ² = ivx+ 7

$$\begin{gathered} \Rightarrow \mathrm{v}{\mathrm{ten}}^{\mathrm{two}}-\mathrm{four}\mathrm{10}-\mathrm{seven}=0 \\ \Rightarrow x^2-\frac{4}{\mathrm{five}}x-\frac{\mathrm{seven}}{5}=0 \\ \Rightarrow x^2-\frac{\mathrm{iv}}{\mathrm{v}}x+{\left(\frac{{\displaystyle \frac{4}{5}}}{\mathrm{two}}\right)}^{\mathrm{two}}-{\left(\frac{{\displaystyle \frac{4}{5}}}{2}\right)}^2-\frac{7}{5}=0 \\ \Rightarrow \left[x^2-\frac{4}{\mathrm{five}}\mathrm{10}+{\left(\frac{\mathrm{two}}{5}\right)}^{\mathrm{two}}\right]-{\left(\frac{\mathrm{two}}{\mathrm{v}}\right)}^2-\frac{\mathrm{seven}}{5}=0 \\ \Rightarrow {\left(\mathrm{10}-\frac{\mathrm{ii}}{5}\right)}^2=\frac{7}{5}+\frac{4}{25}=\frac{35+4}{25}=\frac{39}{25} \\ \Rightarrow {\left(x-\frac{2}{5}\right)}^{\mathrm{two}}=\frac{39}{25} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(\mathrm{ten}-\frac{2}{5}\right)}^{\mathrm{two}}={\left(\frac{\sqrt{39}}{5}\right)}^2 \\ \Rightarrow x-\frac{\mathrm{ii}}{5}=\frac{\sqrt{39}}{5}\mathrm{or}x-\frac{2}{5}=-\frac{\sqrt{39}}{5} \\ \Rightarrow x=\frac{\mathrm{two}+\sqrt{39}}{5}\mathrm{or}\mathrm{ten}=\frac{\mathrm{ii}-\sqrt{39}}{\mathrm{v}} \end{gathered}$$

Folio No 43:

Question 1:

Compare the given quadratic equations to the general form and write values of a,b, c.
(1) x ⁱⁱ – 710 + five = 0
(2) twom ² = 5chiliad – 5
(3) y ² = 7y

Reply:

(1)ten ² – 7x + 5 = 0
General form of the quadratic equation is

$a{x}^2+b\mathrm{10}+c=0$

.
Comparingx ² – 7x + 5 = 0 with the general form we have a = 1, b =

$-$

7 and c = v.

(2) 2yard ² = vgrand – 5

$\Rightarrow 2{\mathrm{yard}}^2-\mathrm{five}m+5=0$

Full general form of the quadratic equation is

$a{x}^{\mathrm{ii}}+bx+c=0$

.
Comparison 2m ^two = 5thou – v with the general form we havea = 2,b =

$-$

v andc = 5.

(3)y ^two = 7y
$\Rightarrow y^{\mathrm{ii}}-7y+0=0$
General form of the quadratic equation is

$a{x}^{\mathrm{two}}+bx+c=0$

.
Comparing $y^2-7y+0=0$  with the full general form nosotros takea = 1,b =

$-$

vii andc = 0.

Page No 43:

Question 2:

Solve using formula.
(1) x ² + half dozenx + five = 0
(two) x ^two – 3x – 2 = 0
(3) 31000 ^two + 2m – 7 = 0
(four) 5m ² – ivchiliad – ii = 0
(5) $y^2+\frac{1}{3}y=\mathrm{ii}$
(half-dozen) 5x ⁱⁱ + 13x + eight = 0

Answer:

(one)x ^two + 6ten + five = 0
On comparing with the equation

$a{\mathrm{ten}}^2+bx+c=0$

,
a =1, b= 6 and c =5
Now

$b^2-4ac=6^2-\mathrm{iv}\times 1\times 5=36-20=16$
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$$\begin{gathered} x=\frac{-6\pm \sqrt{16}}{2\times 1}=\frac{-6\pm 4}{2} \\ \Rightarrow x=\frac{-\mathrm{half-dozen}+4}{2}\mathrm{or}\mathrm{ten}=\frac{-\mathrm{half\; dozen}-4}{\mathrm{ii}} \\ \Rightarrow x=-1\mathrm{or}x=-5 \end{gathered}$$

(ii)x ^two – 3x – 2 = 0
On comparison with the equation

$a{x}^2+bx+c=0$

,
a =1,b=

$-$

three andc =

$-$

2
At present

$b^{\mathrm{two}}-4ac={\left(-3\right)}^2-4\times 1\times \left(-\mathrm{two}\right)=\mathrm{nine}+8=17$
$x=\frac{-b\pm \sqrt{b^2-\mathrm{iv}ac}}{\mathrm{ii}a}$
$$\begin{gathered} x=\frac{3\pm \sqrt{17}}{2\times 1}=\frac{3\pm \sqrt{17}}{2} \\ \Rightarrow x=\frac{\mathrm{iii}+\sqrt{17}}{\mathrm{ii}}\mathrm{or}x=\frac{3-\sqrt{17}}{2} \end{gathered}$$

(three) threem ⁱⁱ + 2k – vii = 0
On comparison with the equation

$a{\mathrm{ten}}^2+bx+c=0$

,
a = 3,b= ii andc =

$-$

vii
Now

$b^2-4ac={\left(2\right)}^2-\mathrm{iv}\times 3\times \left(-7\right)=4+84=88$
$x=\frac{-b\pm \sqrt{b^2-4ac}}{\mathrm{two}a}$
$$\begin{gathered} x=\frac{-\mathrm{ii}\pm \sqrt{88}}{\mathrm{ii}\times 3}=\frac{-2\pm \sqrt{88}}{6} \\ \Rightarrow x=\frac{-2+\sqrt{88}}{6}\mathrm{or}x=\frac{-\mathrm{ii}-\sqrt{88}}{6} \\ \Rightarrow x=\frac{-1+\sqrt{22}}{3}\mathrm{or}\frac{-1-\sqrt{22}}{3} \end{gathered}$$

(4) 5m ² – fourg – two = 0
On comparison with the equation

$a{\mathrm{10}}^{\mathrm{ii}}+bx+c=0$

,
a = 5,b= – 4andc =

$-$

2
Now

$b^{\mathrm{ii}}-4ac={\left(-4\right)}^2-4\times 5\times \left(-2\right)=\mathrm{xvi}+40=56$
$x=\frac{-b\pm \sqrt{b^{\mathrm{ii}}-\mathrm{four}ac}}{2a}$
$$\begin{gathered} x=\frac{-4\pm \sqrt{56}}{2\times \mathrm{v}}=\frac{-\mathrm{four}\pm \sqrt{56}}{10} \\ \Rightarrow x=\frac{-4+\sqrt{56}}{10}\mathrm{or}\mathrm{ten}=\frac{-4-\sqrt{56}}{10} \\ \Rightarrow \mathrm{ten}=\frac{-2+\sqrt{14}}{\mathrm{v}}\mathrm{or}\frac{-\mathrm{two}-\sqrt{\mathrm{xiv}}}{5} \end{gathered}$$

(v)

$y^2+\frac{\mathrm{i}}{\mathrm{three}}y=2$

Multiplying the equation by 3

$$\begin{gathered} \mathrm{iii}{y}^2+y=6 \\ \Rightarrow \mathrm{three}{y}^2+y-6=0 \end{gathered}$$

On comparing with the equation

$a{x}^{\mathrm{two}}+b\mathrm{ten}+c=0$

,
a = 3,b= 1 andc =

$-$

6
Now

$b^{\mathrm{two}}-4ac={\left(1\right)}^{\mathrm{two}}-4\times 3\times \left(-6\right)=1+72=73$
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$$\begin{gathered} x=\frac{-\mathrm{i}\pm \sqrt{73}}{\mathrm{two}\times 3}=\frac{-\mathrm{one}\pm \sqrt{73}}{6} \\ \Rightarrow x=\frac{-\mathrm{i}+\sqrt{73}}{6}\mathrm{or}x=\frac{-\mathrm{one}-\sqrt{73}}{6} \end{gathered}$$

(6) 5x ^two + thirteenten + 8 = 0
On comparing with the equation

$a{\mathrm{10}}^2+bx+c=0$

,
a = 5,b= 13 andc = 8
Now

$b^2-4ac={\left(13\right)}^2-4\times \mathrm{v}\times 8=169-160=9$
$x=\frac{-b\pm \sqrt{b^{\mathrm{two}}-\mathrm{iv}ac}}{\mathrm{ii}a}$
$$\begin{gathered} x=\frac{-\mathrm{thirteen}\pm \sqrt{9}}{\mathrm{ii}\times \mathrm{v}}=\frac{-13\pm 3}{\mathrm{x}} \\ \Rightarrow x=\frac{-13+\mathrm{three}}{10}\mathrm{or}x=\frac{-\mathrm{thirteen}-3}{\mathrm{x}} \\ \Rightarrow x=\frac{-\mathrm{x}}{\mathrm{x}}\mathrm{or}x=\frac{-16}{10} \\ \Rightarrow \mathrm{ten}=-\mathrm{ane}\mathrm{or}\mathrm{10}=\frac{-8}{\mathrm{v}} \end{gathered}$$

Page No 44:

Question iii:

With the assistance of the flow nautical chart given below solve the equation $x^2+2\sqrt{3}\mathrm{10}+\mathrm{iii}=0$ using the formula.

Answer:

$$\begin{gathered} \boxed{\mathrm{Compare}\mathrm{equations} \\ {x}^{\mathrm{two}}+2\sqrt{3}\mathrm{10}+3=0\mathrm{and} \\ a{\mathrm{10}}^{\mathrm{two}}+bx+c=0\mathrm{find} \\ \mathrm{the}\mathrm{values}\mathrm{of}a,b,c.}\to \boxed{\mathrm{Notice}\mathrm{value}\mathrm{of} \\ {b}^2-\mathrm{iv}ac}\to \boxed{\mathrm{Write}\mathrm{formula} \\ \mathrm{to}\mathrm{solve} \\ \mathrm{quadratic} \\ \mathrm{equation}.}\to \boxed{\mathrm{Substitute} \\ \mathrm{values}\mathrm{of} \\ a,b,c\mathrm{and} \\ \mathrm{find}\mathrm{roots}} \end{gathered}$$

comparing

$x^2+2\sqrt{3}\mathrm{ten}+\mathrm{three}=0$

 with

$a{x}^2+b\mathrm{10}+c=0$

 nosotros go

$a=\mathrm{i},b=\mathrm{ii}\sqrt{\mathrm{iii}}\mathrm{and}c=3$
$b^2-4ac={\left(2\sqrt{3}\right)}^{\mathrm{ii}}-4\times \mathrm{ane}\times 3=12-12=0$

Formula to solve a quadratic equation will be

$$\begin{gathered} \mathrm{10}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \Rightarrow x=\frac{-2\sqrt{\mathrm{iii}}\pm \sqrt{0}}{2\times 1}=\frac{-2\sqrt{3}}{2}=-\sqrt{3} \end{gathered}$$

Thus,

$x=-\sqrt{3},-\sqrt{3}$

.

Page No 49:

Question 1:

Fill in the gaps and consummate.


(3)If α, β are roots of quadratic equation,

Respond:

If

$\alpha \mathrm{and}\beta$

 are the roots of the quadratic equation so the quadratic equation can be written equally

$x^{\mathrm{two}}-\left(\alpha +\beta \right)x+\alpha \beta =0$

And so, the quadratic equation with sum of roots every bit

$\alpha +\beta =-7$

 and production of roots as

$\alpha \beta =\mathrm{v}$

 will be

${\mathrm{10}}^{\mathrm{two}}+7\mathrm{10}+\mathrm{v}=0$

Sum of roots =

$\frac{-b}{a}=\frac{-\left(-4\right)}{2}=2$

Product of roots =

$\frac{c}{a}=\frac{-3}{\mathrm{two}}$

Page No 49:

Question ii:

Observe the value of discriminant.
(1) x ² + viix – 1 = 0
(2) 2y ² – fivey + x = 0
(iii) $\sqrt{2}{\mathrm{10}}^2+4x+2\sqrt{2}=0$

Respond:

(1)ten ² + 7x – one = 0
Comparing the given equation with

$a{x}^2+b\mathrm{10}+c=0$

,

$a=1,b=7\mathrm{and}c=-1$

So, the discriminant

$b^2-\mathrm{four}ac={\left(7\right)}^2-\mathrm{four}\times \mathrm{ane}\times \left(-1\right)=49+4=53$

(2) 2y ² – 5y + 10 = 0
Comparison the given equation with

$a{x}^2+b\mathrm{10}+c=0$

,

$a=\mathrm{ii},b=-5\mathrm{and}c=10$

So, the discriminant

$b^2-4ac={\left(-\mathrm{v}\right)}^{\mathrm{ii}}-4\times 2\times \mathrm{x}=25-80=-55$

(3)

$\sqrt{2}{\mathrm{10}}^2+4x+2\sqrt{2}=0$

Comparing the given equation with

$a{x}^2+bx+c=0$

,

$a=\sqrt{\mathrm{ii}},b=4\mathrm{and}c=2\sqrt{\mathrm{ii}}$

So, the discriminant

$b^2-4ac={\left(4\right)}^{\mathrm{two}}-\mathrm{iv}\times \sqrt{2}\times \mathrm{two}\sqrt{2}=16-16=0$

Page No 49:

Question 3:

Determine the nature of roots of the following quadratic equations.
(1) x ^two – 4x + 4 = 0
(2) 2y ⁱⁱ – seveny +2 = 0
(3) m ² + 2m + ix = 0

Answer:

(1)x ² – fourx + 4 = 0
Comparing the given equation with the quadratic equation

$a{x}^{\mathrm{ii}}+bx+c=0$

,

$a=\mathrm{ane},b=-4,c=4$

Discriminant,

$△=b^2-4ac={\left(-4\right)}^2-\mathrm{iv}\times \mathrm{one}\times 4=\mathrm{sixteen}-16=0$

Since the discriminant = 0 and then, the roots of the given quadratic equation are real and equal.

(2) 2y ²– seveny +2 = 0
Comparing the given equation with the quadratic equation

$a{x}^{\mathrm{ii}}+bx+c=0$

,

$a=2,b=-7,c=2$

Discriminant,

$△=b^2-4ac={\left(-7\right)}^2-\mathrm{four}\times 2\times 2=49-\mathrm{sixteen}=33$

Since the discriminant > 0 so, the roots of the given quadratic equation are real and unequal.

(3)m ⁱⁱ + 2m + 9 = 0
Comparing the given equation with the quadratic equation

$a{x}^2+bx+c=0$

,

$a=1,b=2,c=9$

Discriminant,

$△=b^{\mathrm{ii}}-\mathrm{four}ac={\left(\mathrm{two}\right)}^2-4\times \mathrm{ane}\times \mathrm{ix}=4-36=-32$

Since the discriminant < 0 so, the roots of the given quadratic equation are not real.

Page No fifty:

Question four:

Form the quadratic equation from the roots given below.
(1) 0 and 4
(ii) 3 and –10
(3) $\frac{\mathrm{ane}}{\mathrm{two}},-\frac{1}{\mathrm{ii}}$
(4) $2-\sqrt{5},2+\sqrt{5}$

Reply:

(ane) 0 and 4
Sum of roots = 0 + four = iv
Product of roots = 0

$\times$

 four = 0
The general grade of the quadratic equation is

$x^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)\mathrm{ten}+\mathrm{Product}\mathrm{of}\mathrm{roots}=0$

So, the quadratic equation obtained is

$x^2-4x+0=0$
$$\begin{gathered} \Rightarrow x^2-4x=0 \\ \Rightarrow x\left(x-\mathrm{four}\right)=0 \end{gathered}$$

(2) 3 and –10
Sum of roots = 3 + (–10) = −vii
Production of roots = 3

$\times$

 –x = –30
The general form of the quadratic equation is

$x^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)x+\mathrm{Product}\mathrm{of}\mathrm{roots}=0$

And so, the quadratic equation obtained is

$x^2-\left(-7\right)x+\left(-30\right)=0$
$x^{\mathrm{two}}+7x-30=0$

(3)

$\frac{\mathrm{ane}}{\mathrm{two}},-\frac{\mathrm{one}}{2}$

Sum of roots =

$\frac{\mathrm{ane}}{\mathrm{ii}}+\left(-\frac{1}{\mathrm{two}}\right)=0$

Production of roots =

$\frac{1}{2}\times \left(\frac{-1}{2}\right)=\frac{-\mathrm{i}}{4}$

The general form of the quadratic equation is

$x^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)x+\mathrm{Product}\mathrm{of}\mathrm{roots}=0$

And then, the quadratic equation obtained is

$x^{\mathrm{ii}}-\left(0\right)\mathrm{10}+\left(\frac{-1}{\mathrm{four}}\right)=0$
$\Rightarrow x^{\mathrm{ii}}-\frac{\mathrm{ane}}{4}=0$

(iv)

$2-\sqrt{5},2+\sqrt{\mathrm{five}}$

Sum of roots =

$\mathrm{two}-\sqrt{5}+2+\sqrt{5}=4$

Product of roots =

$\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-5=-1$

The full general class of the quadratic equation is

$x^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)x+\mathrm{Product}\mathrm{of}\mathrm{roots}=0$

And then, the quadratic equation obtained is

$x^{\mathrm{two}}-\left(4\right)x+\left(-1\right)=0$
$\Rightarrow {\mathrm{ten}}^{\mathrm{two}}-4x-\mathrm{one}=0$

Page No 50:

Question 5:

Sum of the roots of a quadratic equation is double their production. Detect one thousand if equation ten ² – 4kx + k +three = 0

Answer:

Let the roots of the quadratic equation exist$\alpha \mathrm{and}\beta$.
Sum of roots =$\alpha +\beta$
Production of roots =$\alpha \beta$
Given that sum of roots = double the product
$\Rightarrow \alpha +\beta =\mathrm{ii}\alpha \beta$                               .....(I)
Given:ten ²  – 4one thousand ten  + 1000 +three = 0           .....(2)
General form of the quadratic equation is${\mathrm{10}}^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)x+\mathrm{Product}\mathrm{of}\mathrm{roots}=0$
$$\begin{gathered} \Rightarrow x^{\mathrm{ii}}-\left(\alpha +\beta \right)x+\alpha \beta =0 \\ \Rightarrow x^2-\left(\mathrm{ii}\alpha \beta \right)x+\alpha \beta =0.....\left(\mathrm{Three}\right) \end{gathered}$$
On comparison (Two) and (III) we have
$4\mathrm{one\; thousand}=2\alpha \beta \mathrm{and}m+\mathrm{iii}=\alpha \beta$
Solving these equations nosotros have
$$\begin{gathered} \mathrm{iv}k=2\left(k+\mathrm{three}\right) \\ \Rightarrow 2\mathrm{one\; thousand}=k+\mathrm{three} \\ \Rightarrow 2\mathrm{one\; thousand}-m=\mathrm{iii} \\ \Rightarrow k=3 \end{gathered}$$

Page No 50:

Question 6:

α, β are roots of y ⁱⁱ – 2y –vii = 0 detect,
(1) α² + βⁱⁱ
(2) α³ + β³

Answer:

α, β are roots ofy ² – 2y –vii = 0

$a=1,b=-\mathrm{two},c=-7$ $a=\mathrm{ane},b=-2,c=-7$

(1)

$$\begin{gathered} \alpha +\beta =\frac{-b}{a}=\frac{-\left(-2\right)}{1}=2 \\ \alpha \beta =\frac{c}{a}=\frac{-7}{1}=-7 \end{gathered}$$
$$\begin{gathered} {\alpha }^{\mathrm{ii}}+{\beta }^2={\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-2\alpha \beta \\ ={\left(2\right)}^2-2\left(-\mathrm{vii}\right) \\ =\mathrm{four}+14 \\ =18 \end{gathered}$$

(ii)

$$\begin{gathered} {\alpha }^{\mathrm{three}}+{\beta }^3={\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right) \\ ={\left(2\right)}^3-\mathrm{three}\left(-\mathrm{vii}\right)\left(\mathrm{two}\right) \\ =\mathrm{eight}+42 \\ =50 \end{gathered}$$

Page No 50:

Question 7:

The roots of each of the following quadratic equations are real and equal, discover k.
(ane) threey ² + ky +12 = 0
(2) one thousandx (x – 2) + 6 = 0

Answer:

(1) 3y ²+ ky +12 = 0
The roots of the given quadratic equation are real and equal. So, the discriminant volition exist 0.

$$\begin{gathered} b^2-4ac=0 \\ \Rightarrow k^{\mathrm{two}}-4\times 3\times 12=0 \\ \Rightarrow k^{\mathrm{ii}}-144=0 \\ \Rightarrow k^{\mathrm{two}}=144 \\ \Rightarrow k=\pm 12 \end{gathered}$$

(two) 1000x (10 – two) + half-dozen = 0

$\Rightarrow k{\mathrm{10}}^2-\mathrm{ii}kx+6=0$

The roots of the given quadratic equation are existent and equal. So, the discriminant volition be 0.

$$\begin{gathered} b^{\mathrm{two}}-4ac=0 \\ \Rightarrow {\left(-2g\right)}^2-4\times \mathrm{thou}\times 6=0 \\ \Rightarrow \mathrm{iv}{m}^{\mathrm{two}}-24k=0 \\ \Rightarrow \mathrm{four}\mathrm{one\; thousand}\left(k-6\right)=0 \\ \Rightarrow 4\mathrm{grand}=0\mathrm{or}k-6=0 \\ \Rightarrow k=0\mathrm{or}k=\mathrm{six} \end{gathered}$$

But grand cannot be equal to 0 since then there volition non exist any quadratic equation.
And so, thousand= six

Page No 52:

Question 1:

Production of Pragati'south age two years ago and 3 years hence is 84. Find her nowadays historic period.

Answer:

Allow Pragati'southward present age be 10years.
Her age 2 years ago =$x-\mathrm{ii}$
Her historic period 3 years hence =$\mathrm{10}+\mathrm{three}$
Product of Pragati's age two years ago and 3 years hence is 84.
$$\begin{gathered} \left(x-2\right)\left(\mathrm{10}+3\right)=84 \\ \Rightarrow {\mathrm{10}}^2+\mathrm{10}-\mathrm{six}=84 \\ \Rightarrow x^{\mathrm{two}}+x-90=0 \\ x=\frac{-b\pm \sqrt{b^{\mathrm{two}}-\mathrm{iv}ac}}{\mathrm{two}a} \\ \Rightarrow \mathrm{ten}=\frac{-1\pm \sqrt{1^{\mathrm{ii}}-4\times 1\times \left(-90\right)}}{\mathrm{ii}} \\ =\frac{-\mathrm{i}\pm \sqrt{361}}{2} \\ \mathrm{10}=\frac{-1-19}{\mathrm{two}}\mathrm{or}\frac{-\mathrm{i}+19}{2} \\ \Rightarrow x=-10\mathrm{or}9 \end{gathered}$$
But age cannot be negative then, x= 9.
Thus, Pragati's present historic period is 9 years.

Folio No 52:

Question 2:

The sum of squares of two consecutive natural numbers is 244; notice the numbers.

Answer:

Disclaimer: There is an error in the given question. Hither it should be ii consecutive even natural numbers.

Let the two consecutive natural numbers be

$x\mathrm{and}\mathrm{10}+2$

.
Sum of squares of these 2 consecutive natural numbers is 244

$$\begin{gathered} x^2+{\left(\mathrm{10}+\mathrm{two}\right)}^2=244 \\ \Rightarrow x^{\mathrm{two}}+x^2+\mathrm{four}+\mathrm{four}x=244 \\ \Rightarrow \mathrm{ii}{x}^{\mathrm{two}}+\mathrm{four}x-240=0 \\ \Rightarrow x^2+2x-120=0 \\ \Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{\mathrm{ii}a} \\ \Rightarrow \mathrm{ten}=\frac{-\mathrm{ii}\pm \sqrt{2^2-\mathrm{four}\times 1\times \left(-120\right)}}{\mathrm{two}} \\ \Rightarrow x=\frac{-\mathrm{two}\pm \sqrt{4+480}}{\mathrm{ii}} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\frac{-2\pm \sqrt{484}}{2} \\ \Rightarrow x=\frac{-2\pm 22}{2} \\ \Rightarrow \mathrm{10}=\frac{-2+22}{\mathrm{ii}},\frac{-\mathrm{ii}-22}{2} \\ \Rightarrow \mathrm{ten}=\frac{20}{\mathrm{ii}},\frac{-24}{2} \\ \Rightarrow \mathrm{ten}=10,-12 \end{gathered}$$

Merely the natural number cannot be negative so,
The two numbers are 10 and 10 + 2 = 12.

Page No 52:

Question three:

In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is five more than that in each column. Find the number of trees in each row and each cavalcade with the help of post-obit flow chart.

Answer:

Let the number of trees in each column be x.
Number of trees in a row = x+ 5
Total copse = Number of rows$\times$ Number of columns =$\mathrm{ten}\left(\mathrm{ten}+\mathrm{v}\right)$ = 150
$$\begin{gathered} \Rightarrow x\left(\mathrm{10}+5\right)=150 \\ \Rightarrow {\mathrm{ten}}^2+\mathrm{five}x-150=0 \\ \Rightarrow x^{\mathrm{two}}+15x-\mathrm{ten}\mathrm{10}-150=0 \\ \Rightarrow x\left(x+15\right)-10\left(x+15\right)=0 \\ \Rightarrow \left(\mathrm{10}+\mathrm{fifteen}\right)\left(x-10\right)=0 \\ \Rightarrow x=\mathrm{x},-\mathrm{xv} \end{gathered}$$
But number of columns cannot be negative so, number of columns = ten.
Number of rows = ten + five = 15

Page No 52:

Question 4:

Vivek is older than Kishor past five years. The sum of the reciprocals of their ages is $\frac{1}{\mathrm{half\; dozen}}$. Discover their nowadays ages.

Answer:

Let the present age of Kishor be tenyears.
Vivek is older than Kishor by 5 years.
So, Vivek'due south historic period will be x+ 5.
The sum of the reciprocals of their ages is$\frac{\mathrm{one}}{6}$.
$$\begin{gathered} \frac{\mathrm{i}}{\mathrm{ten}}+\frac{1}{\mathrm{ten}+5}=\frac{\mathrm{i}}{\mathrm{six}} \\ \Rightarrow \frac{\mathrm{10}+5+x}{x\left(x+5\right)}=\frac{\mathrm{one}}{6} \\ \Rightarrow \frac{5+2x}{{\mathrm{ten}}^{\mathrm{two}}+5\mathrm{ten}}=\frac{{\displaystyle \mathrm{one}}}{{\displaystyle \mathrm{six}}} \\ \Rightarrow \mathrm{thirty}+12x=x^2+5\mathrm{10} \\ \Rightarrow x^2-7x-\mathrm{xxx}=0 \\ \Rightarrow x^2-10\mathrm{ten}+\mathrm{iii}\mathrm{ten}-30=0 \\ \Rightarrow \mathrm{ten}\left(x-\mathrm{x}\right)+3\left(x-10\right)=0 \\ \Rightarrow \left(x+3\right)\left(\mathrm{ten}-\mathrm{x}\right)=0 \\ \Rightarrow x=-3,10 \end{gathered}$$
Only age cannot exist negative and then, age of Kishore is x years and that of Vivek is 10 + 5 = 15 years.

Folio No 52:

Question v:

Suyash scored 10 marks more than in second test than that in the outset. five times the score of the second test is the same equally foursquare of the score in the outset test. Discover his score in the first examination.

Answer:

Let the score in the commencement exam be x.
Suyash's score in second test is x more than that in the get-go.
So, his score in the second test =10+ ten
5 times the score of the second test is the same equally square of the score in the kickoff test.
$$\begin{gathered} \mathrm{v}\left(x+\mathrm{x}\right)=x^2 \\ \Rightarrow \mathrm{five}x+\mathrm{l}=x^2 \\ \Rightarrow x^2-\mathrm{v}\mathrm{10}-\mathrm{l}=0 \\ \Rightarrow x^{\mathrm{two}}-10\mathrm{10}+\mathrm{five}x-50=0 \\ \Rightarrow x\left(x-\mathrm{x}\right)+5\left(x-10\right)=0 \\ \Rightarrow \left(x+5\right)\left(\mathrm{10}-10\right)=0 \\ \Rightarrow x=10,-5 \end{gathered}$$
The score is not negative and so, Suyash scored 10 marks in offset test.

Folio No 52:

Question 6:

Mr. Kasam runs a small business organisation of making earthen pots. He makes certain number of pots on daily basis. Product cost of each pot is Rs xl more than 10 times total number of pots, he makes in 1 day. If production cost of all pots per mean solar day is Rs 600, discover production cost of one pot and number of pots he makes per twenty-four hours.

Answer:

Let the number of pots Mr. Kasam makes in one solar day be x.
Product toll of each pot is Rs forty more than 10 times total number of pots, he makes in one solar day = 10ten+ 40
Product price of all pots per twenty-four hour period is Rs 600
(xx+ 40)ten= 600
$$\begin{gathered} 10x^2+40x=600 \\ \Rightarrow x^2+4x-60=0 \\ \Rightarrow x^{\mathrm{ii}}+10x-\mathrm{half\; dozen}\mathrm{ten}-\mathrm{sixty}=0 \\ \Rightarrow \mathrm{10}\left(x+10\right)-6\left(x+10\right)=0 \\ \Rightarrow \left(\mathrm{ten}-\mathrm{half-dozen}\right)\left(\mathrm{ten}+10\right)=0 \\ \Rightarrow x=6,-10 \end{gathered}$$
But the number of pots cannot be negative so,
$$\begin{gathered} x\ne -\mathrm{x} \\ \Rightarrow x=6 \end{gathered}$$
Production cost of 1 pot =$\mathrm{x}\times 6+40=60+40=\mathrm{Rs}100$

Page No 52:

Question vii:

Pratik takes 8 hours to travel 36 km down stream and render to the same spot. The speed of boat in even so water is 12 km. per hr. Discover the speed of water current.

Answer:

Let the speed of water electric current = y km/h
Speed of boat in even so water = 12 km/h
Speed upstream =$12-y$
Speed downstream = 12 + y
Distance travelled = 36 km downstream and 36 km upstream
$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}} \\ \Rightarrow \mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}} \end{gathered}$$
$$\begin{gathered} \frac{36}{12-y}+\frac{36}{12+y}=8 \\ \Rightarrow 36\left[\frac{1}{12-y}+\frac{1}{12+y}\right]=8 \\ \Rightarrow \frac{12+y+12-y}{144-y^2}=\frac{1}{\mathrm{iv}} \\ \Rightarrow \frac{24}{144-y^{\mathrm{ii}}}=\frac{{\displaystyle 1}}{{\displaystyle 4}} \\ \Rightarrow 144-y^{\mathrm{ii}}=24\times 4 \\ \Rightarrow y^2=36 \\ \Rightarrow y^2-36=0 \\ \Rightarrow y=\pm \mathrm{half\; dozen} \end{gathered}$$
Speed cannot exist negative and so, the speed of water current = 6 km/h.

Page No 52:

Question 8:

Pintu takes half dozen days more than those of Nishu to complete certain piece of work. If they piece of work together they finish it in 4 days. How many days would it accept to consummate the work if they piece of work lonely.

Answer:

Let the number of days taken past Nishu exist x.
So, number of days taken by Pintu will existx+ 6
Nishu'due south ane day work =$\frac{1}{\mathrm{10}}$
Pintu'due south one twenty-four hour period piece of work =$\frac{\mathrm{one}}{x+6}$
Piece of work washed together in 4 days =$\frac{1}{\mathrm{iv}}$
$$\begin{gathered} \frac{1}{\mathrm{10}}+\frac{\mathrm{i}}{x+\mathrm{vi}}=\frac{\mathrm{ane}}{4} \\ \Rightarrow \frac{\mathrm{10}+6+\mathrm{ten}}{x\left(\mathrm{10}+\mathrm{vi}\right)}=\frac{\mathrm{ane}}{4} \\ \Rightarrow \frac{6+\mathrm{two}\mathrm{ten}}{{\mathrm{ten}}^{\mathrm{ii}}+\mathrm{half-dozen}x}=\frac{1}{4} \\ \Rightarrow 24+8x={\mathrm{ten}}^2+\mathrm{six}\mathrm{ten} \\ \Rightarrow x^2-\mathrm{ii}x-24=0 \\ \Rightarrow x^2-6x+4\mathrm{ten}-24=0 \\ \Rightarrow \left(x-\mathrm{half-dozen}\right)\left(x+4\right)=0 \\ \Rightarrow x=6,-4 \end{gathered}$$
Number of days cannot exist negative so, ten = vi.
Thus, number of days taken by Nishu = 6 and that by Pintu = vi + 6 = 12.

Folio No 52:

Question 9:

If 460 is divided by a natural number, quotient is half-dozen more than five times the divisor and remainder is i. Find quotient and diviser.

Reply:

Let the divisor be x.
Dividend = 460
Quotient = 5x + 6
Residual = 1
We know
$\mathrm{Divident}=\mathrm{Divisor}\times \mathrm{Quotient}+\mathrm{Remainder}$
$$\begin{gathered} 460=x\left(5x+6\right)+\mathrm{one} \\ \Rightarrow 460=\mathrm{five}{x}^2+\mathrm{six}\mathrm{10}+1 \\ \Rightarrow \mathrm{five}{x}^2+6x-459=0 \\ \Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \Rightarrow x=\frac{-\mathrm{vi}\pm \sqrt{{\left(6\right)}^2-4\times 5\times \left(-459\right)}}{2\times 5} \\ \Rightarrow \mathrm{10}=\frac{-\mathrm{six}\pm \sqrt{36+9180}}{10} \\ \Rightarrow x=\frac{-\mathrm{half-dozen}\pm \sqrt{9216}}{\mathrm{x}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \mathrm{10}=\frac{-\mathrm{six}\pm 96}{\mathrm{ten}} \\ \Rightarrow x=\frac{-6+96}{\mathrm{ten}},\frac{-6-96}{\mathrm{ten}} \\ \Rightarrow \mathrm{ten}=9,-10.2 \end{gathered}$$
Thus, divisor = 9, quotient = v10 + 6 =$\mathrm{five}\times \mathrm{ix}+6=45+\mathrm{half\; dozen}=51$

Page No 52:

Question 10:

In the adjoining fig. $\square$ABCD is a trapezium AB || CD and its area is 33 cm^two . From the data given in the figure find the lengths of all sides of the $\square$ABCD. Fill in the empty boxes to get the solution.

Answer:

$\square$ABCD is a trapezium.
AB || CD
A($\square$ABCD) =$\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)\times \boxed{\mathrm{AM}}$
$$\begin{gathered} 33=\frac{1}{2}\left(x+2\mathrm{ten}+1\right)\times \boxed{\left(x-\mathrm{iv}\right)} \\ \therefore \boxed{33\times 2}=\left(3x+1\right)\times \boxed{\left(x-4\right)} \\ \therefore 3x^{\mathrm{two}}+\boxed{11x}-\boxed{70}=0 \\ \therefore 3x\left(\mathrm{10}-7\right)+10\left(\mathrm{10}-7\right)=0 \\ \therefore \left(\mathrm{three}x+10\right)\left(\mathrm{ten}-\mathrm{seven}\right)=0 \\ \therefore \left(3x+\mathrm{ten}\right)=0\mathrm{or}\boxed{x-7}=0 \\ \therefore \mathrm{10}=\frac{-10}{\mathrm{three}}\mathrm{or}x=\boxed{\mathrm{seven}} \end{gathered}$$
Simply length is never negative.
$\therefore x\ne \frac{-10}{3}\therefore x=\boxed{\mathrm{vii}}$
AB = 7, CD = xv, BC = 5

Page No 53:

Question ane:

Choose the correct answers for the following questions.
(1) Which one is the quadratic equation ?

(A) $\frac{5}{\mathrm{ten}}-3=x^2$

(B) x(x + 5) = 2

(C) northward – 1 = 2n

(D) $\frac{1}{x^2}\left(\mathrm{10}+\mathrm{ii}\right)=x$

(2) Out of the following equations which one is non a quadratic equation ?

(A) x ii + fourx = 11 + x 2

(B) 10 2 = 410

(C) fivex 2 = 90

(D) iiten – 10 2 = ten ii + five

(3) The roots of x ² + kx + g = 0 are real and equal, discover yard.

(A) 0

(B) 4

(C) 0 or 4

(D) ii

(4) For

$\sqrt{\mathrm{two}}{x}^2-5x+\sqrt{2}=0$

detect the value of the discriminant.

(A) –5

(B) 17

(C) $\sqrt{2}$

(D) $2\sqrt{2}-5$

(v) Which of the following quadratic equations has roots three,v ?

(A) ten two – 15x + 8 = 0	(B) 10 ii – 8ten + xv = 0
(C) 10 ii + 3x + 5 = 0	(D) x two + eightx – 15 = 0

(six) Out of the following equations, discover the equation having the sum of its roots –5.

(A) 3x 2 – 1510 + iii = 0	(B) x 2 – v10 + 3 = 0
(C) x 2 + iii10 – five = 0	(D) 3x two + 15x + 3 = 0

(7)

$\sqrt{5}{m}^2-\sqrt{5}\mathrm{thousand}+\sqrt{5}=0$

 which of the following statement is true for this given equation ?

(A) Real and uneual roots	(B) Real and equal roots
(C) Roots are not existent	(D) Three roots.

(viii) One of the roots of equation 10 ² + mx – 5 = 0 is 2; find 1000.

(A) –two

(B) $-\frac{1}{\mathrm{ii}}$

(C) $\frac{1}{\mathrm{two}}$

(D) 2

Respond:

(i) The general form of a quadratic equation is

$a{\mathrm{ten}}^{\mathrm{ii}}+b\mathrm{10}+c=0$

.
For (A),

$\frac{5}{x}-3=x^{\mathrm{two}}$
$5-3x={\mathrm{ten}}^3$

Thus, (A) is not quadratic equation.
For (B)x(x + five) = 2

$x^2+5\mathrm{ten}-\mathrm{two}=0$

Thus, (B) is a qudratic equation.
(C) is as well non a quadratic equation.
(D)

$\frac{1}{x^2}\left(x+\mathrm{two}\right)=x$
$x+\mathrm{two}=x^3$

 which is likewise not quadratic.
Hence, the correct respond is option (B).

(2) (A)x ^two + 4x = xi +x ²

$\Rightarrow 4x=\mathrm{xi}$

Thus, (A) is not a quadratic equation.
(B) and (C) can be written as

$x^2-\mathrm{four}x+0=0$

 and

$\mathrm{five}{\mathrm{10}}^2-\mathrm{ninety}+0=0$

 respectively.
And then, B and C are quadratic equations.
(D) iix –10 ² =ten ² + five tin be written equally

$2x^{\mathrm{ii}}-2\mathrm{10}+\mathrm{five}=0$

.
And so, it as well forms a quadratic equation.
Hence, the correct answer is option (A).

(iii) Given quadratic equation isten ² + kx + one thousand = 0
For existent and equal roots, D = 0

$$\begin{gathered} b^2-\mathrm{iv}ac=0 \\ \Rightarrow k^{\mathrm{ii}}-4\times 1\times k=0 \\ \Rightarrow {\mathrm{one\; thousand}}^2-4k=0 \\ \Rightarrow g\left(k-4\right)=0 \\ \Rightarrow k=0,k=\mathrm{four} \end{gathered}$$

Hence, the correct respond is pick C.

(4)

$\sqrt{2}{x}^2-\mathrm{v}\mathrm{ten}+\sqrt{\mathrm{two}}=0$
$$\begin{gathered} a=\sqrt{2},b=-5,c=\sqrt{2} \\ D=b^2-4ac \\ ={\left(-5\right)}^{\mathrm{ii}}-4\times \sqrt{2}\times \sqrt{2} \\ =25-8 \\ =17 \end{gathered}$$

Hence, the right answer is selection B.

(5) The roots of the quadratic equationten ^two – 810 + 15 = 0 are

$$\begin{gathered} x^2-\mathrm{five}x-3\mathrm{10}+15=0 \\ \Rightarrow x\left(x-\mathrm{five}\right)-\mathrm{three}\left(x-5\right)=0 \\ \Rightarrow \left(x-3\right)\left(\mathrm{10}-\mathrm{five}\right)=0 \\ \Rightarrow \mathrm{10}=3,5 \end{gathered}$$

Hence, the correct reply is option B.

(six) For the equation 3x ² + 15x + three = 0

$\mathrm{Sum}\mathrm{of}\mathrm{roots}=\frac{-b}{a}=\frac{-15}{3}=-5$

Hence, the right answer is option D.

(7) For the given quadratic equation

$\sqrt{5}{m}^2-\sqrt{5}\mathrm{yard}+\sqrt{5}=0$
$D=b^2-4ac={\left(-\sqrt{\mathrm{v}}\right)}^2-\mathrm{iv}\times \sqrt{\mathrm{v}}\times \sqrt{\mathrm{v}}=5-\mathrm{xx}=-15$

Since D < 0 so, the roots are not real.
Hence, the correct respond is selection C.

(8) For the equation,x ^two +mx – 5 = 0
1 of the roots is 2. So, 2 should satisfy the given equation.

$$\begin{gathered} {\left(2\right)}^{\mathrm{ii}}+m\times \left(2\right)-5=0 \\ \Rightarrow \mathrm{iv}+\mathrm{two}k-5=0 \\ \Rightarrow \mathrm{ii}m-1=0 \\ \Rightarrow m=\frac{\mathrm{i}}{2} \end{gathered}$$

Hence, the right answer is option C.

Page No 54:

Question ii:

Which of the following equations is quadratic ?
(1) ${\mathrm{10}}^{\mathrm{two}}+2x+11=0$
(2) $x^2-\mathrm{ii}x+5=x^{\mathrm{two}}$
(three) ${\left(x+\mathrm{ii}\right)}^2=2x^2$

Answer:

General form of a quadratic equation is
$a{x}^2+bx+c=0$
(i)$x^2+\mathrm{ii}x+11=0$ is in the form of the full general quadratic equation.
(2)$x^2-2x+\mathrm{v}=x^2$
$\Rightarrow -2x+5=0$
It is non in the form of quadratic equation.
(three)${\left(x+\mathrm{two}\right)}^{\mathrm{ii}}=\mathrm{ii}{\mathrm{ten}}^{\mathrm{ii}}$
$$\begin{gathered} \Rightarrow {\mathrm{ten}}^2+\mathrm{iv}+4x=\mathrm{two}{x}^{\mathrm{ii}} \\ \Rightarrow x^2-4\mathrm{10}-4=0 \end{gathered}$$
It is also in the form of a quadratic equation.
Hence, (1) and (3) are quadratic equations.

Page No 54:

Question 3:

Find the value of discriminant for each of the following equations.
(1) $2y^{\mathrm{ii}}-y+2=0$
(2) $\mathrm{v}{m}^2-\mathrm{chiliad}=0$
(iii) $\sqrt{5}{x}^2-\mathrm{10}-\sqrt{\mathrm{v}}=0$

Respond:

(1)

$\mathrm{two}{y}^{\mathrm{ii}}-y+2=0$

For the quadratic equation,

$a{x}^{\mathrm{ii}}+bx+c=0$
$D=b^2-4ac$

Here,

$$\begin{gathered} a=2,b=-\mathrm{ane},c=2 \\ D={\left(-1\right)}^{\mathrm{ii}}-4\times 2\times \mathrm{two}=1-16=-15 \end{gathered}$$

(2)

$5m^2-m=0$

For the quadratic equation,

$a{\mathrm{10}}^2+bx+c=0$
$D=b^{\mathrm{two}}-\mathrm{four}ac$

Here,

$$\begin{gathered} m=5,b=-1,c=0 \\ D={\left(-1\right)}^{\mathrm{two}}-4\times 5\times 0=1 \end{gathered}$$

(three)

$\sqrt{\mathrm{five}}{x}^2-x-\sqrt{5}=0$

For the quadratic equation,

$a{x}^2+bx+c=0$
$D=b^2-\mathrm{four}ac$

Hither,

$$\begin{gathered} a=\sqrt{5},b=-\mathrm{ane},c=-\sqrt{5} \\ D={\left(-1\right)}^2-4\times \left(\sqrt{5}\right)\times \left(-\sqrt{5}\right) \\ =\mathrm{one}+20=21 \end{gathered}$$

Page No 54:

Question 4:

One of the roots of quadratic equation $\mathrm{two}{\mathrm{ten}}^2+\mathrm{thousand}x-2=0$ is –2. find k.

Respond:

$\mathrm{two}{x}^{\mathrm{two}}+k\mathrm{10}-\mathrm{ii}=0$
One of the roots is –ii then, information technology will satisfy the given equation.
$$\begin{gathered} 2{\left(-\mathrm{two}\right)}^2+k\left(-\mathrm{ii}\right)-2=0 \\ \Rightarrow 8-2k-2=0 \\ \Rightarrow 6-2\mathrm{grand}=0 \\ \Rightarrow 6=2k \\ \Rightarrow k=3 \end{gathered}$$

Folio No 54:

Question 5:

Two roots of quadratic equations are given ; frame the equation.
(1) 10 and –10
(two) $1-3\sqrt{5}\mathrm{and}1+3\sqrt{\mathrm{five}}$
(iii) 0 and seven

Respond:

(1) x and –10
Sum of roots = x + (–ten) = 0
Production of roots =

$10\times \left(-10\right)=-100$

The general form of the quadratic equation is

$x^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)x+\mathrm{product}\mathrm{of}\mathrm{roots}=0$

So, the quadratic equation will be

$$\begin{gathered} x^2-0x-100=0 \\ \Rightarrow x^{\mathrm{ii}}-100=0 \end{gathered}$$

(2)

$1-3\sqrt{5}\mathrm{and}\mathrm{one}+3\sqrt{5}$

Sum of roots =

$1-\mathrm{iii}\sqrt{5}+\mathrm{one}+3\sqrt{5}=2$

Product of roots =

$\left(1-3\sqrt{5}\right)\left(1+3\sqrt{\mathrm{v}}\right)=1-45=-44$

The general class of the quadratic equation is

$x^2-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)x+\mathrm{product}\mathrm{of}\mathrm{roots}=0$

And so, the quadratic equation will be

${\mathrm{ten}}^2-2x-44=0$

(3) 0 and seven
Sum of roots = 0 + 7 = 7
Product of roots =

$0\times \mathrm{vii}=0$

The full general form of the quadratic equation is

${\mathrm{10}}^{\mathrm{ii}}-\left(\mathrm{Sum}\mathrm{of}\mathrm{roots}\right)x+\mathrm{product}\mathrm{of}\mathrm{roots}=0$

And then, the quadratic equation will be

$$\begin{gathered} x^{\mathrm{two}}-7\mathrm{10}+0=0 \\ \Rightarrow {\mathrm{ten}}^{\mathrm{ii}}-7x=0 \end{gathered}$$

Page No 54:

Question six:

Make up one's mind the nature of roots for each of the quadratic equation.
(1) $3x^{\mathrm{ii}}-\mathrm{five}\mathrm{10}+7=0$
(2) $\sqrt{\mathrm{iii}}{x}^{\mathrm{two}}+\sqrt{2}x-2\sqrt{\mathrm{iii}}=0$
(3) $m^{\mathrm{two}}-4x-3=0$

Answer:

(1)

$3{\mathrm{ten}}^2-5\mathrm{ten}+7=0$

In social club to detect the nature of the roots we need to find the discriminant.

$$\begin{gathered} D=b^2-4ac \\ a=\mathrm{iii},b=-\mathrm{five},c=7 \\ D={\left(-5\right)}^{\mathrm{two}}-4\times \mathrm{iii}\times 7=25-84=-59 \end{gathered}$$

Since, the D < 0 so, at that place is no real root.

(2)

$\sqrt{3}{x}^{\mathrm{two}}+\sqrt{2}x-\mathrm{ii}\sqrt{3}=0$

In club to find the nature of the roots we demand to find the discriminant.

$$\begin{gathered} D=b^2-\mathrm{four}ac \\ a=\sqrt{3},b=\sqrt{\mathrm{two}},c=-\mathrm{two}\sqrt{3} \\ D={\left(\sqrt{2}\right)}^2-\mathrm{iv}\times \sqrt{3}\times \left(-2\sqrt{3}\right)=2+24=26>0 \end{gathered}$$

Since, the D > 0 and so, real and unequal roots.

(three)

$m^2-4\mathrm{10}-3=0$

In order to notice the nature of the roots nosotros need to find the discriminant.

$$\begin{gathered} D=b^2-\mathrm{four}ac \\ a=\mathrm{one},b=-4,c=-\mathrm{three} \\ D={\left(-4\right)}^{\mathrm{ii}}-4\times \left(1\right)\times \left(-\mathrm{iii}\right)=16+12=28 \end{gathered}$$

Since, the D > 0 so, real and unequal roots.

Page No 54:

Question 7:

Solve the following quadratic equations.
(1) $\frac{\mathrm{one}}{x+5}=\frac{\mathrm{one}}{{\mathrm{10}}^2}$
(2) $x^{\mathrm{two}}-\frac{3x}{10}-\frac{\mathrm{one}}{10}=0$
(three) ${\left(2x+\mathrm{iii}\right)}^2=25$
(4) ${\mathrm{chiliad}}^2+\mathrm{v}k+5=0$
(five) $\mathrm{v}{m}^2+2\mathrm{1000}+1=0$
(six) ${\mathrm{10}}^2-4x-3=0$

Answer:

(i)

$\frac{1}{\mathrm{ten}+5}=\frac{\mathrm{ane}}{x^2}$
$$\begin{gathered} \Rightarrow x^2=x+5 \\ \Rightarrow x^2-x-5=0 \\ \Rightarrow \mathrm{10}=\frac{-\left(-\mathrm{one}\right)\pm \sqrt{{\left(-\mathrm{i}\right)}^2-4\times \mathrm{one}\times \left(-\mathrm{five}\right)}}{\mathrm{ii}\times \mathrm{ane}} \\ \Rightarrow x=\frac{1\pm \sqrt{1+20}}{\mathrm{two}}=\frac{\mathrm{ane}\pm \sqrt{21}}{2} \end{gathered}$$

(2)

$x^{\mathrm{ii}}-\frac{3\mathrm{ten}}{10}-\frac{1}{10}=0$
$$\begin{gathered} \Rightarrow 10{\mathrm{10}}^2-3x-\mathrm{one}=0 \\ \Rightarrow x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^{\mathrm{ii}}-\mathrm{iv}\times \mathrm{ten}\times \left(-\mathrm{i}\right)}}{2\times \mathrm{ten}} \\ \Rightarrow x=\frac{3\pm \sqrt{9+40}}{20}=\frac{3\pm \sqrt{49}}{20} \\ \Rightarrow x=\frac{3\pm 7}{20} \\ \Rightarrow \mathrm{ten}=\frac{\mathrm{iii}+\mathrm{seven}}{20},\frac{3-7}{\mathrm{twenty}} \\ \Rightarrow x=\frac{\mathrm{x}}{20},\frac{-4}{20} \\ \Rightarrow x=\frac{\mathrm{i}}{2},\frac{-\mathrm{ane}}{5} \end{gathered}$$

(3)

${\left(2\mathrm{ten}+\mathrm{iii}\right)}^{\mathrm{ii}}=25$
$$\begin{gathered} \Rightarrow \mathrm{four}{x}^2+9+12x=25 \\ \Rightarrow 4{\mathrm{10}}^{\mathrm{ii}}+12x-16=0 \\ \Rightarrow x^2+\mathrm{three}x-4=0 \\ \Rightarrow {\mathrm{10}}^2+4x-\mathrm{10}-4=0 \\ \Rightarrow \mathrm{10}\left(\mathrm{ten}+4\right)-1\left(\mathrm{ten}+4\right)=0 \\ \Rightarrow \left(\mathrm{ten}-1\right)\left(x+\mathrm{iv}\right)=0 \\ \Rightarrow \mathrm{10}=1,-4 \end{gathered}$$

(iv)

$m^2+5k+\mathrm{v}=0$
$$\begin{gathered} \Rightarrow m=\frac{-\mathrm{five}\pm \sqrt{{\left(5\right)}^2-4\times 1\times \mathrm{v}}}{2} \\ \Rightarrow m=\frac{-\mathrm{v}\pm \sqrt{25-20}}{\mathrm{ii}} \\ \Rightarrow g=\frac{-\mathrm{v}\pm \sqrt{5}}{2} \end{gathered}$$

(5)

$5{\mathrm{chiliad}}^2+2\mathrm{thou}+\mathrm{ane}=0$
$$\begin{gathered} \Rightarrow \mathrm{yard}=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\times \mathrm{i}\times 5}}{2} \\ \Rightarrow m=\frac{-2\pm \sqrt{4-\mathrm{xx}}}{\mathrm{ii}} \\ \Rightarrow \mathrm{thou}=\frac{-2\pm \sqrt{-\mathrm{xvi}}}{2} \end{gathered}$$

So, the roots are non real every bit the discriminant is negative.

(6)

${\mathrm{ten}}^{\mathrm{two}}-4\mathrm{10}-3=0$
$$\begin{gathered} \Rightarrow x=\frac{-\left(-\mathrm{four}\right)\pm \sqrt{{\left(-4\right)}^{\mathrm{ii}}-\mathrm{four}\times 1\times \left(-3\right)}}{2\times 1} \\ \Rightarrow x=\frac{4\pm \sqrt{16+12}}{2} \\ \Rightarrow \mathrm{10}=\frac{4\pm \sqrt{28}}{2} \\ \Rightarrow \mathrm{10}=\frac{\mathrm{four}\pm 2\sqrt{7}}{2} \\ \Rightarrow x=\mathrm{two}\pm \sqrt{7} \end{gathered}$$

Folio No 54:

Question 8:

Detect grand if (m – 12) ten ² + 2(m –12) x + 2 = 0 has real and equal roots.

Reply:

For real and equal roots, the Discriminant should exist 0.
(m – 12)ten ² + 2(m–12)x+ two = 0
$$\begin{gathered} ∆=b^2-4ac=0 \\ a=k-12,b=2\left(m-12\right),c=\mathrm{ii} \end{gathered}$$
$$\begin{gathered} {\left[2\left(\mathrm{thousand}-12\right)\right]}^2-\mathrm{four}\left(m-12\right)\times \mathrm{ii}=0 \\ \Rightarrow 4m^{\mathrm{two}}+576-96m-8m+96=0 \\ \Rightarrow \mathrm{iv}{m}^{\mathrm{two}}-104k+672=0 \\ \Rightarrow {\mathrm{1000}}^2-26\mathrm{yard}+168=0 \\ \Rightarrow m^2-14\mathrm{yard}-12\mathrm{thousand}+168=0 \\ \Rightarrow m\left(m-14\right)-12\left(m-\mathrm{xiv}\right)=0 \\ \Rightarrow \left(\mathrm{1000}-14\right)\left(g-12\right)=0 \\ m=12,\mathrm{xiv} \end{gathered}$$
But if yard = 12, the quadratic equation volition be formed hence, the vaalue of mis xiv.

Page No 54:

Question nine:

The sum of ii roots of a quadratic equation is v and sum of their cubes is 35, observe the equation.

Answer:

Let the roots be$\alpha \mathrm{and}\beta$.
Sum of roots = v
$\alpha +\beta =5$
Likewise,
$$\begin{gathered} {\alpha }^3+{\beta }^3=35 \\ \Rightarrow {\alpha }^3+{\beta }^{\mathrm{three}}={\left(\alpha +\beta \right)}^{\mathrm{three}}-3\alpha \beta \left(\alpha +\beta \right) \\ \Rightarrow 35=5^3-3\alpha \beta \left(\mathrm{five}\right) \\ \Rightarrow \mathrm{fifteen}\alpha \beta =125-35 \\ \Rightarrow \alpha \beta =\mathrm{vi} \end{gathered}$$
The general quadratic equation is
$x^2-\left(\alpha +\beta \right)\mathrm{ten}+\alpha \beta =0$
So, the required quadratic equation will be
$$\begin{gathered} x^2-\left(5\right)\mathrm{10}+\mathrm{six}=0 \\ \Rightarrow x^2-\mathrm{five}x+\mathrm{half\; dozen}=0 \end{gathered}$$

Folio No 54:

Question 10:

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation $2{\mathrm{ten}}^{\mathrm{two}}+2\left(p+q\right)x+p^2+q^2=0$

Answer:

$\mathrm{ii}{\mathrm{ten}}^2+\mathrm{ii}\left(p+q\right)x+p^2+q^2=0$
Let the roots of the given quadratic equation be$\alpha$ and$\beta$.
Sum of roots,$\alpha$ +$\beta$ =$\frac{-2\left(p+q\right)}{2}=-\left(p+q\right)$
$\Rightarrow {\left(\alpha +\beta \right)}^{\mathrm{ii}}={\left(p+q\right)}^2$                              .....(A)
$\alpha \beta =\frac{p^2+q^2}{2}$
$$\begin{gathered} {\left(\alpha -\beta \right)}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta \\ ={\left(p+q\right)}^2-4\left(\frac{p^2+q^{\mathrm{two}}}{2}\right) \\ ={\left(p+q\right)}^2-2\left(p^2+q^2\right) \\ =-{\left(p-q\right)}^2.....\left(B\right) \end{gathered}$$
$$\begin{gathered} A+B={\left(p+q\right)}^2-{\left(p-q\right)}^{\mathrm{two}} \\ =\left(p+q-p+q\right)\left(p+q+p-q\right)\left[\because x^2-y^2=\left(x-y\right)\left(x+y\right)\right] \\ =4pq \end{gathered}$$
$$\begin{gathered} AB={\left(p+q\right)}^{\mathrm{two}}\left[-{\left(p-q\right)}^2\right] \\ =-\left(p^2+q^2+2pq\right)\left(p^2+q^{\mathrm{two}}-\mathrm{two}pq\right) \\ =-\left[{\left(p^{\mathrm{two}}\right)}^2+{\left(q^2\right)}^{\mathrm{two}}-2p^2{q}^2\right] \\ =-{\left(p^{\mathrm{ii}}-q^{\mathrm{ii}}\right)}^{\mathrm{ii}} \end{gathered}$$
Full general grade of quadratic equation is
${\mathrm{ten}}^{\mathrm{ii}}-\left(A+B\right)\mathrm{ten}+AB=0$
Putting the value of A and B we become
$x^{\mathrm{ii}}-\mathrm{iv}pq-{\left(p^{\mathrm{two}}-q^{\mathrm{ii}}\right)}^2=0$.

Folio No 54:

Question 11:

Mukund possesses Rs 50 more than what Sagar possesses. The product of the amount they take is 15,000. Find the amount each one has.

Answer:

Let amount with Sagar be Rs x.
Amount with Mukund = Rs ten+ 50
The product of the amount they have is fifteen,000.
$$\begin{gathered} x\left(x+50\right)=15000 \\ \Rightarrow {\mathrm{10}}^2+50x=15000 \\ \Rightarrow x^{\mathrm{two}}+50\mathrm{ten}-15000=0 \\ \Rightarrow x^{\mathrm{two}}+150\mathrm{10}-100x-15000=0 \\ \Rightarrow \mathrm{ten}\left(x+150\right)-100\left(\mathrm{ten}+150\right)=0 \\ \Rightarrow \left(x-100\right)\left(x+150\right)=0 \\ \Rightarrow x=100,-150 \end{gathered}$$
Simply corporeality cannot be negative so,
Amount with Sagar = Rs 100 and that with Mukund is Rs 150.

Page No 54:

Question 12:

The divergence between squares of 2 numbers is 120. The square of smaller number is twice the greater number. Discover the numbers.

Answer:

Allow the smaller number be x.
Larger number be y.
The foursquare of smaller number is twice the greater number.
$x^{\mathrm{two}}=2y$
The departure between squares of two numbers is 120.
$$\begin{gathered} y^{\mathrm{ii}}-x^2=120 \\ \Rightarrow y^{\mathrm{two}}-\mathrm{two}y-120=0 \\ \Rightarrow y^2-12y+10y-120=0 \\ \Rightarrow y\left(y-12\right)+10\left(y-12\right)=0 \\ \Rightarrow \left(y+10\right)\left(y-12\right)=0 \\ \Rightarrow y=12,-\mathrm{x} \end{gathered}$$
And then, the numbers are 12 as the larger number and the smaller number is
$$\begin{gathered} x^2=2\times 12=24 \\ \Rightarrow x=\pm \sqrt{24} \end{gathered}$$

Folio No 54:

Question thirteen:

Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Notice the number of students.

Answer:

Full oranges = 540
Allow the total number of students be 10.
Orange each educatee gets volition exist$\frac{540}{x}$
If 30 more students are there and then, the total number of students will existten+ thirty
Number of oranges per person will be$\frac{540}{x+30}$.
If thirty students were more each would go 3 oranges less = $\frac{540}{x}-\mathrm{three}$
$\frac{540}{x+\mathrm{thirty}}$ =$\frac{540}{\mathrm{ten}}-3$
$$\begin{gathered} \Rightarrow 3=540\left(\frac{1}{x}-\frac{\mathrm{i}}{\mathrm{ten}+30}\right) \\ \Rightarrow 1=180\left(\frac{x+30-x}{\mathrm{ten}\left(x+30\right)}\right) \\ \Rightarrow 1=180\left(\frac{30}{x^{\mathrm{ii}}+\mathrm{xxx}\mathrm{10}}\right) \\ \Rightarrow x^{\mathrm{two}}+\mathrm{xxx}\mathrm{ten}=5400 \\ \Rightarrow {\mathrm{10}}^2+30\mathrm{ten}-5400=0 \\ \Rightarrow x^2+90x-60x-5400=0 \\ \Rightarrow \mathrm{10}\left(x+90\right)-60\left(\mathrm{ten}+\mathrm{xc}\right)=0 \\ \Rightarrow \left(x-60\right)\left(\mathrm{ten}+90\right)=0 \\ \Rightarrow x=\mathrm{lx},-90 \end{gathered}$$
But number of students cannot be negative so, the number of students is threescore.

Page No 54:

Question 14:

Mr. Dinesh owns an agricultural subcontract at village Talvel. The length of the farm is 10 meter more than twice the breadth. In social club to harvest rain water, he dug a foursquare shaped pond within the subcontract. The side of swimming is$\frac{1}{3}$ of the latitude of the farm. The area of the farm is 20 times the area of the pond. Notice the length and breadth of the farm and of the pond.

Answer:

Breadth of subcontract = ten
Length of subcontract =$2x+10$
Surface area of farm =$lb=x\left(2x+10\right)=2x^{\mathrm{two}}+10x$
Side of pond =$\frac{\mathrm{10}}{\mathrm{three}}$
Area of pond =${\mathrm{south}}^2={\left(\frac{x}{3}\right)}^{\mathrm{two}}$
Surface area of farm = 2(Area of swimming)
$$\begin{gathered} \Rightarrow \mathrm{ii}{\mathrm{10}}^2+10\mathrm{ten}=20\times \frac{x^2}{9} \\ \Rightarrow 18x^2+90x=\mathrm{twenty}{x}^2 \\ \Rightarrow \mathrm{ix}{x}^2+45x=10x^2 \\ \Rightarrow {\mathrm{ten}}^2-45\mathrm{10}=0 \\ \Rightarrow \mathrm{ten}\left(x-45\right)=0 \\ \Rightarrow x=0,45 \end{gathered}$$
Length of farm =$\mathrm{two}\times 45+\mathrm{x}=100\mathrm{units}$
Breadth of farm = 45 units
Side of pond =$\frac{45}{\mathrm{iii}}=\mathrm{fifteen}$ units

View NCERT Solutions for all chapters of Class x

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Source: https://www.meritnation.com/maharashtra-class-10/math/mathematics-part---i-%28solutions%29/quadratic-equations/textbook-solutions/91_1_3379_24383

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