Solve the following quadratic equations:
(i) $ x^{2}+x-20=0 $
(ii) $ 9 y^{2}-12 y+2=0 $

Given:

(i) \( x^{2}+x-20=0 \)
(ii) \( 9 y^{2}-12 y+2=0 \)

To do:

We have to solve the given quadratic equations.

Solution:

(i) $x^{2}+x-20=0$

$x^2+5x-4x-20=0$                   [Since $5x-4x=x$ and $5x \times (-4x)=-20x^2$]
$x(x+5)-4(x+5)=0$

$(x+5)(x-4)=0$

$\Rightarrow (x+5)=0$ or $x-4=0$

$\Rightarrow x=-5$ or $x=4$.

(ii) $9y^{2}-12y+2=0$

$y=\frac{-(-12) \pm \sqrt{(-12)^2-4\times9\times2}}{2\times9}$

$=\frac{12 \pm \sqrt{144-72}}{18}$

$=\frac{12 \pm \sqrt{72}}{18}$

$=\frac{12 \pm \sqrt{36\times2}}{18}$

$=\frac{6\times2 \pm 6\sqrt{2}}{6\times3}$

$=\frac{2 \pm \sqrt{2}}{3}$

$y=\frac{2 + \sqrt{2}}{3}$ or $y=\frac{2-\sqrt{2}}{3}$.

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Updated on: 10-Oct-2022

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