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#### Find the roots of the quadratic equations by using the quadratic formula in each of the following:$(i)2x^{2}-3x-5=0$$(ii)5x^{2}+13x+8=0$       $(iii)-3x^{2}+5x+12=0$$(iv)-x^{2}+7x-10=0$$(v)x^{2}+2\sqrt{2}x-6=0$$(vi)x^{2}+3\sqrt{5}x+10=0$$(vii)\frac{1}{2}x^{2}-\sqrt{11}x+1=0$

(i) $\frac{5}{2},-1$

$2x^{2}-3x-5=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

Here    a=2, b=-3, c=-5
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-(-3)\pm \sqrt{(-3)^{2}-4(2)(-5)}}{2 \times 2}\\ x=\frac{3 \pm \sqrt{9+40}}{4}\\ x=\frac{3 \pm 7}{4}\\ x=\frac{3+7}{4}=\frac{10}{4}=\frac{5}{2 }\;\; \; \;\;\; x=\frac{3-7}{4} =\frac{-4}{4} =-1$

$x=\frac{5}{2},-1$

$\left (\frac{5}{2},-1 \right )$  are the roots of the equation $2x^{2}-3x-5=0$

(ii) $-1,\frac{-8}{5}$

$5x^{2}+13x+8=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

Here    a=5, b=13, c=8
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-13\pm \sqrt{(13)^{2}-4(5)(8)}}{2 \times 5}\\ x=\frac{-13 \pm \sqrt{169-160}}{10}\\ x=\frac{-13 \pm 3}{10}\\ x=\frac{-13+3}{10}=\frac{-10}{10}=-1\;\; \; \;\;\; x=\frac{-13-3}{10} =\frac{-16}{10} =\frac{-8}{5}$

$x=-1,\frac{-8}{5}$

$\left (-1,\frac{-8}{5} \right )$  are the roots of the equation $5x^{2}+13x+8=0$

(iii) $\frac{-4}{3},3$

$-3x^{2}+5x+12=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

Here    a=-3, b=5, c=12

$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-5\pm \sqrt{(5)^{2}-4(-3)(12)}}{2 \times (-3)}\\ x=\frac{-5 \pm \sqrt{25+144}}{-6}\\ x=\frac{-5 \pm 13}{-6}\\ x=\frac{-5+13}{-6}=\frac{8}{-6}=\frac{-4}{3 }\;\; \; \;\;\; x=\frac{-5-13}{-6} =\frac{-18}{-6} =3$

$x=\frac{-4}{3},3$

$\left (\frac{-4}{3},3 \right )$  $\text{are the roots of the equation}$ $-3x^{2}+5x+12=0$

(iv) 5,2

$-x^{2}+7x-10=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

Here    a=-1, b=7, c=-10

$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-7\pm \sqrt{49-4(-1)(-10)}}{2 \times (-1)}\\ x=\frac{-7 \pm \sqrt{9}}{-2}\\ x=\frac{-7 \pm 3}{-2}\\ x=\frac{-7+3}{-2}=\frac{-4}{-2}=2\;\; \; \;\;\; x=\frac{-7-3}{-2} =\frac{-10}{-2} =5$

x=5,2

(5,2)  are the roots of the equation $-x^{2}+7x-10=0$

(v) $\sqrt{2},-3\sqrt{2}$

$x^{2}+2\sqrt{2}x-6=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

Here    $a=1, b=2\sqrt{2}, c=-6$

$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-2\sqrt{2}\pm \sqrt{(2\sqrt{2})^{2}-4(1)(-6)}}{2 \times 1}\\ x=\frac{-2\sqrt{2} \pm \sqrt{8+24}}{2}\\ x=\frac{2\sqrt{2} \pm 4\sqrt{2}}{2}\\ x=\frac{2\left (\sqrt{2} \pm 2 \sqrt{2} \right )}{2}\\ x=-\sqrt{2}\pm2\sqrt{2}\\ x=-\sqrt{2}+2\sqrt{2}=\sqrt{2}\;\; \; \;\;\; x=-\sqrt{2}-2\sqrt{2}=-3\sqrt{2}$

$x=\sqrt{2},-3\sqrt{2}$

$\sqrt{2},-3\sqrt{2}$  are the roots of the equation $x^{2}+2\sqrt{2}x-6=0$

(vi) $2\sqrt{5},\sqrt{5}$

$x^{2}+3\sqrt{5}x+10=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

$a=1, b=3\sqrt{5}, c=10$

$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{3\sqrt{5}\pm \sqrt{(3\sqrt{5})^{2}-4(1)(10)}}{2 \times 1}\\ x=\frac{3\sqrt{5} \pm \sqrt{45-40}}{2}\\ x=\frac{3\sqrt{5} \pm \sqrt{5}}{2}\\ x=\frac{3\sqrt{5} + \sqrt{5}}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}\;\; \; \;\;\; x=\frac{3\sqrt{5} - \sqrt{5}}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5}$

$x=2\sqrt{5},\sqrt{5}$

$2\sqrt{5},\sqrt{5}$  are the roots of the equation $x^{2}+3\sqrt{5}x+10=0$

(vii) $\sqrt{11}+3,\sqrt{11}-3$

$\frac{1}{2}x^{2}-\sqrt{11}x+1=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

Here $a=\frac{1}{2},b=-\sqrt{11},c=1$

$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{\sqrt{11}\pm \sqrt{(-\sqrt{11})^{2}-4\left ( \frac{1}{2} \right )(1)}}{2 \times \frac{1}{2}}\\ x=\frac{\sqrt{11}\pm\sqrt{11-2}}{1}\\ x=\sqrt{11}\pm 3\\ x=\sqrt{11}+ 3\;\;\;\;\;\;\;x=\sqrt{11}- 3$

$x=\left (\sqrt{11}+ 3 \right ),\left (\sqrt{11}- 3 \right )$

hence $\left (\sqrt{11}+ 3 \right ),\left (\sqrt{11}- 3 \right )$ are the root of the equation $\frac{1}{2}x^{2}-\sqrt{11}x+1=0$