# Find the roots of the quadratic equations

Question:

Find the roots of the quadratic equations by using the quadratic formula irt each of the following

Solution:

(i) Given equation is $2 x^{2}-3 x-5=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=2, b=-3$ and $c=-5$

By quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-3) \pm \sqrt{(-3)^{2}-4(2)(-5)}}{2(2)}=\frac{3 \pm \sqrt{9+40}}{4}$

$=\frac{3 \pm \sqrt{49}}{4}=\frac{3 \pm 7}{4}=\frac{10}{4}, \frac{-4}{4}=\frac{5}{2},-1$

So, $\frac{5}{2}$ and $-1$ are the roots of the given equation.

(ii) Given equation is $5 x^{2}+13 x+8=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=5, b=13$ and $c=8$

By quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(13) \pm \sqrt{(13)^{2}-4(5)(8)}}{2(5)}$

$=\frac{-13 \pm \sqrt{169-160}}{10}=\frac{-13 \pm \sqrt{9}}{10}$

$=\frac{-13 \pm 3}{10}=-\frac{10}{10},-\frac{16}{10}=-1,-\frac{8}{5}$

(iii) Given equation is $-3 x^{2}+5 x+12=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=-3, b=5$ and $c=12$

By quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(5) \pm \sqrt{(5)^{2}-4(-3)(12)}}{2(-3)}$

$=\frac{-5 \pm \sqrt{25+144}}{-6}=\frac{-5 \pm \sqrt{169}}{-6}$

$=\frac{-5 \pm 13}{-6}=\frac{8}{-6}, \frac{-18}{-6}=-\frac{4}{3}, 3$

So, $-\frac{4}{3}$ and 3 are two roots of the given equation.

(iv) Given equation is $-x^{2}+7 x-10=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=-1, b=7$ and $c=-10$

By quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(7) \pm \sqrt{(7)^{2}-4(-1)(-10)}}{2(-1)}=\frac{-7 \pm \sqrt{49-40}}{-2}$

$=\frac{-7 \pm \sqrt{9}}{-2}=\frac{-7 \pm 3}{-2}=\frac{-4}{-2}, \frac{-10}{-2}=2,5$

So, 2 and 5 are two roots of the given equation.

(v) Given equation is $x^{2}+2 \sqrt{2} x-6=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=1, b=2 \sqrt{2}$ and $c=-6$

By quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(2 \sqrt{2}) \pm \sqrt{(2 \sqrt{2})^{2}-4(1)(-6)}}{2(1)}=\frac{-2 \sqrt{2} \pm \sqrt{8+24}}{2}$

$=\frac{-2 \sqrt{2} \pm \sqrt{32}}{2}=\frac{-2 \sqrt{2} \pm 4 \sqrt{2}}{2}$

$=\frac{-2 \sqrt{2}+4 \sqrt{2}}{2}, \frac{-2 \sqrt{2}-4 \sqrt{2}}{2}=\sqrt{2},-3 \sqrt{2}$

So, $\sqrt{2}$ and $-3 \sqrt{2}$ are the roots of the given equation.

(vi) Given equation is $x^{2}-3 \sqrt{5} x+10=0$

On comparing with $a x^{2}+b x+c=0$, we have

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

By quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-3 \sqrt{5}) \pm \sqrt{(-3 \sqrt{5})^{2}-4(1)(10)}}{2(1)}$

$=\frac{3 \sqrt{5} \pm \sqrt{45-40}}{2}=\frac{3 \sqrt{5} \pm \sqrt{5}}{2}$

$=\frac{3 \sqrt{5}+\sqrt{5}}{2}, \frac{3 \sqrt{5}-\sqrt{5}}{2}=2 \sqrt{5}, \sqrt{5}$

So, $2 \sqrt{5}$ and $\sqrt{5}$ are the roots of the given equation.

(vii) Given equation is $\frac{1}{2} x^{2}-\sqrt{11} x+1=0$.

On comparing with $a x^{2}+b x+c=0$, we get

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

$\therefore$ By quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-\sqrt{11}) \pm \sqrt{(-\sqrt{11})^{2}-4 \times \frac{1}{2} \times 1}}{2\left(\frac{1}{2}\right)}$

$=\frac{\sqrt{11} \pm \sqrt{11-2}}{1}=\sqrt{11} \pm \sqrt{9}$

$=\sqrt{11} \pm 3=3+\sqrt{11}, \sqrt{11}-3$

So, $3+\sqrt{11}$ and $\sqrt{11}-3$ are the roots of the given equation.