# Which of the following equations

Question:

Which of the following equations has the sum of its roots as 3 ?

(a) $2 x^{2}-3 x+6=0$

(b) $-x^{2}+3 x-3=0$

(c) $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+1=0$

(d) $3 x^{2}-3 x+3=0$

Solution:

(b)

(a) Given that, $2 x^{2}-3 x+6=0$

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

$a=2, b=-3$ and $c=6$

$\therefore \quad$ Sum of the roots $=\frac{-b}{a}=\frac{-(-3)}{2}=\frac{3}{2}$

So, sum of the roots of the quadratic equation $2 x^{2}-3 x+6=0$ is not $3 ;$ so it is not the answer.

(b) Given that, $-x^{2}+3 x-3=0$

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

$a=-1, b=3$ and $c=-3$

$\therefore$ Sum of the roots $=\frac{-b}{a}=\frac{-(3)}{-1}=3$

So, sum of the roots of the quadratic equation $-x^{2}+3 x-3=0$ is 3 , so it is the answer.

(c) Given that, $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+1=0$

$\Rightarrow \quad 2 x^{2}-3 x+\sqrt{2}=0$

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

$a=2, b=-3$ and $c=\sqrt{2}$

$\therefore \quad$ Sum of the roots $=\frac{-b}{a}=\frac{-(-3)}{2}=\frac{3}{2}$

So, sum of the roots of the quadratic equation $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+1=0$ is not 3 , so it is not

(d) Given that, $3 x^{2}-3 x+3=0$

$\Rightarrow \quad x^{2}-x+1=0$

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

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

$\therefore \quad$ Sum of the roots $=\frac{-b}{a}=\frac{-(-1)}{1}=1$

So, sum of the roots of the quadratic equation $3 x^{2}-3 x+3=0$ is not 3, so it is not the answer.