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# Choose the correct answer from the given four options in the following questions:

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 $

,b>(C) $ \sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+1=0 $

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

To do:

We have to find the correct answer.

Solution:

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

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

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

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.

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

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

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

Sum of the roots $=\frac{-3}{-1}$

$=3$

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

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

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

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

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

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.

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

$x^{2}-x+1=0$

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

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

Sum of the roots $=\frac{-b}{a}$

$=\frac{-(-1)}{1}$

$=1$

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