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