#### Which of the following equations has the sum of its roots as 3?          $\\(A)2x^{2}-3x+6=0\\ (B)-x^{2}+3x-3=0\\ (C)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0\\ (D)3x^{2}-3x+3=0$

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

If $(B)ax^{2}+bx+c=0$  is a quadratic equation the sum of its roots are $-\frac{b}{a}$

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

Here b=-3,a=2

Sum of its roots =$-\frac{b}{a}$$-\left ( \frac{-3}{2} \right )=\frac{3}{2}$

(B)

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

Here b=-3,a=3

Sum of its roots =$-\frac{b}{a}=\frac{-3}{-1}=3$

(C)

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

Here $b=\frac{-3}{\sqrt{2}}, a=\sqrt{2}$

Sum of its roots = $-\frac{b}{a}=-\left (\frac{-3}{\sqrt{2}} \right )\times \sqrt{2}=\frac{3}{2}$

(D)

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

Here b=-3,a=3

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

Hence only B has sum of its roots 3