Question

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) $3x^2-3x+3=0$

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Answer 1

The correct option is B

(B)

Explanation for the correct option:

Option (B): Determine whether the equation has the sum of its roots as $3$.

Consider $-x^2+3x-3=0$.

The sum of roots of a quadratic equation $a{x}^2+bx+c=0$ is given by $-\frac{b}{a}$.

$\begin{array}{rcl}\mathrm{Sum}\mathrm{of}\mathrm{roots}& =& -\frac{b}{a}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=-\frac{3}{\left(-1\right)}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=3\end{array}$

Hence, option (B) is correct.

Explanation for the incorrect options:

Option (A): Consider $2x^2-3x+6=0$.

$\begin{array}{rcl}\mathrm{Sum}\mathrm{of}\mathrm{roots}& =& -\frac{b}{a}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=-\frac{\left(-3\right)}{2}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=\frac{3}{2}\end{array}$

Therefore, option (A) is incorrect.

Option (C): Consider $\sqrt{2}{x}^2-\frac{3}{\sqrt{2}x}+1=0$.

$\begin{array}{rcl}\mathrm{Sum}\mathrm{of}\mathrm{roots}& =& -\frac{b}{a}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=-\frac{\left(-{\displaystyle \frac{3}{\sqrt{2}}}\right)}{\sqrt{2}}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=\frac{3}{2}\end{array}$

Therefore, option (C) is incorrect.

Option (D): Consider $3x^2-3x+3=0$.

$\begin{array}{rcl}\mathrm{Sum}\mathrm{of}\mathrm{roots}& =& -\frac{b}{a}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=-\frac{\left(-3\right)}{3}\\ & \Rightarrow & \mathrm{Sum}\mathrm{of}\mathrm{roots}=1\end{array}$

Therefore, option (D) is incorrect.

Thus, options (A), (C), and (D) are incorrect.

Hence, option (B) is correct.