Question

Question 5
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) $3x^2-3x+3=0$​​​​​​​

Answer 1

a) Given that, $2x^2-3x+6=0$
On comparing with $a{x}^2$ + bx + 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 $2x^2$ – 3x + 6 = 0 is not 3, so it not the answer
b) Given that, $-x^2$ + 3x – 3 = 0
On compare with $a{x}^2$ + bx + c = 0, we get
a = -1, b = 3 and c = - 3
$\therefore$ sum of the roots $=\frac{-b}{a}=\frac{-\left(3\right)}{-1}=3$
So, sum of the roots of the quadratic equation $-x^2+3x–3$ is 3, so it is the answer.
c) Given that $\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+1=0$
$\Rightarrow 2x^2-3x+\sqrt{2}=0$
On comparing with $a{x}^2$ + bx + 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, so it is not the answer.
d) Given that, $3x^2$ – 3x + 3 = 0
$\Rightarrow x^2-x+1=0$
On comparing with $a{x}^2$ + bx + c = 0 get
a = 1, b = - 1 and c = 1
sum of the roots $=\frac{-b}{a}=\frac{-(-1)}{1}=1$
So, sum of the roots of the quadratic equation $3x^2$ - 3x + 3 = 0 is not 3, so it is not the answer.