Q. If the roots of the equation $3x^{2} - 6x + 5 = 0$ are $\alpha$ and $\beta$, then the equation whose roots are $\alpha + \beta$ and $\frac{2}{\alpha + \beta}$ will be
A) $x^{2} + 3x - 1 = 0$
B) $x^{2} + 3x - 2 = 0$
C) $x^{2} + 3x + 2 = 0$
D) $x^{2} - 3x + 2 = 0$
Explanation
(d)
Here, $\alpha + \beta = 2$ and $\alpha\beta = \frac{5}{3}$
Now, $\alpha + \beta + \frac{2}{\alpha + \beta} = 2 + \frac{2}{2} = 3$
And $(\alpha + \beta) \times \frac{2}{\alpha + \beta} = 2$
$\therefore$ Required equation is
$x^{2} - \left( (\alpha + \beta) + \frac{2}{(\alpha + \beta)} \right)x + \left( (\alpha + \beta) \times \frac{2}{(\alpha + \beta)} \right) = 0$
$\Rightarrow \ \ \ x^{2} - 3x + 2 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.