Question

If $\alpha \text{and}\beta$ are the roots of $x^2-5x+6=0$ then $\frac{1}{\alpha }+\frac{1}{\beta }$ is

• A. $\frac{4}{5}$
• B. $\frac{3}{5}$
• C. $\frac{6}{5}$
• D. $\frac{5}{6}$

Answer 1

The correct option is D $\frac{5}{6}$

Given, $x^2-5x+6=0$
$\Rightarrow x^2-2x-3x+6=0$
$\Rightarrow x(x-2)-3(x-2)=0\Rightarrow (x-2)(x-3)=0$
So, roots are say, $\alpha =2,\beta =3$
Thus, $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$

Alternate:

For equation $x^2-5x+6=0$.
$\text{Sum of roots}=\frac{-b}{a}=\frac{-(-5)}{1}=5$
$\text{products of roots}=\frac{c}{a}=\frac{6}{1}=6$
$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \cdot \beta }=\frac{\text{Sum of roots}}{\text{Product of roots}}=\frac{5}{6}$