Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=x^2-x-2$, find the value of $\frac{1}{\alpha }-\frac{1}{\beta }.$

• A. $\frac{-3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $1$

Answer 1

Answer: (A)

$\frac{-3}{2}$

$f\left(x\right)=x^2-x-2$

$a=1$

$b=-1$

$c=-2$

$D=b^2-4ac={(-1)}^2-4\times 1\times (-2)=1+8=9$

$\because \alpha$ and $\beta$ are the zeroes of above polynomial.

$\therefore$ Sum of roots $=\frac{-b}{a}$

$\Rightarrow \alpha +\beta =-\frac{-1}{1}$

$\Rightarrow \alpha +\beta =1⟶\left(1\right)$

Difference of roots $=\frac{\sqrt{D}}{a}$

$\Rightarrow |\beta -\alpha |=\frac{\sqrt{9}}{1}=3$

Product of roots $=\frac{c}{a}$

$\Rightarrow \alpha \beta =\frac{-2}{1}$

$\Rightarrow \alpha \beta =-2⟶\left(2\right)$

$\therefore \frac{1}{\alpha }-\frac{1}{\beta }=\frac{|\beta -\alpha |}{\alpha \beta }$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\Rightarrow \frac{1}{\alpha }-\frac{1}{\beta }=\frac{3}{-2}=-\frac{3}{2}$

Or

Simply find the roots of given equation & hence solve.

Hence, (A) is the correct answer.