Question

If $\alpha \&\beta$ be the zeroes of $2x^2-9x+4$ then $\frac{1}{\alpha }+\frac{1}{\beta }$ =

• A. $\frac{9}{4}$
• B. $\frac{-9}{4}$
• C. $\frac{-1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{9}{4}$

We can simplify as,
$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\beta +\alpha }{\alpha \beta }---\left(1\right)$

So, Lets find out Sum and Product of Zeroes for $2x^2-9x+4$

Compare with standard form $a{x}^2+bx+c$
$a=2,b=-9,c=4$

Sum of Zeroes, $\alpha +\beta =\frac{-b}{a}=\frac{-(-9)}{2}=\frac{9}{2}$

Product of Zeroes, $\alpha \times \beta =\frac{c}{a}=\frac{4}{2}=2$

Substitute in $\left(1\right)$

$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\frac{9}{2}}{2}=\frac{9}{2}\times \frac{1}{2}=\frac{9}{4}$