Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $2x^2-4x+5$. Form the polynomial where zeros are $\frac{1}{{\alpha }^2}$ and $\frac{1}{{\beta }^2}$.

• A. $\frac{1}{2}(25x^2+4x+4)$
• B. $\frac{1}{2}(45x^2+9x+4)$
• C. $\frac{1}{2}(92x^2+16x+4)$
• D. $\frac{1}{2}(44x^2+5x+4)$

Answer 1

The correct option is A $\frac{1}{2}(25x^2+4x+4)$

Given, $2x^2-4x+5$

$\therefore \alpha +\beta =\frac{4}{2}=2$

and

$\alpha \beta =\frac{5}{2}$

we have to find

$\frac{1}{(\alpha {)}^2}+\frac{1}{(\beta {)}^2}=\frac{{\alpha }^2+{\beta }^2}{(\alpha \beta {)}^2}=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}$

$=\frac{2^2-2\left(\frac{5}{2}\right)}{(\frac{5}{2}{)}^2}$

$=\frac{-1}{\frac{25}{4}}=\frac{-4}{25}$

and

$\frac{1}{(\alpha {)}^2}\times \frac{1}{(\beta {)}^2}=\frac{1}{(\alpha \beta {)}^2}=\frac{1}{(\frac{5}{2}{)}^2}=\frac{4}{25}$

So, polynomial whose zeros are $\frac{1}{(\alpha {)}^2}$ and $\frac{1}{(\beta {)}^2}$ is

$x^2-\left(sumofzeros\right)x+\left(productofzeros\right)$

$\Rightarrow x^2+\frac{4}{25}x+\frac{4}{25}$

$\Rightarrow \frac{25x^2+4x+4}{25}$

$\Rightarrow \frac{1}{25}(25x^2+4x+4)$

Option A is correct.