Question

If $\alpha$ and $\beta$ are the two zeros of the polynomial ${25p}^2-15p+2,$ find a quadratic polynomial whose zeros are $\frac{1}{2\alpha }$ and $\frac{1}{2\beta }.$

• A. $\frac{1}{8}({6p}^2+25p-30)$
• B. $\frac{1}{8}({6p}^2-30p+25)$
• C. $\frac{1}{8}({8p}^2+30p-25)$
• D. $\frac{1}{8}({8p}^2-30p+25)$

Answer 1

Answer: (D)

$\frac{1}{8}({8p}^2-30p+25)$

We have,

Polynomial $25p^2-15p+2$

On comparing that,

$A{p}^2+Bp+C$

Then,

$A=25,B=-15,C=2$

Given that,

Sum of roots

$=\alpha +\beta =\frac{-B}{A}$

$\alpha +\beta =\frac{15}{25}$

$\alpha +\beta =\frac{3}{5}$

Product of roots

$\alpha .\beta =\frac{C}{A}$

$\alpha .\beta =\frac{2}{25}$

Now,

$\frac{1}{2\alpha }and\frac{1}{2\beta }$

Then,

Sum of roots

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

$=2\frac{(\alpha +\beta )}{4\alpha \beta }$

$=\frac{(\alpha +\beta )}{2\alpha \beta }=\frac{\frac{3}{5}}{\frac{2\times 2}{25}}=\frac{3}{5}\times \frac{25}{4}=\frac{15}{4}$

$\frac{1}{2\alpha }+\frac{1}{2\beta }=\frac{15}{4}$

Product of roots

$=\frac{1}{2\alpha }\times \frac{1}{2\beta }=\frac{1}{4\alpha \beta }=\frac{1}{\frac{4\times 2}{25}}$

$=\frac{25}{8}$

So, the equation of polynomial is

$p^2-\left(Sumofroots\right)p+productofroots$

$\Rightarrow p^2-\frac{15}{4}p+\frac{25}{8}$

$\Rightarrow \frac{8p^2-30p+25}{8}$

Hence, this is the answer.