Question

Let $\alpha$ , $\beta$ be the zeros of the polynomial $x^2-px+r$ and $\frac{\alpha }{2}$, $2\beta$ be the zeros of $x^2-qx+r$. Then the value of $r$ is

• A. $\frac{2}{9}$ $(p-q)(2q-p)$
• B. $\frac{2}{9}$ $(q-p)(2q-p)$
• C. $\frac{2}{9}$ $(q-2p)(2q-p)$
• D. $\frac{2}{9}$ $(2p-q)(2q-p)$

Answer 1

The correct option is A $\frac{2}{9}$ $(p-q)(2q-p)$

According to the problem,

$\alpha ,\beta$ are the roots of the equation $x^2-px+r=0$, then

$\alpha +\beta =p$......(1) and $\alpha .\beta =r$.......(2).

Again $\frac{\alpha }{2},2\beta$ are the roots of the equation $x^2-qx+r=0$, then

$\frac{\alpha }{2}+2\beta =q\Rightarrow \alpha +4\beta =2q$.........(3) and $\alpha .\beta =r$.

Now, subtracting (1) from (3) we get,

$3\beta =(2q-p)$

or, $\beta =\frac{(2q-p)}{3}$

Then $\alpha =\frac{2(p-q)}{3}$

Now, using $\alpha ,\beta$ in (2) we get

$r=\frac{2(p-q)(2q-p)}{9}$