Question

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f\left(x\right)=2x^2-5x+7$,then a polynomial whose zeroes are $2\alpha +3\beta$ and $3\alpha +2\beta$ is

• A. $k(x^2-\frac{25}{2}x-41)$
• B. $k(x^2+\frac{25}{2}x+41)$
• C. $k(x^2-\frac{25}{2}x+41)$
• D. $-k(x^2-\frac{25}{2}x+41)$

Answer 1

The correct option is C $k(x^2-\frac{25}{2}x+41)$

Since $\alpha$ and $\beta$ are the roots of $f\left(x\right)=2x^2-5x+7$

So, $\alpha +\beta =-(-\frac{5}{2})=\frac{5}{2}$ and $\alpha \beta =\frac{7}{2}$

Let $S$ and $P$ denote respectively the sum and product of zeroes of the required polynomial .Then,polynomial is

$p\left(x\right)=k(x^2-Sx+P)$

Now, $S=(2\alpha +3\beta )+(3\alpha +2\beta )$

$=5(\alpha +\beta )$

$=5\times \frac{5}{2}$

$=\frac{25}{2}$

and $P=(2\alpha +3\beta )(3\alpha +2\beta )$

$=6({\alpha }^2+{\beta }^2)+13\alpha \beta$

$=6{\alpha }^2+6{\beta }^2+12\alpha \beta +\alpha \beta$

$=6\times (\frac{5}{2}{)}^2+\frac{7}{2}$

$=\frac{75}{2}+\frac{7}{2}$

$=41$

Hence,the required polynomial is given by

$p\left(x\right)=k(x^2-Sx+P)$

$=>p\left(x\right)=k(x^2-\frac{25}{2}x+41)$