Question

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

Answer 1

Given: $p\left(x\right)=2x²-5x+7$ and its zeros are denoted by $\alpha$, $\beta$.
Compare $p\left(x\right)=2x²-5x+7$ with $ax²+bx+c$
$a=2,b=-5,c=7$
we know that ,
sum of zeros $=\alpha +\beta$
$=-\frac{b}{a}$
$=\frac{5}{2}$

product of zeros $=\frac{c}{a}$
$=\frac{7}{2}$
$2\alpha +3\beta$ and $3\alpha +2\beta$ are zeros of required polynomial.

sum of zeros $=2\alpha +3\beta +3\alpha +2\beta$
$=5\alpha +5\beta$
$=5[\alpha +\beta ]$
$=5\times \frac{5}{2}$
$=\frac{25}{2}$

product of zeros $=(2\alpha +3\beta )(3\alpha +2\beta )$
$=2\alpha [3\alpha +2\beta ]+3\beta [3\alpha +2\beta ]$
$=6\alpha ²+4\alpha \beta +9\alpha \beta +6\beta ²$
$=6\alpha ²+13\alpha \beta +6\beta ²$
$=6[\alpha ²+\beta ²]+13\alpha \beta$
$=6\left[\right(\alpha +\beta )²-2\alpha \beta ]+13\alpha \beta$
$=6\left[\right(\frac{5}{2})²-2\times \frac{7}{2}]+13\times \frac{7}{2}$
$=6[\frac{25}{4}-7]+\frac{91}{2}$
$=6\left[\frac{25-28}{4}\right]+\frac{91}{2}$
$=6\left[\frac{-3}{4}\right]+\frac{91}{2}$
$=\frac{-18}{4}+\frac{91}{2}$
$=\frac{-9}{2}+\frac{91}{2}$
$=\frac{82}{2}$
$=41$

A quadratic polynomial is given by :-$k[x²-(sumofzeros)x+(productofzeros\left)\right]$

$=k[x^2-\frac{25}{2}x+41]$

Take $k=2$

$=2[x²-\frac{25}{2}x+41]$

$=2x^2-25x+82$ is the required polynomial.