Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=x^2-3x-2$, find a quadratic polynomial whose zeros are $\frac{1}{2\alpha +\beta }$ and $\frac{1}{2\beta +\alpha }$.

Answer 1

$f\left(x\right)=x^2-3x-2=0$

Sum of roots $=\alpha +\beta =3$

Product of roots $=\alpha \beta =-2$

New Roots are $\frac{1}{2\alpha +\beta }$ and $\frac{1}{2\beta +\alpha }$

Sum$=\frac{1}{2\alpha +\beta }$ and $\frac{1}{2\beta +\alpha }$

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

$\Rightarrow \frac{3(\alpha +\beta )}{2(\alpha +\beta {)}^2+\alpha \beta }$ Subtituting the values of $\alpha$ and $\beta$ from the given quadractic polynomial

$\Rightarrow \frac{9}{16}$

Product $=\frac{1}{2\alpha +\beta }\times \frac{1}{2\beta +\alpha }=\frac{1}{4\alpha \beta +2({\alpha }^2+{\beta }^2{)}^2+\alpha \beta }=\frac{1}{2(\alpha +\beta {)}^2+\alpha \beta }=\frac{1}{16}$

$\therefore f\left(x\right)=(x^2-\frac{9}{16}x+\frac{1}{16})$