Question

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

Answer 1

$f\left(x\right)=x^2-2x+3\dots \left(1\right)$

$\cdot$ Let the roots of $f\left(x\right)=\alpha ,\beta$

Now, sum of roots $\alpha +\beta =-\frac{b}{a}=2\dots \left(2\right)$

Product of roots $\alpha \beta =\frac{c}{a}=3\dots \left(3\right)$

$\cdot$ General form of Quadratic equation in sum of roots & product of roots is

$p\left(x\right)=x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)\dots \left(4\right)$

$\cdot$ Polynomial with roots $\alpha +2,\beta +2$

$\to \text{Sum of roots}=\alpha +2+\beta +2=\alpha +\beta +4=2+4=6$

$\to \text{Product of roots}=(\alpha +2)(\beta +2)=\alpha \beta +2(\alpha +\beta )+4=\left(3\right)+2\left(2\right)+4=11$

Polynomial $\Rightarrow k(x^2-6x+11)=0$(k is constant).

$\cdot$ Polynomial with roots$\frac{\alpha -1}{\alpha +1},\frac{\beta -1}{\beta +1}$

$\to \text{Sum of roots}=\frac{\alpha \beta +\alpha -\beta -1+\alpha \beta -\alpha +\beta -1}{\beta \alpha +\alpha +\beta +1}=\frac{2\left(3\right)-2}{3+1+2}=\frac{4}{6}=\frac{2}{3}$

$\to \text{Product of roots}=\left(\frac{\alpha -1}{\alpha +1}\right)\left(\frac{\beta -1}{\beta +1}\right)=\frac{\alpha \beta -(\alpha +\beta )+1}{\alpha \beta +(\alpha +\beta )+1}=\frac{3-2+1}{3+2+1}=\frac{2}{6}=\frac{1}{3}$

Polynomial : $k(x^2-\frac{2}{3}x+\frac{1}{3})=0$ (K is constant).