Question

If $\alpha ,\beta$ are roots of $x^2-px+q=0$ and $\alpha -2,\beta +2$ are roots of $x^2-px+r=0$, then prove that $16q+(r+4-q{)}^2=4p^2$.

Answer 1

Given that,

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

Sum of roots $=\alpha +\beta =\frac{-b}{a}=p$

Product of roots $=\alpha \beta =\frac{c}{a}=q$

Difference of roots $=\alpha -\beta =\frac{\sqrt{b^2-4ac}}{a}=\sqrt{p^2-4q}$

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

Sum of roots $=\alpha +\beta =\frac{-b}{a}=p$

Product of roots $=(\alpha -2)(\beta +2)=\frac{c}{a}=r$

$⟹\alpha \beta +2(\alpha -\beta )-4=r$

$⟹q+2\left(\sqrt{p^2-4q}\right)-4=r$

$⟹2\left(\sqrt{p^2-4q}\right)=r+4-q$

Squaring on both sides, we get

$4(p^2-4q)=(r+4-q{)}^2$

$⟹4p^2-16q=(r+4-q{)}^2$

$\therefore 4p^2=16q+(r+4-q{)}^2$

Hence proved.