Question

If the equations $x^3-x^2+bx+c=0$ and $x^3+c{x}^2+bx-d=0$ have two common roots then show that $b^2=d$

Answer 1

Let ${}^{\prime }{\alpha }^{\prime }$ be the common root of Both the equation then

${\alpha }^3-{\alpha }^2+b\alpha +c={\alpha }^3+c{\alpha }^2+b\alpha -d$

$\Rightarrow -{\alpha }^2+c=c{\alpha }^2-d$

$\Rightarrow (c+1){\alpha }^2-(c+d)=0$

$\therefore (c+1)x^2-(c+d)=0$

$\alpha +\beta =0$

Hence, sum of common root is zero

$x^3-x^2+bx+c=0$ , $x^3+c{x}^2+bx-d=0$

$\alpha +\beta +\gamma =1$ , $\alpha +\beta +\gamma =-c$

$\Rightarrow \gamma =1$ $\Rightarrow \gamma =-c$

$1-1+b+c=0$ , $-c^3+c^3-bc-d=0$

$\Rightarrow b=-c$ $-bc=d$

$\Rightarrow d=-b(-b)$

$\Rightarrow b^2=d$ (proved)