Question

If the equations $x^2+abx+c=0$ and $x^2+acx+b=0$ have a common root, then establish that their other roots are the roots of the equation $x^2-a(b+c)x+a^{2}bc=0.$

Answer 1

Let the roots of the equation be $\alpha ,\beta$ and $\alpha ,\gamma$ as one root is common.
$\alpha +\beta =-ab,\alpha \beta =c$ ...(1)
$\alpha +\gamma =-ac,\alpha \gamma =b$ ...(2)
we have to find the equation whose roots are $\beta$ and $\gamma$ for which we must kmow the value of $\beta +\gamma ,\beta \gamma .$
$\because x^2+abx+c=0$ and $x^2+acx+b=0$ haw a common root.
$\therefore \frac{x^2}{a(b^2-c^2)}=\frac{x}{c-b}=\frac{1}{a(c-b)}$
$\Rightarrow \frac{x^2}{-a(b+c)}=\frac{x}{1}=\frac{1}{a}$
$\Rightarrow x^2=-(b+c)$ and $x=a$
$\Rightarrow a^2=-(b+c)\Rightarrow a^2+b+c=0$ ...(3)
Also the common root $x=a^2$
Putting $\Rightarrow \alpha$ in (1) and (2), we get
$\beta =ab$ and $\gamma =ac$
$\therefore S=\beta +\gamma =ab+ac=a(b+c)$
$P=\beta \gamma =a^{2}bc$