Question

If the quadratic equation $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have a common root prove that the equation containing their other roots is $x^2+ax+bc=0$

Answer 1

Given the quadratic equation is

$x^2+bx+ac=0$___(1)

$x^2+cx+ab=0$___(2)

Let roots of (1) be $\alpha$ and $\beta$ and roots of (2)

be $\alpha .\gamma .$

common root $=\alpha$

$\therefore {\alpha }^2+b\alpha +ac={\alpha }^2+c\alpha +ab$

$(b-c)\alpha =ab-ac=a(b-c)$

$\therefore \alpha =a$

Now $\alpha +\beta =-b$ $\alpha \beta =ac$

$\alpha +\gamma =-c$ $\alpha \gamma =ab$

$\frac{\alpha \beta }{a\gamma }=\frac{ac}{ab}\Rightarrow \frac{b}{\gamma }=\frac{c}{b}$

$\therefore \beta =c,\gamma =b$ ____(3)

$Since,\alpha$ is a common root $\therefore f\left(\alpha \right)=0$

$\Rightarrow {\alpha }^2+\alpha b+ac=0$

$\alpha (\alpha +b)+ac=0$ $since\alpha =a$

$a^2+ab+ac=0$

$\Rightarrow a(a+b+c)=0\Rightarrow a+b+c=0$ of $b+c=-a$ ___(4)

Equation having roots $\beta$ and $\gamma$ is

$x^2-(\beta +\gamma )x+\beta \gamma =0$

$x^2-(c+b)x+bc=0$ from (3)

$x^2-(-a)x+bc=0$ from (4)

$x^2+ax+bc=0$ is the equation