Question

If equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$ have only one non-zero common root, then find the equation for which other roots must satisfy.

• A. $x^2+bx+bc=0$
• B. None of the above
• C. $x^2+ax+bc=0$
• D. $x^2+cx+ac=0$

Answer 1

The correct option is C $x^2+ax+bc=0$

Let $\alpha ,\beta$ be the roots of $x^2+bx+ca=0$ and $\alpha ,\gamma$ be the roots of $x^2+cx+ab=0$

Now, $\alpha +\beta =-b$ and $\alpha .\beta =ca$

Similarly, $\alpha +\gamma =-c$ and $\alpha .\gamma =ab$

Since, $\alpha$ be a common root for the equations $x^2+bx+ca=0$ and $x^2+cx+ab=0$

$\therefore {\alpha }^2+b\alpha +ca=0$ ............ (1)

And, ${\alpha }^2+c\alpha +ab=0$ ............... (2)

On substracting from (1) to (2), we get

$({\alpha }^2+b\alpha +ca)-({\alpha }^2+c\alpha +ab)=0$

$\Rightarrow (b-c)\alpha -a(b-c)=0$

$\Rightarrow (b-c)(\alpha -a)=0$

$\Rightarrow b=c$ or $\alpha =a$ (b=c not possible)

Putting $\alpha =a$ in equation (1), we get

$a^2+ba+ca=0\Rightarrow a(a+b+c)=0$

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

$\alpha +\beta +\alpha +\gamma =-(b+c)$

$\Rightarrow 2\alpha +\beta +\gamma =a$

$\Rightarrow \beta +\gamma =a-2a=-a$

Now, $\alpha \beta \times \alpha \gamma =\left(ca\right)\left(ab\right)$

$\Rightarrow {\alpha }^2\beta \gamma =a^{2}bc$

$\Rightarrow \beta \gamma =bc$ [since, $\alpha =a$]

Here, $\beta +\gamma =-a$ and $\beta \gamma =bc$

So, Quadratic equation whose roots are $\beta ,\gamma$

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

$\Rightarrow x^2+ax+bc=0$