Question

If the equations $x^2+ax+12=0$, $x^2+bx+15=0$ and $x^2+(a+b)x+36=0$ have a common root then the possible values of $a,b$ is (are)

• A. $a=7$
• B. $a=-7$
• C. $b=8$
• D. $b=-8$

Answer 1

Answer: (B), (D)

The correct options are
B $a=-7$
D $b=-8$

We have $x^2+ax+12=0$ .......$\left(1\right)$

$x^2+bx+15=0$ .......$\left(2\right)$

$x^2+(a+b)x+36=0$ .......$\left(3\right)$

Let the common root be $m$

So,$m$ will satisfy all three equations

$m^2+am+12=0$ .......$\left(4\right)$

$m^2+bm+15=0$ .......$\left(5\right)$

$m^2+(a+b)m+36=0$ .......$\left(6\right)$

Adding $\left(5\right)$ and $\left(6\right)$, we get

$2m^2+(a+b)m+27=0$ .......$\left(7\right)$

Subtrating $\left(6\right)$ from $\left(7\right)$ we get

$m^2-9=0$

$\Rightarrow m=3$ or $m=-3$(which is not possible)

Put $m=3$ in eqn$\left(4\right)$

$9+3a+12=0$

$\Rightarrow 3a+21=0$

$\Rightarrow 3a=-21$

$\Rightarrow a=\frac{-21}{3}=-7$

$\therefore a=-7$

Put $m=3$ in eqn$\left(5\right)$ we get

$9+3b+15=0$

$\Rightarrow 3b=-24$

$\Rightarrow b=\frac{-24}{3}=-8$

$\therefore a=-7$,$b=-8$