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 common roots is (are),

• A. $4$
• B. $3$
• C. $-3$
• D. $-4$

Answer 1

The correct option is B $3$

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

$9+3b+15=0$

$\Rightarrow 3b=-24$

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

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

Substituting $a=-7$,$b=-8$ in $\left(1\right)$ we get

$x^2-7x+12=x^2-4x-3x+12=x(x-4)-3(x-4)=(x-3)(x-4)=0$

Roots are $3$ and $4$

Substituting $a=-7$,$b=-8$ in $\left(2\right)$ we get

$x^2-8x+15=x^2-5x-3x+15=x(x-5)-3(x-5)=(x-3)(x-5)=0$

Roots are $3$ and $5$

Substituting $a=-7$,$b=-8$ in $\left(3\right)$ we get

$x^2+(-7-8)x+36=x^2-12x-3x+36=x(x-12)-3(x-12)=(x-12)(x-3)=0$

Roots are $12$ and $3$

$\therefore \{3,4\}\cap \{3,5\}\cap \{3,12\}=\left\{3\right\}$

$\therefore$ the possible common root is $\left\{3\right\}$