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 positive root, then the value of $a+b$ is equal to

• A. $15$
• B. $1$
• C. $-1$
• D. $-15$

Answer 1

The correct option is D $-15$

$x^2+ax+12=0\cdots \left(1\right)x^2+bx+15=0\cdots \left(2\right)x^2+(a+b)x+36=0\cdots \left(3\right)$

Adding $\left(1\right)\text{and}\left(2\right)$, we get
$2x^2+(a+b)x+27=0\cdots \left(4\right)$

Now, subtracting $\left(3\right)$ from $\left(4\right)$, we get
$x^2-9=0$
$\Rightarrow x=3,-3$
$\therefore$ Common positive root is $3$.
From equation $\left(3\right)$,
$9+3(a+b)+36=0\Rightarrow a+b=-15$