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 integral root, then the value of $a-b$ is

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

Answer 1

The correct option is A $1$

Let $\alpha$ be the common root of all the given equations, so
${\alpha }^2+a\alpha +12=0\cdots \left(1\right){\alpha }^2+b\alpha +15=0\cdots \left(2\right){\alpha }^2+(a+b)\alpha +36=0\cdots \left(3\right)$
From equation $\left(1\right)$ and $\left(2\right),$ we get
$2{\alpha }^2+(a+b)\alpha +27=0$
Using equation $\left(3\right),$
${\alpha }^2-9=0$
As the common root is a positive integer,
$\Rightarrow \alpha =3$
Using equation $\left(1\right),$
$a=-\left(\frac{{\alpha }^2+12}{\alpha }\right)\Rightarrow a=-\left(\frac{9+12}{3}\right)=-7b=-\left(\frac{{\alpha }^2+15}{\alpha }\right)\Rightarrow b=-\left(\frac{9+15}{3}\right)=-8\therefore a-b=-7+8=1$