Question

If $x^2-11x+a$ and $x^2-14x+2a$ have common factor, then value(s) of $a$ is/are

• A. $a=0$
• B. $a=16$
• C. $a=24$
• D. $a=4$

Answer 1

Answer: (A), (C)

$a=24$

Let $x-\alpha$ be a common factor of $x^2-11x+a$ and $x^2-14x+2a.$
Then $x=\alpha$ is root of
$x^2-11x+a=0$ and $x^2-14x+2a=0.$

Now, putting $x=\alpha$
${\alpha }^2-11\alpha +a=0{\alpha }^2-14\alpha +2a=0$
From above equation, we get
$3\alpha -a=0\Rightarrow \alpha =\frac{a}{3}$
Putting $\alpha$ in any one the equation, we get
$\frac{a^2}{9}-\frac{11a}{3}+a=0\Rightarrow a(a-33+9)=0\Rightarrow a(a-24)=0\therefore a=0,24$