Question

For non-zero distinct real numbers $a_1$ and $a_2$, let $f\left(x\right)=a_1{x}^2+b_{1}x+c_1,$$g\left(x\right)=a_2{x}^2+b_{2}x+c_2$ and $p\left(x\right)=f\left(x\right)-g\left(x\right)$. If $p\left(x\right)=0$ only at $x=-1$ and $p(-2)=2$, then the value of $p\left(2\right)$ is

• A. $9$
• B. $6$
• C. $18$
• D. $3$

Answer 1

The correct option is C $18$

$p\left(x\right)=f\left(x\right)-g\left(x\right)=(a_1-a_2)x^2+(b_1-b_2)x+(c_1-c_2)$
Let $p\left(x\right)=a{x}^2+bx+c$ where $a=a_1-a_2\ne 0$
It is given that $p\left(x\right)=0$ only for $x=-1$.
Therefore, $p\left(x\right)=0$ has equal roots, both roots are $x=-1$
So,
$p\left(x\right)=a(x+1{)}^2\cdots (1)$
Now, we know that
$p(-2)=2$
Putting $x=-2$ in equation $\left(1\right)$,
$\Rightarrow a=2$
$\therefore p\left(x\right)=2(x+1{)}^2$

Hence, the value of
$p\left(2\right)=18.$