Question

Let for $a\ne a_1\ne 0,f\left(x\right)=a{x}^2+bx+c,g\left(x\right)=a_1{x}^2+b_{1}x+c_1$ and $p\left(x\right)=f\left(x\right)-g\left(x\right)$. If $p\left(x\right)=0$ only for $x=-1$ and $p(-2)=2$, then the value of $p\left(2\right)$ is

• A. $6$
• B. $18$
• C. $3$
• D. $9$

Answer 1

Answer: (B)

$18$

Given, $f\left(x\right)=a{x}^2+bx+c,g\left(x\right)=a{x}^2+bx+c,$

and $p\left(x\right)=f\left(x\right)-g\left(x\right)$

$p\left(x\right)=0$ for only $x=-1,p(-2)=2$ and $a\ne a_1\ne 0$

$\therefore p\left(x\right)=(a-a_1)x^2+(b-b_1)x+(c-c_1)$

The standard quadratic equation is $a{x}^2+bx+c=0$

For roots to be equal $\Rightarrow b^2-4ac=0$

Then Sum of roots = $-\frac{b}{a}$

Here, Only root is $-1\Rightarrow D=0\Rightarrow (b-b_1{)}^2=4(a-a_1\left)\right(c-c_1)\to (1)$

Sum of roots $\Rightarrow (-1)+(-1)=\frac{-(b-b_1)}{(a-a_1)}\Rightarrow (b-b_1)=2(a-a_1)\to \left(2\right)$

From (1) and (2), $c-c_1=(a-a_1)\to \left(3\right)$

$\Rightarrow p\left(x\right)=(a-a_1)x^2+2(a-a_1)x+(a-a_1)$

Now, $p(-2)=2$

$\Rightarrow 2=(a-a_1)(-2{)}^2-2(a-a_1\left)\right(-2)+(a-a_1)$

$\Rightarrow (a-a_1)=2$

Now $p\left(2\right)=2(2{)}^2+2(2\left)\right(2)+2=18$