Question

For $p,q\in R$, consider the real valued function $f(x)=(x-p{)}^2-q,x\in R$ and $q>0$. Let $a_1,a_2,a_3$ and $a_4$ be in an arithmetic progression with mean $p$ and positive common difference. If $\left|f\left(a_i\right)\right|=500$ for all $i=1,2,3,4$, then the absolute difference between the roots of $f(x)=0$ is

Answer 1

Answer: 50

$f(x)=0\Rightarrow (x-p{)}^2-q=0$
Roots are $p+\sqrt{q},p-\sqrt{q}$ absolute difference between roots $2\sqrt{q}$.
Now, $\left|f\left(a_i\right)\right|=500$
Let $a_1,a_2,a_3,a_4$ are $a_{1}a+d,a+2d,a+3d$
$\left|f\left(a_4\right)\right|=500$
$\left|{\left(a_1-p\right)}^2-q\right|=500$
$\Rightarrow {\left(a_1-p\right)}^2-q=500$
$\Rightarrow \frac{9}{4}{d}^2-q=500$ ....(1)
$\text{ and }{\left|f\left(a_1\right)\right|}^2={\left|f\left(a_2\right)\right|}^2$
${\left({\left(a_1-p\right)}^2-q\right)}^2={\left({\left(a_2-p\right)}^2-q\right)}^2$
$\Rightarrow \left({\left(a_1-p\right)}^2-{\left(a_2-p\right)}^2\right)\left({\left(a_1-p\right)}^2-q+{\left(a_2-p\right)}^2-q\right)=0$
$\Rightarrow \frac{9}{4}{d}^2-q+\frac{d^2}{4}-q=0$
$2q=\frac{10d^2}{4}\Rightarrow q=\frac{5d^2}{4}$
$\Rightarrow d^2=\frac{4q}{5}$
From equation (1) $\frac{9}{4}\cdot \frac{4\cdot q}{5}-q=500$
$\frac{4q}{5}=500$
$\frac{4q}{5}=500$
and $2\sqrt{q}=2\times \frac{50}{2}=50$