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

$f(x)=0 \Rightarrow(x-p)^2-q=0$
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 $a_1 a+d, a+2 d, a+3 d$
$ \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 $
$ 2 q =\frac{10 d^2}{4} \Rightarrow q=\frac{5 d^2}{4} $
$\Rightarrow d^2=\frac{4 q}{5}$
From equation (1) $\frac{9}{4} \cdot \frac{4 \cdot q}{5}-q=500$
$ \frac{4 q}{5}=500 $
$\frac{4 q}{5}=500$
and $2 \sqrt{ q }=2 \times \frac{50}{2}=50$