Question

For p, q, ∈ R, consider the real valued function $f(x) = (x – p)^2 – q$, x ∈ 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 |f(ai)| = 500 for all i = 1, 2, 3, 4, then the absolute difference between the roots of f(x) = 0 is

Answer 1

$f(x)=0 ⇒ (x−p) ^2−q=0$

Roots are $p+\sqrt q$, $p− \sqrt q$ absolute difference between roots $2\sqrt q$.
Now, ∣f(a_i)$$∣=500

Let $a_1,a_2,a_3,a_4 \space are \space a_1a+d,a+2d,a+3d$
$∣f(a_4)∣=500$

$∣(a_1−p)^2−q ∣=500$

⇒ $(a_1−p)^2−q=500$
⇒ $\frac{9}{4}d^2−q=500 ....(1)$

and $∣f(a_1 )∣^2=∣f(a_2)∣ ^2$

$((a_1−p)^2−q)^2=((a_2−p)^2−q)^2$

⇒ $((a_1​−p)^2−(a_2−p)^2)((a_1−p)^2−q+(a_2−p)^2−q)=0$
⇒ $\frac{9}{4}d^2−q+ \frac{d^2}{4}−q=0$
$2q= \frac{10d^2}{4}$ ⇒ $q= \frac{5d^2}{4}$

⇒ $d^2= \frac{4q}{5}$
​From equation (1) $\frac{9}{4}⋅ \frac{4⋅q}{5}−q=500$

$\frac{4q}{5}=500$

and $2\sqrt q=2× \frac{50}{2}=50$