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(a_i)| = 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 \(2\sqrt q=2× \frac{50}{2}=50\)