If the roots of the equation x2 + px + q = 0 differ by 1, prove that p2 = 1 + 4q.

4. If the roots of the equation x² + px + q = 0 differ by 1, prove that p² = 1 + 4q.

Sol.  x² + px + q = 0

Comparing with ax² + bx + c = 0 we have a = 1, b = p, c = q

Let α and β  be the roots of given quadratic equation.

α  – β  = 1 ......(i) [Given]

 α + β  = ^-b/_a = -p/1 = -p

Also, α . β = c/a = q/1 = q

We know that,

(α  – β )² = (α  + β )² – 4αβ

∴  (1)² = (– p)² – 4 (q)

∴  1 = p² – 4q

∴  1 + 4q = p²

∴  p² = 1 + 4q

Hence proved.