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.