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Question

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

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Solution

Given: x2 + px + q = 0
On comparing this equation with ax2 + bx + c = 0, we get:
a = 1, b = p and c = q
Let α and β be the roots of the quadratic equation x2 + px + q = 0. Then,
α – β = 1 …(1)
We know that α + β = –ba and αβ = ca.
Thus,
α + β = –p1 = – p …(2)
αβ = q1 = q …(3)
Now, on using the identity (a + b)2 = (a – b)2 + 4ab, we get:
(α + β)2 = (α – β)2 + 4 αβ
From equations (1), (2) and (3), we get:
(–p)2 = (1)2 + 4(q)
p2 = 1 + 4q

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