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

Given: x2 + px + q = 0On comparing this equation with ax2 + bx + c = 0, we get:a = 1, b = p and c = qLet alpha; and beta; be the roots of the quadratic eq ...

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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 = – $\frac{b}{a}$ and αβ = $\frac{c}{a}$ .
Thus,
α + β = – $\frac{p}{1}$ = – p …(2)
αβ = $\frac{q}{1}$ = 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)
$\Rightarrow$ p² = 1 + 4q

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

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