Question

If the roots of the equation x² – px + q = 0 differ by unity then

• A. p² = 1 – 4q
• B. p² = 1 + 4q
• C. q² = 1 – 4p
• D. q² = 1 + 4p

Answer 1

Answer: (B)

p² = 1 + 4q

Answer 2

Suppose the equation x² – px + q = 0 has the roots α + 1

and α then α + 1+ α = p ⇒ 2α = p – 1 . . . . (1)

and (α+1) α = q ⇒ α² + α = q. . . . .. (2)

Putting the value of α from (1) in (2), we get

$\frac{(p - 1)^{2}}{4} + \frac{p - 1}{2} = q$ ⇒ (p – 1)² + 2(p – 1) = 4q

⇒ p² – 1 = 4q ⇒ p² = 4q + 1.

Alternative: Let α and β be the roots. |α – β| = 1

⇒ (α + β)² – 4αβ = 1

⇒ p² – 4q = 1, or p² = 1+ 4q