If the sum of the roots of the equation ax2 + bx + c = 0 is equal to t

1.	If the sum of the roots of the equation ax2 + bx + c = 0 is equal to the sum of their squares, then which one of the following is correct ?(a) a2 + b2 = c2 (b) a2 + b2 = a + b (c) 2ac = ab + b2 (d) 2c + b = 0
Answer» (c) ab + b² = 2ac. Let α, β be the roots of the equation ax² + bx + c = 0. Then, α + β = –\(\frac{b}{a}\), αβ = \(\frac{c}{a}\) Given, α + β = α² + β² i.e., α + β = (α² + β)² – 2αβ ⇒ \(\frac{-b}{a}\) = \(\frac{b^2}{a^2}\) - \(\frac{2c}{a}\) ⇒ \(-\frac{ab}{a^2}\) = \(\frac{b^2}{a^2}\) - \(\frac{2ac}{a^2}\) ⇒ ab + b² = 2ac.

If the sum of the roots of the equation ax2 + bx + c = 0 is equal to the sum of their squares, then which one of the following is correct ?(a) a2 + b2 = c2 (b) a2 + b2 = a + b (c) 2ac = ab + b2 (d) 2c + b = 0

Answer»

(c) ab + b² = 2ac.

Let α, β be the roots of the equation ax² + bx + c = 0. Then,

α + β = –\(\frac{b}{a}\), αβ = \(\frac{c}{a}\)

Given, α + β = α² + β²

i.e., α + β = (α² + β)² – 2αβ

⇒ \(\frac{-b}{a}\) = \(\frac{b^2}{a^2}\) - \(\frac{2c}{a}\)

⇒ \(-\frac{ab}{a^2}\) = \(\frac{b^2}{a^2}\) - \(\frac{2ac}{a^2}\) ⇒ ab + b² = 2ac.