Question

If the sum of the roots of the equation ax²+bx+c=0 is equal to the sum of squares of their reciprocals, then prove that 2a²c = c²b + b²a

Answer 1

Question :-----

• sum of roots of equation ax² + bx + c = 0 is equal to the sum of square to their reciprocals .

To Prove :----

• 2a²c = c²b + b²a

Formula used :----

• sum of roots of equation ax² + bx + c = 0 is -b/a
• product of roots = c/a
• a² + b² = (a+b)² - 2ab

Solution :-----

let roots of Equation be P and Q ,

so,

p + q = -b/a = 1/p² + 1/q²

p×q = c/a

solving 1/p² + 1/q² first we get,

1/p² + 1/q² =(q² + p²) /p²q²

putting formula in numerator a² + b² = (a+b)² - 2ab we get,

→ 1/p² + 1/q² = (p+q)² - 2pq/(pq)²

putting value of (p+q) and pq now we get,

→ 1/p² + 1/q² = (-b/a)² - 2(c/a) /(c/a)²

→ -b/a × (c/a)² = (-b/a)² - 2(c/a)

→ -bc²/a³ = b²/a² - 2c/a

→ 2c/a = b²/a² + bc²/a³

→ 2c/a = (ab²+bc²) /a³

→ 2c/a × (a³) = b(ab+c²)

→ 2ca² = ab² + bc² or ,,,

→ 2a²c = c²b + b²a

Hence ,,, Proved

(Hope it helps you)