Question

Suppose x1,x2 be the roots of ax^2+bx+c=0 and x3,x4 be the roots of the equation px^2+qx+c=0

Answer 1

Answer:

i) Applying properties of roots of Q.E.,

x₁ + x₂ = -b/a and x₁*x₂ = c/a --------- (1)

x₃ + x₄ = -q/p and x₃*x₄ = r/p ----------- (2)

ii) Since, x₁, x₂, 1/x₃, 1/x₄ are in A.P.,

x₂ - x₁ = 1/x₄ - 1/x₃ [Definition of AP - Difference of successive terms remain same]

==> x₂ - x₁ = (x₃ - x₄)/(x₃*x₄)

Squaring both sides, (x₂ - x₁)² = (x₃ - x₄)²/(x₃*x₄)²

Applying algebraic identity, (a - b)² = (a + b)² - 4ab to the above,

(x₂ + x₁)² - 4x₁*x₂ = {(x₃ + x₄)² - 4x₃*x₄}/(x₃*x₄)²

Substituting the values from equations (1) & (2) to the above,

(b²/a²) - 4c/a = [(q²/p²) - 4r/p]/(r/p)²

This simplifies as: (b² - 4ac)/a² = (q² - 4pr)/r²

So, (b² - 4ac)/(q² - 4pr) = (a/r)²

Hope it helps you....