Question

Find the nature of roots of the equation \(4x^2 - 4a^2x + a^4 - b^4 = 0\), where \(b \ne 0\).

Hint:

Use the discriminant \(D = b^2 - 4ac\) to determine the nature of roots: If \(D>0\) → real and distinct.

Answer 1

Given: \(4x^2 - 4a^2x + a^4 - b^4 = 0\) Compare with \(ax^2 + bx + c = 0\), we have: \[ a = 4,\ b = -4a^2,\ c = a^4 - b^4 \] Discriminant: \[ D = b^2 - 4ac = (-4a^2)^2 - 4(4)(a^4 - b^4) = 16a^4 - 16(a^4 - b^4) \] \[ = 16a^4 - 16a^4 + 16b^4 = 16b^4 \] Since \(b \ne 0 \Rightarrow b^4>0 \Rightarrow D>0\) Therefore, the quadratic has **real and distinct roots**.