Solve the following quadratic equation for x: `4x^2 + 4bx – (a^2 – b^2) = 0

Answer: \(x = \dfrac{a-b}{2}\) or \(x = -\dfrac{a+b}{2}\)

Compare \(4x^2 + 4bx - (a^2 - b^2) = 0\) with \(px^2 + qx + r = 0\): \(p = 4\), \(q = 4b\), \(r = -(a^2 - b^2)\).

Discriminant:

\[q^2 - 4pr = 16b^2 + 16(a^2 - b^2) = 16a^2\]

So \(\sqrt{q^2 - 4pr} = 4a\). By the formula,

\[x = \frac{-4b \pm 4a}{8} = \frac{-b \pm a}{2}\]

Hence \(x = \dfrac{a-b}{2}\) or \(x = \dfrac{-(a+b)}{2}\).

(Factorisation also works: \(4x^2 + 4bx + b^2 - a^2 = (2x+b)^2 - a^2 = (2x+b-a)(2x+b+a) = 0\).)