Question

Find the roots of the following equation, if they exist, by applying the quadratic formula:

4x² – 4a²x + (a⁴ – b⁴) = 0

Answer 1

The given equation is 4x² – 4a²x + (a⁴ – b⁴) = 0

Comparing it with Ax² + Bx + C = 0, we get

A = 4, B = –4a² and C = a⁴ – b⁴ 

∴ Discriminant, B² – 4AC = (–4a²)² – 4 × 4 × (a² – b⁴)

= 16a⁴ – 16a⁴ + 16b⁴

= 16b⁴ > 0 

So, the given equation has real roots.

Now, `sqrt(D) = sqrt(16b^4) = 4b^2` 

∴ `α = (-B + sqrt(D))/(2A)`

= `(-(-4a^2) + 4b^2)/(2 xx 4)`

= `(4(a^2 + b^2))/8`

= `(a^2 + b^2)/2`

`β = (-B - sqrt(D))/(2A)`

= `(-(-4a^2) - 4b^2)/(2 xx 4)`

= `(4(a^2 - b^2))/8`

= `(a^2 - b^2)/2`

Hence, `1/2(a^2 + b^2)` and `1/2(a^2 - b^2)` are the roots of the given equation.