Solve the following quadratic equation by …

We have,

9x² - 6b²x - (a⁴ - b⁴) = 0
Here, Constant term = a⁴ - b⁴ = (a² - b²)(a² + b²)
Also, the coefficient of middle term is - 6b² = [3(a² + b²) - 3(a² - b²)]
Now using the above two values in given equation, 9x² - 6b²x - (a⁴ - b⁴) = 0
{tex}{/tex} we have, 9x²  - [3(a² + b²) - 3(a² - b²)]x - (a² - b²)(a² + b²) = 0
{tex}\Rightarrow{/tex} 9x² - 3(a² + b²)x + 3(a² - b²)x - (a² - b²)(a² + b²) = 0
{tex}\Rightarrow{/tex} 3x[3x - (a² + b²)] + (a² - b²)[3x - (a² + b²)] = 0
{tex}\Rightarrow{/tex} [3x + (a² - b²)][3x - (a² + b²)] = 0
{tex}\Rightarrow{/tex}either [3x + (a²  - b²)] = 0 or, [3x - (a² + b²)] = 0
{tex}\Rightarrow{/tex} 3x = -(a² - b²) or 3x = a² + b²
{tex}\Rightarrow x = -(\frac{{{a^2} - {b^2}}}{3}){/tex} or {tex}x = \frac{{{a^2} + {b^2}}}{3}{/tex}
{tex}\Rightarrow x = \frac{{{b^2} - {a^2}}}{3}{/tex} or {tex}x = \frac{{{a^2} + {b^2}}}{3}{/tex}

Hence, the roots of given quadratic equation are {tex}\frac{{b^2 - a^2}}{3}{/tex}and {tex}\frac{{a^2 + b^2}}{3}{/tex}