Question 204232
SOlve for x:
You can expect to get four solutions to this "quartic" equation.  First, rewrite the equation so as to appear like a "quadratic" equation in {{{x^2}}}:
{{{8(x^2)^2-6(x^2)-5 = 0}}} Factor this "quadratic"equation.
{{{(4x^2-5)(2x^2+1) = 0}}} Apply the zero product rule:
{{{4x^2-5 = 0}}} or {{{2x^2+1 = 0}}} so that...
{{{4x^2 = 5}}} or {{{2x^2 = -1}}} and...
{{{highlight(x^2 = 5/4)}}} or {{{highlight_green(x^2 = -1/2)}}} Now take the square root of both sides of these two intermdediate solutions.
{{{highlight(x = sqrt(5)/2)}}} or {{{highlight(x = -sqrt(5)/2)}}} or
{{{highlight_green(x = sqrt(-1/2))}}} or {{{highlight_green(x = -sqrt(-1/2))}}}