Solve 4(q^2+q^4)=24q

$$4 ( q ^ { 2 } + q ^ { 4 } ) = 24 q$$

Solve for q

$$q = \frac{9 ^ {\frac{2}{3}} {(\sqrt[3]{2 \sqrt{183} + 27} + \sqrt[3]{27 - 2 \sqrt{183}})}}{9} \approx 1.634365293$$

$$q=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by q^{2}+q^{4}.

$$4q^{2}+4q^{4}=24q$$

Subtract 24q from both sides.

$$4q^{2}+4q^{4}-24q=0$$

Substitute t for q^{2}.

$$4t^{2}+4t-24=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, 4 for b, and -24 for c in the quadratic formula.

$$t=\frac{-4±\sqrt{4^{2}-4\times 4\left(-24\right)}}{2\times 4}$$

Do the calculations.

$$t=\frac{-4±20}{8}$$

Solve the equation t=\frac{-4±20}{8} when ± is plus and when ± is minus.

$$t=2 t=-3$$

Since q=t^{2}, the solutions are obtained by evaluating q=±\sqrt{t} for positive t.

$$q=\sqrt{2} q=-\sqrt{2}$$