Solve 3x^4+84=52x^2

$$3 x ^ { 4 } + 84 = 52 x ^ { 2 }$$

Solve for x

$$x = -\frac{\sqrt{78 - 6 \sqrt{106}}}{3} \approx -1.342726048$$

$$x = \frac{\sqrt{78 - 6 \sqrt{106}}}{3} \approx 1.342726048$$

$$x = \frac{\sqrt{6 \sqrt{106} + 78}}{3} \approx 3.940865399$$

$$x = -\frac{\sqrt{6 \sqrt{106} + 78}}{3} \approx -3.940865399$$

Steps Using the Quadratic Formula

Subtract 52x^{2} from both sides.

$$3x^{4}+84-52x^{2}=0$$

Substitute t for x^{2}.

$$3t^{2}-52t+84=0$$

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 3 for a, -52 for b, and 84 for c in the quadratic formula.

$$t=\frac{-\left(-52\right)±\sqrt{\left(-52\right)^{2}-4\times 3\times 84}}{2\times 3}$$

Do the calculations.

$$t=\frac{52±4\sqrt{106}}{6}$$

Solve the equation t=\frac{52±4\sqrt{106}}{6} when ± is plus and when ± is minus.

$$t=\frac{2\sqrt{106}+26}{3} t=\frac{26-2\sqrt{106}}{3}$$

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

$$x=\frac{\sqrt{\frac{8\sqrt{106}+104}{3}}}{2} x=-\frac{\sqrt{\frac{8\sqrt{106}+104}{3}}}{2} x=\frac{\sqrt{\frac{104-8\sqrt{106}}{3}}}{2} x=-\frac{\sqrt{\frac{104-8\sqrt{106}}{3}}}{2}$$