Solve 0=01x^4-2x^2+64

$$0 = 01 x ^ { 4 } - 2 x ^ { 2 } + 64$$

Solve for x

$$x=4\sqrt{2}\approx 5.656854249$$

$$x=-4\sqrt{2}\approx -5.656854249$$

Steps by Finding Square Root

Multiply 0 and 1 to get 0.

$$0=0x^{4}-2x^{2}+64$$

Anything times zero gives zero.

$$0=0-2x^{2}+64$$

Add 0 and 64 to get 64.

$$0=64-2x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$64-2x^{2}=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}=-64$$

Divide both sides by -2.

$$x^{2}=\frac{-64}{-2}$$

Divide -64 by -2 to get 32.

$$x^{2}=32$$

Take the square root of both sides of the equation.

$$x=4\sqrt{2} x=-4\sqrt{2}$$

Steps Using the Quadratic Formula

Multiply 0 and 1 to get 0.

$$0=0x^{4}-2x^{2}+64$$

Anything times zero gives zero.

$$0=0-2x^{2}+64$$

Add 0 and 64 to get 64.

$$0=64-2x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$64-2x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-2x^{2}+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 0 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 64}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 64}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 64}}{2\left(-2\right)}$$

Multiply 8 times 64.

$$x=\frac{0±\sqrt{512}}{2\left(-2\right)}$$

Take the square root of 512.

$$x=\frac{0±16\sqrt{2}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±16\sqrt{2}}{-4}$$

Now solve the equation x=\frac{0±16\sqrt{2}}{-4} when ± is plus.

$$x=-4\sqrt{2}$$

Now solve the equation x=\frac{0±16\sqrt{2}}{-4} when ± is minus.

$$x=4\sqrt{2}$$

The equation is now solved.

$$x=-4\sqrt{2} x=4\sqrt{2}$$