Solve x^2=2^9

$$x ^ { 2 } = 2 ^ { 9 }$$

Solve for x

$$x=16\sqrt{2}\approx 22.627416998$$

$$x=-16\sqrt{2}\approx -22.627416998$$

Steps by Finding Square Root

Calculate 2 to the power of 9 and get 512.

$$x^{2}=512$$

Take the square root of both sides of the equation.

$$x=16\sqrt{2} x=-16\sqrt{2}$$

Steps Using the Quadratic Formula

Calculate 2 to the power of 9 and get 512.

$$x^{2}=512$$

Subtract 512 from both sides.

$$x^{2}-512=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -512 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-512\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-512\right)}}{2}$$

Multiply -4 times -512.

$$x=\frac{0±\sqrt{2048}}{2}$$

Take the square root of 2048.

$$x=\frac{0±32\sqrt{2}}{2}$$

Now solve the equation x=\frac{0±32\sqrt{2}}{2} when ± is plus.

$$x=16\sqrt{2}$$

Now solve the equation x=\frac{0±32\sqrt{2}}{2} when ± is minus.

$$x=-16\sqrt{2}$$

The equation is now solved.

$$x=16\sqrt{2} x=-16\sqrt{2}$$