Solve (2x)^2=x^2+9^2

$$( 2 x ) ^ { 2 } = x ^ { 2 } + 9 ^ { 2 }$$

Solve for x

$$x=3\sqrt{3}\approx 5.196152423$$

$$x=-3\sqrt{3}\approx -5.196152423$$

Steps by Finding Square Root

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=x^{2}+9^{2}$$

Calculate 9 to the power of 2 and get 81.

$$4x^{2}=x^{2}+81$$

Subtract x^{2} from both sides.

$$4x^{2}-x^{2}=81$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}=81$$

Divide both sides by 3.

$$x^{2}=\frac{81}{3}$$

Divide 81 by 3 to get 27.

$$x^{2}=27$$

Take the square root of both sides of the equation.

$$x=3\sqrt{3} x=-3\sqrt{3}$$

Steps Using the Quadratic Formula

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=x^{2}+9^{2}$$

Calculate 9 to the power of 2 and get 81.

$$4x^{2}=x^{2}+81$$

Subtract x^{2} from both sides.

$$4x^{2}-x^{2}=81$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}=81$$

Subtract 81 from both sides.

$$3x^{2}-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-81\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-81\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-81\right)}}{2\times 3}$$

Multiply -12 times -81.

$$x=\frac{0±\sqrt{972}}{2\times 3}$$

Take the square root of 972.

$$x=\frac{0±18\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±18\sqrt{3}}{6}$$

Now solve the equation x=\frac{0±18\sqrt{3}}{6} when ± is plus.

$$x=3\sqrt{3}$$

Now solve the equation x=\frac{0±18\sqrt{3}}{6} when ± is minus.

$$x=-3\sqrt{3}$$

The equation is now solved.

$$x=3\sqrt{3} x=-3\sqrt{3}$$