Solve (2x)^2=x^2+(5sqrt{3})^2

$$( 2 x ) ^ { 2 } = x ^ { 2 } + ( 5 \sqrt { 3 } ) ^ { 2 }$$

Solve for x

$$x=-5$$

$$x=5$$

Steps Using Difference of Squares

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

$$2^{2}x^{2}=x^{2}+\left(5\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=x^{2}+\left(5\sqrt{3}\right)^{2}$$

Expand \left(5\sqrt{3}\right)^{2}.

$$4x^{2}=x^{2}+5^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$4x^{2}=x^{2}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$4x^{2}=x^{2}+25\times 3$$

Multiply 25 and 3 to get 75.

$$4x^{2}=x^{2}+75$$

Subtract x^{2} from both sides.

$$4x^{2}-x^{2}=75$$

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

$$3x^{2}=75$$

Subtract 75 from both sides.

$$3x^{2}-75=0$$

Divide both sides by 3.

$$x^{2}-25=0$$

Consider x^{2}-25. Rewrite x^{2}-25 as x^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-5\right)\left(x+5\right)=0$$

To find equation solutions, solve x-5=0 and x+5=0.

$$x=5 x=-5$$

Steps by Finding Square Root

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

$$2^{2}x^{2}=x^{2}+\left(5\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=x^{2}+\left(5\sqrt{3}\right)^{2}$$

Expand \left(5\sqrt{3}\right)^{2}.

$$4x^{2}=x^{2}+5^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$4x^{2}=x^{2}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$4x^{2}=x^{2}+25\times 3$$

Multiply 25 and 3 to get 75.

$$4x^{2}=x^{2}+75$$

Subtract x^{2} from both sides.

$$4x^{2}-x^{2}=75$$

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

$$3x^{2}=75$$

Divide both sides by 3.

$$x^{2}=\frac{75}{3}$$

Divide 75 by 3 to get 25.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

Steps Using the Quadratic Formula

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

$$2^{2}x^{2}=x^{2}+\left(5\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}=x^{2}+\left(5\sqrt{3}\right)^{2}$$

Expand \left(5\sqrt{3}\right)^{2}.

$$4x^{2}=x^{2}+5^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$4x^{2}=x^{2}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$4x^{2}=x^{2}+25\times 3$$

Multiply 25 and 3 to get 75.

$$4x^{2}=x^{2}+75$$

Subtract x^{2} from both sides.

$$4x^{2}-x^{2}=75$$

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

$$3x^{2}=75$$

Subtract 75 from both sides.

$$3x^{2}-75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-75\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-75\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-75\right)}}{2\times 3}$$

Multiply -12 times -75.

$$x=\frac{0±\sqrt{900}}{2\times 3}$$

Take the square root of 900.

$$x=\frac{0±30}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±30}{6}$$

Now solve the equation x=\frac{0±30}{6} when ± is plus. Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{0±30}{6} when ± is minus. Divide -30 by 6.

$$x=-5$$

The equation is now solved.

$$x=5 x=-5$$