Solve sqrt{2^2+x^2}=2x

$$\sqrt{ { 2 }^{ 2 } + { x }^{ 2 } } =2x$$

Solve for x

$$x = \frac{2 \sqrt{3}}{3} \approx 1.154700538$$

Solution Steps

Square both sides of the equation.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{4+x^{2}} to the power of 2 and get 4+x^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

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

$$4-3x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{2}=-4$$

Divide both sides by -3.

$$x^{2}=\frac{-4}{-3}$$

Fraction \frac{-4}{-3} can be simplified to \frac{4}{3} by removing the negative sign from both the numerator and the denominator.

$$x^{2}=\frac{4}{3}$$

Take the square root of both sides of the equation.

$$x=\frac{2\sqrt{3}}{3} x=-\frac{2\sqrt{3}}{3}$$

Substitute \frac{2\sqrt{3}}{3} for x in the equation \sqrt{2^{2}+x^{2}}=2x.

$$\sqrt{2^{2}+\left(\frac{2\sqrt{3}}{3}\right)^{2}}=2\times \frac{2\sqrt{3}}{3}$$

Simplify. The value x=\frac{2\sqrt{3}}{3} satisfies the equation.

$$\frac{4}{3}\times 3^{\frac{1}{2}}=\frac{4}{3}\times 3^{\frac{1}{2}}$$

Substitute -\frac{2\sqrt{3}}{3} for x in the equation \sqrt{2^{2}+x^{2}}=2x.

$$\sqrt{2^{2}+\left(-\frac{2\sqrt{3}}{3}\right)^{2}}=2\left(-\frac{2\sqrt{3}}{3}\right)$$

Simplify. The value x=-\frac{2\sqrt{3}}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{4}{3}\times 3^{\frac{1}{2}}=-\frac{4}{3}\times 3^{\frac{1}{2}}$$

Equation \sqrt{x^{2}+4}=2x has a unique solution.

$$x=\frac{2\sqrt{3}}{3}$$