Solve {left(frac{2sqrt{3}}{3}right)}^2={left(frac{3sqrt{13}}{13}right)}^2+x^2

$${ \left( \frac{ 2 \sqrt{ 3 } }{ 3 } \right) }^{ 2 } = { \left( \frac{ 3 \sqrt{ 13 } }{ 13 } \right) }^{ 2 } + { x }^{ 2 }$$

Solve for x

$$x=\frac{5\sqrt{39}}{39}\approx 0.800640769$$

$$x=-\frac{5\sqrt{39}}{39}\approx -0.800640769$$

Steps by Finding Square Root

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\left(\frac{3\sqrt{13}}{13}\right)^{2}+x^{2}$$

To raise \frac{3\sqrt{13}}{13} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}}{13^{2}}+x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{13^{2}}{13^{2}}.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}}{13^{2}}+\frac{x^{2}\times 13^{2}}{13^{2}}$$

Since \frac{\left(3\sqrt{13}\right)^{2}}{13^{2}} and \frac{x^{2}\times 13^{2}}{13^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

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

$$\frac{2^{2}\left(\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4\left(\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{4\times 3}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{12}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{12}{9}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{4}{3}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Expand \left(3\sqrt{13}\right)^{2}.

$$\frac{4}{3}=\frac{3^{2}\left(\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{4}{3}=\frac{9\left(\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

The square of \sqrt{13} is 13.

$$\frac{4}{3}=\frac{9\times 13+x^{2}\times 13^{2}}{13^{2}}$$

Multiply 9 and 13 to get 117.

$$\frac{4}{3}=\frac{117+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 13 to the power of 2 and get 169.

$$\frac{4}{3}=\frac{117+x^{2}\times 169}{13^{2}}$$

Calculate 13 to the power of 2 and get 169.

$$\frac{4}{3}=\frac{117+x^{2}\times 169}{169}$$

Divide each term of 117+x^{2}\times 169 by 169 to get \frac{9}{13}+x^{2}.

$$\frac{4}{3}=\frac{9}{13}+x^{2}$$

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

$$\frac{9}{13}+x^{2}=\frac{4}{3}$$

Subtract \frac{9}{13} from both sides.

$$x^{2}=\frac{4}{3}-\frac{9}{13}$$

Subtract \frac{9}{13} from \frac{4}{3} to get \frac{25}{39}.

$$x^{2}=\frac{25}{39}$$

Take the square root of both sides of the equation.

$$x=\frac{5\sqrt{39}}{39} x=-\frac{5\sqrt{39}}{39}$$

Steps Using the Quadratic Formula

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\left(\frac{3\sqrt{13}}{13}\right)^{2}+x^{2}$$

To raise \frac{3\sqrt{13}}{13} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}}{13^{2}}+x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{13^{2}}{13^{2}}.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}}{13^{2}}+\frac{x^{2}\times 13^{2}}{13^{2}}$$

Since \frac{\left(3\sqrt{13}\right)^{2}}{13^{2}} and \frac{x^{2}\times 13^{2}}{13^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

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

$$\frac{2^{2}\left(\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4\left(\sqrt{3}\right)^{2}}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{4\times 3}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{12}{3^{2}}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{12}{9}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{4}{3}=\frac{\left(3\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Expand \left(3\sqrt{13}\right)^{2}.

$$\frac{4}{3}=\frac{3^{2}\left(\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{4}{3}=\frac{9\left(\sqrt{13}\right)^{2}+x^{2}\times 13^{2}}{13^{2}}$$

The square of \sqrt{13} is 13.

$$\frac{4}{3}=\frac{9\times 13+x^{2}\times 13^{2}}{13^{2}}$$

Multiply 9 and 13 to get 117.

$$\frac{4}{3}=\frac{117+x^{2}\times 13^{2}}{13^{2}}$$

Calculate 13 to the power of 2 and get 169.

$$\frac{4}{3}=\frac{117+x^{2}\times 169}{13^{2}}$$

Calculate 13 to the power of 2 and get 169.

$$\frac{4}{3}=\frac{117+x^{2}\times 169}{169}$$

Divide each term of 117+x^{2}\times 169 by 169 to get \frac{9}{13}+x^{2}.

$$\frac{4}{3}=\frac{9}{13}+x^{2}$$

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

$$\frac{9}{13}+x^{2}=\frac{4}{3}$$

Subtract \frac{4}{3} from both sides.

$$\frac{9}{13}+x^{2}-\frac{4}{3}=0$$

Subtract \frac{4}{3} from \frac{9}{13} to get -\frac{25}{39}.

$$-\frac{25}{39}+x^{2}=0$$

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.

$$x^{2}-\frac{25}{39}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{25}{39} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{25}{39}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{25}{39}\right)}}{2}$$

Multiply -4 times -\frac{25}{39}.

$$x=\frac{0±\sqrt{\frac{100}{39}}}{2}$$

Take the square root of \frac{100}{39}.

$$x=\frac{0±\frac{10\sqrt{39}}{39}}{2}$$

Now solve the equation x=\frac{0±\frac{10\sqrt{39}}{39}}{2} when ± is plus.

$$x=\frac{5\sqrt{39}}{39}$$

Now solve the equation x=\frac{0±\frac{10\sqrt{39}}{39}}{2} when ± is minus.

$$x=-\frac{5\sqrt{39}}{39}$$

The equation is now solved.

$$x=\frac{5\sqrt{39}}{39} x=-\frac{5\sqrt{39}}{39}$$