Solve {left(10/3right)}^2+{left(frac{2sqrt{73}}{3}right)}^2=2{left(frac{sqrt{52}}{3}right)}^2+2x^2

$${ \left( \frac{ 10 }{ 3 } \right) }^{ 2 } + { \left( \frac{ 2 \sqrt{ 73 } }{ 3 } \right) }^{ 2 } =2 { \left( \frac{ \sqrt{ 52 } }{ 3 } \right) }^{ 2 } +2 { x }^{ 2 }$$

Solve for x

$$x=4$$

$$x=-4$$

Steps Using Difference of Squares

Calculate \frac{10}{3} to the power of 2 and get \frac{100}{9}.

$$\frac{100}{9}+\left(\frac{2\sqrt{73}}{3}\right)^{2}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

To raise \frac{2\sqrt{73}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{100}{9}+\frac{\left(2\sqrt{73}\right)^{2}}{3^{2}}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 3^{2}.

$$\frac{100}{9}+\frac{\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

Since \frac{100}{9} and \frac{\left(2\sqrt{73}\right)^{2}}{9} have the same denominator, add them by adding their numerators.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

Factor 52=2^{2}\times 13. Rewrite the square root of the product \sqrt{2^{2}\times 13} as the product of square roots \sqrt{2^{2}}\sqrt{13}. Take the square root of 2^{2}.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{2\sqrt{13}}{3}\right)^{2}+2x^{2}$$

To raise \frac{2\sqrt{13}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \frac{\left(2\sqrt{13}\right)^{2}}{3^{2}}+2x^{2}$$

Express 2\times \frac{\left(2\sqrt{13}\right)^{2}}{3^{2}} as a single fraction.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}}{3^{2}}+2x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x^{2} times \frac{3^{2}}{3^{2}}.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}}{3^{2}}+\frac{2x^{2}\times 3^{2}}{3^{2}}$$

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

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Expand \left(2\sqrt{73}\right)^{2}.

$$\frac{100+2^{2}\left(\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{100+4\left(\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

The square of \sqrt{73} is 73.

$$\frac{100+4\times 73}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 4 and 73 to get 292.

$$\frac{100+292}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Add 100 and 292 to get 392.

$$\frac{392}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

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

$$\frac{392}{9}=\frac{2\times 2^{2}\left(\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{392}{9}=\frac{2\times 4\left(\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

The square of \sqrt{13} is 13.

$$\frac{392}{9}=\frac{2\times 4\times 13+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 4 and 13 to get 52.

$$\frac{392}{9}=\frac{2\times 52+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 2 and 52 to get 104.

$$\frac{392}{9}=\frac{104+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{392}{9}=\frac{104+2x^{2}\times 9}{3^{2}}$$

Multiply 2 and 9 to get 18.

$$\frac{392}{9}=\frac{104+18x^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{392}{9}=\frac{104+18x^{2}}{9}$$

Divide each term of 104+18x^{2} by 9 to get \frac{104}{9}+2x^{2}.

$$\frac{392}{9}=\frac{104}{9}+2x^{2}$$

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

$$\frac{104}{9}+2x^{2}=\frac{392}{9}$$

Subtract \frac{392}{9} from both sides.

$$\frac{104}{9}+2x^{2}-\frac{392}{9}=0$$

Subtract \frac{392}{9} from \frac{104}{9} to get -32.

$$-32+2x^{2}=0$$

Divide both sides by 2.

$$-16+x^{2}=0$$

Consider -16+x^{2}. Rewrite -16+x^{2} as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

To find equation solutions, solve x-4=0 and x+4=0.

$$x=4 x=-4$$

Steps by Finding Square Root

Calculate \frac{10}{3} to the power of 2 and get \frac{100}{9}.

$$\frac{100}{9}+\left(\frac{2\sqrt{73}}{3}\right)^{2}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

To raise \frac{2\sqrt{73}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{100}{9}+\frac{\left(2\sqrt{73}\right)^{2}}{3^{2}}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 3^{2}.

$$\frac{100}{9}+\frac{\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

Since \frac{100}{9} and \frac{\left(2\sqrt{73}\right)^{2}}{9} have the same denominator, add them by adding their numerators.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

Factor 52=2^{2}\times 13. Rewrite the square root of the product \sqrt{2^{2}\times 13} as the product of square roots \sqrt{2^{2}}\sqrt{13}. Take the square root of 2^{2}.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{2\sqrt{13}}{3}\right)^{2}+2x^{2}$$

To raise \frac{2\sqrt{13}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \frac{\left(2\sqrt{13}\right)^{2}}{3^{2}}+2x^{2}$$

Express 2\times \frac{\left(2\sqrt{13}\right)^{2}}{3^{2}} as a single fraction.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}}{3^{2}}+2x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x^{2} times \frac{3^{2}}{3^{2}}.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}}{3^{2}}+\frac{2x^{2}\times 3^{2}}{3^{2}}$$

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

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Expand \left(2\sqrt{73}\right)^{2}.

$$\frac{100+2^{2}\left(\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{100+4\left(\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

The square of \sqrt{73} is 73.

$$\frac{100+4\times 73}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 4 and 73 to get 292.

$$\frac{100+292}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Add 100 and 292 to get 392.

$$\frac{392}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

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

$$\frac{392}{9}=\frac{2\times 2^{2}\left(\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{392}{9}=\frac{2\times 4\left(\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

The square of \sqrt{13} is 13.

$$\frac{392}{9}=\frac{2\times 4\times 13+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 4 and 13 to get 52.

$$\frac{392}{9}=\frac{2\times 52+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 2 and 52 to get 104.

$$\frac{392}{9}=\frac{104+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{392}{9}=\frac{104+2x^{2}\times 9}{3^{2}}$$

Multiply 2 and 9 to get 18.

$$\frac{392}{9}=\frac{104+18x^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{392}{9}=\frac{104+18x^{2}}{9}$$

Divide each term of 104+18x^{2} by 9 to get \frac{104}{9}+2x^{2}.

$$\frac{392}{9}=\frac{104}{9}+2x^{2}$$

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

$$\frac{104}{9}+2x^{2}=\frac{392}{9}$$

Subtract \frac{104}{9} from both sides.

$$2x^{2}=\frac{392}{9}-\frac{104}{9}$$

Subtract \frac{104}{9} from \frac{392}{9} to get 32.

$$2x^{2}=32$$

Divide both sides by 2.

$$x^{2}=\frac{32}{2}$$

Divide 32 by 2 to get 16.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Steps Using the Quadratic Formula

Calculate \frac{10}{3} to the power of 2 and get \frac{100}{9}.

$$\frac{100}{9}+\left(\frac{2\sqrt{73}}{3}\right)^{2}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

To raise \frac{2\sqrt{73}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{100}{9}+\frac{\left(2\sqrt{73}\right)^{2}}{3^{2}}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 3^{2}.

$$\frac{100}{9}+\frac{\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

Since \frac{100}{9} and \frac{\left(2\sqrt{73}\right)^{2}}{9} have the same denominator, add them by adding their numerators.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{\sqrt{52}}{3}\right)^{2}+2x^{2}$$

Factor 52=2^{2}\times 13. Rewrite the square root of the product \sqrt{2^{2}\times 13} as the product of square roots \sqrt{2^{2}}\sqrt{13}. Take the square root of 2^{2}.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \left(\frac{2\sqrt{13}}{3}\right)^{2}+2x^{2}$$

To raise \frac{2\sqrt{13}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=2\times \frac{\left(2\sqrt{13}\right)^{2}}{3^{2}}+2x^{2}$$

Express 2\times \frac{\left(2\sqrt{13}\right)^{2}}{3^{2}} as a single fraction.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}}{3^{2}}+2x^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x^{2} times \frac{3^{2}}{3^{2}}.

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}}{3^{2}}+\frac{2x^{2}\times 3^{2}}{3^{2}}$$

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

$$\frac{100+\left(2\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Expand \left(2\sqrt{73}\right)^{2}.

$$\frac{100+2^{2}\left(\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{100+4\left(\sqrt{73}\right)^{2}}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

The square of \sqrt{73} is 73.

$$\frac{100+4\times 73}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 4 and 73 to get 292.

$$\frac{100+292}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Add 100 and 292 to get 392.

$$\frac{392}{9}=\frac{2\times \left(2\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

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

$$\frac{392}{9}=\frac{2\times 2^{2}\left(\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{392}{9}=\frac{2\times 4\left(\sqrt{13}\right)^{2}+2x^{2}\times 3^{2}}{3^{2}}$$

The square of \sqrt{13} is 13.

$$\frac{392}{9}=\frac{2\times 4\times 13+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 4 and 13 to get 52.

$$\frac{392}{9}=\frac{2\times 52+2x^{2}\times 3^{2}}{3^{2}}$$

Multiply 2 and 52 to get 104.

$$\frac{392}{9}=\frac{104+2x^{2}\times 3^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{392}{9}=\frac{104+2x^{2}\times 9}{3^{2}}$$

Multiply 2 and 9 to get 18.

$$\frac{392}{9}=\frac{104+18x^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{392}{9}=\frac{104+18x^{2}}{9}$$

Divide each term of 104+18x^{2} by 9 to get \frac{104}{9}+2x^{2}.

$$\frac{392}{9}=\frac{104}{9}+2x^{2}$$

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

$$\frac{104}{9}+2x^{2}=\frac{392}{9}$$

Subtract \frac{392}{9} from both sides.

$$\frac{104}{9}+2x^{2}-\frac{392}{9}=0$$

Subtract \frac{392}{9} from \frac{104}{9} to get -32.

$$-32+2x^{2}=0$$

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.

$$2x^{2}-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-32\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-32\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-32\right)}}{2\times 2}$$

Multiply -8 times -32.

$$x=\frac{0±\sqrt{256}}{2\times 2}$$

Take the square root of 256.

$$x=\frac{0±16}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±16}{4}$$

Now solve the equation x=\frac{0±16}{4} when ± is plus. Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{0±16}{4} when ± is minus. Divide -16 by 4.

$$x=-4$$

The equation is now solved.

$$x=4 x=-4$$