Solve (x+7/3)^2+(14/3sqrt{2})^2=(2sqrt{2}x)^2

$$( x + \frac { 7 } { 3 } ) ^ { 2 } + ( \frac { 14 } { 3 } \sqrt { 2 } ) ^ { 2 } = ( 2 \sqrt { 2 } x ) ^ { 2 }$$

Solve for x

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$x=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{7}{3}\right)^{2}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\left(\frac{14}{3}\sqrt{2}\right)^{2}=\left(2\sqrt{2}x\right)^{2}$$

Expand \left(\frac{14}{3}\sqrt{2}\right)^{2}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\left(\frac{14}{3}\right)^{2}\left(\sqrt{2}\right)^{2}=\left(2\sqrt{2}x\right)^{2}$$

Calculate \frac{14}{3} to the power of 2 and get \frac{196}{9}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\frac{196}{9}\left(\sqrt{2}\right)^{2}=\left(2\sqrt{2}x\right)^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\frac{196}{9}\times 2=\left(2\sqrt{2}x\right)^{2}$$

Multiply \frac{196}{9} and 2 to get \frac{392}{9}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\frac{392}{9}=\left(2\sqrt{2}x\right)^{2}$$

Add \frac{49}{9} and \frac{392}{9} to get 49.

$$x^{2}+\frac{14}{3}x+49=\left(2\sqrt{2}x\right)^{2}$$

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

$$x^{2}+\frac{14}{3}x+49=2^{2}\left(\sqrt{2}\right)^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}+\frac{14}{3}x+49=4\left(\sqrt{2}\right)^{2}x^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}+\frac{14}{3}x+49=4\times 2x^{2}$$

Multiply 4 and 2 to get 8.

$$x^{2}+\frac{14}{3}x+49=8x^{2}$$

Subtract 8x^{2} from both sides.

$$x^{2}+\frac{14}{3}x+49-8x^{2}=0$$

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}+\frac{14}{3}x+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, \frac{14}{3} for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{14}{3}±\sqrt{\left(\frac{14}{3}\right)^{2}-4\left(-7\right)\times 49}}{2\left(-7\right)}$$

Square \frac{14}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{14}{3}±\sqrt{\frac{196}{9}-4\left(-7\right)\times 49}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-\frac{14}{3}±\sqrt{\frac{196}{9}+28\times 49}}{2\left(-7\right)}$$

Multiply 28 times 49.

$$x=\frac{-\frac{14}{3}±\sqrt{\frac{196}{9}+1372}}{2\left(-7\right)}$$

Add \frac{196}{9} to 1372.

$$x=\frac{-\frac{14}{3}±\sqrt{\frac{12544}{9}}}{2\left(-7\right)}$$

Take the square root of \frac{12544}{9}.

$$x=\frac{-\frac{14}{3}±\frac{112}{3}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-\frac{14}{3}±\frac{112}{3}}{-14}$$

Now solve the equation x=\frac{-\frac{14}{3}±\frac{112}{3}}{-14} when ± is plus. Add -\frac{14}{3} to \frac{112}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{98}{3}}{-14}$$

Divide \frac{98}{3} by -14.

$$x=-\frac{7}{3}$$

Now solve the equation x=\frac{-\frac{14}{3}±\frac{112}{3}}{-14} when ± is minus. Subtract \frac{112}{3} from -\frac{14}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{42}{-14}$$

Divide -42 by -14.

$$x=3$$

The equation is now solved.

$$x=-\frac{7}{3} x=3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{7}{3}\right)^{2}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\left(\frac{14}{3}\sqrt{2}\right)^{2}=\left(2\sqrt{2}x\right)^{2}$$

Expand \left(\frac{14}{3}\sqrt{2}\right)^{2}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\left(\frac{14}{3}\right)^{2}\left(\sqrt{2}\right)^{2}=\left(2\sqrt{2}x\right)^{2}$$

Calculate \frac{14}{3} to the power of 2 and get \frac{196}{9}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\frac{196}{9}\left(\sqrt{2}\right)^{2}=\left(2\sqrt{2}x\right)^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\frac{196}{9}\times 2=\left(2\sqrt{2}x\right)^{2}$$

Multiply \frac{196}{9} and 2 to get \frac{392}{9}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}+\frac{392}{9}=\left(2\sqrt{2}x\right)^{2}$$

Add \frac{49}{9} and \frac{392}{9} to get 49.

$$x^{2}+\frac{14}{3}x+49=\left(2\sqrt{2}x\right)^{2}$$

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

$$x^{2}+\frac{14}{3}x+49=2^{2}\left(\sqrt{2}\right)^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}+\frac{14}{3}x+49=4\left(\sqrt{2}\right)^{2}x^{2}$$

The square of \sqrt{2} is 2.

$$x^{2}+\frac{14}{3}x+49=4\times 2x^{2}$$

Multiply 4 and 2 to get 8.

$$x^{2}+\frac{14}{3}x+49=8x^{2}$$

Subtract 8x^{2} from both sides.

$$x^{2}+\frac{14}{3}x+49-8x^{2}=0$$

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}+\frac{14}{3}x+49=0$$

Subtract 49 from both sides. Anything subtracted from zero gives its negation.

$$-7x^{2}+\frac{14}{3}x=-49$$

Divide both sides by -7.

$$\frac{-7x^{2}+\frac{14}{3}x}{-7}=-\frac{49}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\frac{\frac{14}{3}}{-7}x=-\frac{49}{-7}$$

Divide \frac{14}{3} by -7.

$$x^{2}-\frac{2}{3}x=-\frac{49}{-7}$$

Divide -49 by -7.

$$x^{2}-\frac{2}{3}x=7$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=7+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=7+\frac{1}{9}$$

Add 7 to \frac{1}{9}.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{64}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{8}{3} x-\frac{1}{3}=-\frac{8}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=3 x=-\frac{7}{3}$$