Solve 1/2(16+x^2)=4/5(2+x)^2

$$\frac { 1 } { 2 } ( 16 + x ^ { 2 } ) = \frac { 4 } { 5 } ( 2 + x ) ^ { 2 }$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=-12$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2} by 16+x^{2}.

$$8+\frac{1}{2}x^{2}=\frac{4}{5}\left(2+x\right)^{2}$$

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

$$8+\frac{1}{2}x^{2}=\frac{4}{5}\left(4+4x+x^{2}\right)$$

Use the distributive property to multiply \frac{4}{5} by 4+4x+x^{2}.

$$8+\frac{1}{2}x^{2}=\frac{16}{5}+\frac{16}{5}x+\frac{4}{5}x^{2}$$

Subtract \frac{16}{5} from both sides.

$$8+\frac{1}{2}x^{2}-\frac{16}{5}=\frac{16}{5}x+\frac{4}{5}x^{2}$$

Subtract \frac{16}{5} from 8 to get \frac{24}{5}.

$$\frac{24}{5}+\frac{1}{2}x^{2}=\frac{16}{5}x+\frac{4}{5}x^{2}$$

Subtract \frac{16}{5}x from both sides.

$$\frac{24}{5}+\frac{1}{2}x^{2}-\frac{16}{5}x=\frac{4}{5}x^{2}$$

Subtract \frac{4}{5}x^{2} from both sides.

$$\frac{24}{5}+\frac{1}{2}x^{2}-\frac{16}{5}x-\frac{4}{5}x^{2}=0$$

Combine \frac{1}{2}x^{2} and -\frac{4}{5}x^{2} to get -\frac{3}{10}x^{2}.

$$\frac{24}{5}-\frac{3}{10}x^{2}-\frac{16}{5}x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-\frac{3}{10}x^{2}-\frac{16}{5}x+\frac{24}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{10} for a, -\frac{16}{5} for b, and \frac{24}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{16}{5}\right)±\sqrt{\left(-\frac{16}{5}\right)^{2}-4\left(-\frac{3}{10}\right)\times \frac{24}{5}}}{2\left(-\frac{3}{10}\right)}$$

Square -\frac{16}{5} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{16}{5}\right)±\sqrt{\frac{256}{25}-4\left(-\frac{3}{10}\right)\times \frac{24}{5}}}{2\left(-\frac{3}{10}\right)}$$

Multiply -4 times -\frac{3}{10}.

$$x=\frac{-\left(-\frac{16}{5}\right)±\sqrt{\frac{256}{25}+\frac{6}{5}\times \frac{24}{5}}}{2\left(-\frac{3}{10}\right)}$$

Multiply \frac{6}{5} times \frac{24}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{16}{5}\right)±\sqrt{\frac{256+144}{25}}}{2\left(-\frac{3}{10}\right)}$$

Add \frac{256}{25} to \frac{144}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{16}{5}\right)±\sqrt{16}}{2\left(-\frac{3}{10}\right)}$$

Take the square root of 16.

$$x=\frac{-\left(-\frac{16}{5}\right)±4}{2\left(-\frac{3}{10}\right)}$$

The opposite of -\frac{16}{5} is \frac{16}{5}.

$$x=\frac{\frac{16}{5}±4}{2\left(-\frac{3}{10}\right)}$$

Multiply 2 times -\frac{3}{10}.

$$x=\frac{\frac{16}{5}±4}{-\frac{3}{5}}$$

Now solve the equation x=\frac{\frac{16}{5}±4}{-\frac{3}{5}} when ± is plus. Add \frac{16}{5} to 4.

$$x=\frac{\frac{36}{5}}{-\frac{3}{5}}$$

Divide \frac{36}{5} by -\frac{3}{5} by multiplying \frac{36}{5} by the reciprocal of -\frac{3}{5}.

$$x=-12$$

Now solve the equation x=\frac{\frac{16}{5}±4}{-\frac{3}{5}} when ± is minus. Subtract 4 from \frac{16}{5}.

$$x=-\frac{\frac{4}{5}}{-\frac{3}{5}}$$

Divide -\frac{4}{5} by -\frac{3}{5} by multiplying -\frac{4}{5} by the reciprocal of -\frac{3}{5}.

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

The equation is now solved.

$$x=-12 x=\frac{4}{3}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2} by 16+x^{2}.

$$8+\frac{1}{2}x^{2}=\frac{4}{5}\left(2+x\right)^{2}$$

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

$$8+\frac{1}{2}x^{2}=\frac{4}{5}\left(4+4x+x^{2}\right)$$

Use the distributive property to multiply \frac{4}{5} by 4+4x+x^{2}.

$$8+\frac{1}{2}x^{2}=\frac{16}{5}+\frac{16}{5}x+\frac{4}{5}x^{2}$$

Subtract \frac{16}{5}x from both sides.

$$8+\frac{1}{2}x^{2}-\frac{16}{5}x=\frac{16}{5}+\frac{4}{5}x^{2}$$

Subtract \frac{4}{5}x^{2} from both sides.

$$8+\frac{1}{2}x^{2}-\frac{16}{5}x-\frac{4}{5}x^{2}=\frac{16}{5}$$

Combine \frac{1}{2}x^{2} and -\frac{4}{5}x^{2} to get -\frac{3}{10}x^{2}.

$$8-\frac{3}{10}x^{2}-\frac{16}{5}x=\frac{16}{5}$$

Subtract 8 from both sides.

$$-\frac{3}{10}x^{2}-\frac{16}{5}x=\frac{16}{5}-8$$

Subtract 8 from \frac{16}{5} to get -\frac{24}{5}.

$$-\frac{3}{10}x^{2}-\frac{16}{5}x=-\frac{24}{5}$$

Divide both sides of the equation by -\frac{3}{10}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{10}x^{2}-\frac{16}{5}x}{-\frac{3}{10}}=-\frac{\frac{24}{5}}{-\frac{3}{10}}$$

Dividing by -\frac{3}{10} undoes the multiplication by -\frac{3}{10}.

$$x^{2}+\left(-\frac{\frac{16}{5}}{-\frac{3}{10}}\right)x=-\frac{\frac{24}{5}}{-\frac{3}{10}}$$

Divide -\frac{16}{5} by -\frac{3}{10} by multiplying -\frac{16}{5} by the reciprocal of -\frac{3}{10}.

$$x^{2}+\frac{32}{3}x=-\frac{\frac{24}{5}}{-\frac{3}{10}}$$

Divide -\frac{24}{5} by -\frac{3}{10} by multiplying -\frac{24}{5} by the reciprocal of -\frac{3}{10}.

$$x^{2}+\frac{32}{3}x=16$$

Divide \frac{32}{3}, the coefficient of the x term, by 2 to get \frac{16}{3}. Then add the square of \frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{32}{3}x+\left(\frac{16}{3}\right)^{2}=16+\left(\frac{16}{3}\right)^{2}$$

Square \frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{32}{3}x+\frac{256}{9}=16+\frac{256}{9}$$

Add 16 to \frac{256}{9}.

$$x^{2}+\frac{32}{3}x+\frac{256}{9}=\frac{400}{9}$$

Factor x^{2}+\frac{32}{3}x+\frac{256}{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}.

$$\left(x+\frac{16}{3}\right)^{2}=\frac{400}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{16}{3}\right)^{2}}=\sqrt{\frac{400}{9}}$$

Simplify.

$$x+\frac{16}{3}=\frac{20}{3} x+\frac{16}{3}=-\frac{20}{3}$$

Subtract \frac{16}{3} from both sides of the equation.

$$x=\frac{4}{3} x=-12$$