Solve (4/3x-5)^2+x^2=25

$$( \frac { 4 } { 3 } x - 5 ) ^ { 2 } + x ^ { 2 } = 25$$

Solve for x

$$x = \frac{24}{5} = 4\frac{4}{5} = 4.8$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}x-5\right)^{2}.

$$\frac{16}{9}x^{2}-\frac{40}{3}x+25+x^{2}=25$$

Combine \frac{16}{9}x^{2} and x^{2} to get \frac{25}{9}x^{2}.

$$\frac{25}{9}x^{2}-\frac{40}{3}x+25=25$$

Subtract 25 from both sides.

$$\frac{25}{9}x^{2}-\frac{40}{3}x+25-25=0$$

Subtract 25 from 25 to get 0.

$$\frac{25}{9}x^{2}-\frac{40}{3}x=0$$

Factor out x.

$$x\left(\frac{25}{9}x-\frac{40}{3}\right)=0$$

To find equation solutions, solve x=0 and \frac{25x}{9}-\frac{40}{3}=0.

$$x=0 x=\frac{24}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}x-5\right)^{2}.

$$\frac{16}{9}x^{2}-\frac{40}{3}x+25+x^{2}=25$$

Combine \frac{16}{9}x^{2} and x^{2} to get \frac{25}{9}x^{2}.

$$\frac{25}{9}x^{2}-\frac{40}{3}x+25=25$$

Subtract 25 from both sides.

$$\frac{25}{9}x^{2}-\frac{40}{3}x+25-25=0$$

Subtract 25 from 25 to get 0.

$$\frac{25}{9}x^{2}-\frac{40}{3}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{9} for a, -\frac{40}{3} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{40}{3}\right)±\sqrt{\left(-\frac{40}{3}\right)^{2}}}{2\times \frac{25}{9}}$$

Take the square root of \left(-\frac{40}{3}\right)^{2}.

$$x=\frac{-\left(-\frac{40}{3}\right)±\frac{40}{3}}{2\times \frac{25}{9}}$$

The opposite of -\frac{40}{3} is \frac{40}{3}.

$$x=\frac{\frac{40}{3}±\frac{40}{3}}{2\times \frac{25}{9}}$$

Multiply 2 times \frac{25}{9}.

$$x=\frac{\frac{40}{3}±\frac{40}{3}}{\frac{50}{9}}$$

Now solve the equation x=\frac{\frac{40}{3}±\frac{40}{3}}{\frac{50}{9}} when ± is plus. Add \frac{40}{3} to \frac{40}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{80}{3}}{\frac{50}{9}}$$

Divide \frac{80}{3} by \frac{50}{9} by multiplying \frac{80}{3} by the reciprocal of \frac{50}{9}.

$$x=\frac{24}{5}$$

Now solve the equation x=\frac{\frac{40}{3}±\frac{40}{3}}{\frac{50}{9}} when ± is minus. Subtract \frac{40}{3} from \frac{40}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{50}{9}}$$

Divide 0 by \frac{50}{9} by multiplying 0 by the reciprocal of \frac{50}{9}.

$$x=0$$

The equation is now solved.

$$x=\frac{24}{5} x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}x-5\right)^{2}.

$$\frac{16}{9}x^{2}-\frac{40}{3}x+25+x^{2}=25$$

Combine \frac{16}{9}x^{2} and x^{2} to get \frac{25}{9}x^{2}.

$$\frac{25}{9}x^{2}-\frac{40}{3}x+25=25$$

Subtract 25 from both sides.

$$\frac{25}{9}x^{2}-\frac{40}{3}x=25-25$$

Subtract 25 from 25 to get 0.

$$\frac{25}{9}x^{2}-\frac{40}{3}x=0$$

Divide both sides of the equation by \frac{25}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{25}{9}x^{2}-\frac{40}{3}x}{\frac{25}{9}}=\frac{0}{\frac{25}{9}}$$

Dividing by \frac{25}{9} undoes the multiplication by \frac{25}{9}.

$$x^{2}+\left(-\frac{\frac{40}{3}}{\frac{25}{9}}\right)x=\frac{0}{\frac{25}{9}}$$

Divide -\frac{40}{3} by \frac{25}{9} by multiplying -\frac{40}{3} by the reciprocal of \frac{25}{9}.

$$x^{2}-\frac{24}{5}x=\frac{0}{\frac{25}{9}}$$

Divide 0 by \frac{25}{9} by multiplying 0 by the reciprocal of \frac{25}{9}.

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

Divide -\frac{24}{5}, the coefficient of the x term, by 2 to get -\frac{12}{5}. Then add the square of -\frac{12}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{24}{5}x+\left(-\frac{12}{5}\right)^{2}=\left(-\frac{12}{5}\right)^{2}$$

Square -\frac{12}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=\frac{144}{25}$$

Factor x^{2}-\frac{24}{5}x+\frac{144}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{12}{5}\right)^{2}=\frac{144}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{12}{5}\right)^{2}}=\sqrt{\frac{144}{25}}$$

Simplify.

$$x-\frac{12}{5}=\frac{12}{5} x-\frac{12}{5}=-\frac{12}{5}$$

Add \frac{12}{5} to both sides of the equation.

$$x=\frac{24}{5} x=0$$