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

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

Solve for x

$$x = \frac{64}{35} = 1\frac{29}{35} \approx 1.828571429$$

$$x=0$$

Steps Using Factoring

Calculate \frac{2}{5} to the power of 2 and get \frac{4}{25}.

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

Expand \left(\frac{3}{4}x\right)^{2}.

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

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{4}{25}+\frac{9}{16}x^{2}=\left(x-\frac{2}{5}\right)^{2}$$

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

$$\frac{4}{25}+\frac{9}{16}x^{2}=x^{2}-\frac{4}{5}x+\frac{4}{25}$$

Subtract x^{2} from both sides.

$$\frac{4}{25}+\frac{9}{16}x^{2}-x^{2}=-\frac{4}{5}x+\frac{4}{25}$$

Combine \frac{9}{16}x^{2} and -x^{2} to get -\frac{7}{16}x^{2}.

$$\frac{4}{25}-\frac{7}{16}x^{2}=-\frac{4}{5}x+\frac{4}{25}$$

Add \frac{4}{5}x to both sides.

$$\frac{4}{25}-\frac{7}{16}x^{2}+\frac{4}{5}x=\frac{4}{25}$$

Subtract \frac{4}{25} from both sides.

$$\frac{4}{25}-\frac{7}{16}x^{2}+\frac{4}{5}x-\frac{4}{25}=0$$

Subtract \frac{4}{25} from \frac{4}{25} to get 0.

$$-\frac{7}{16}x^{2}+\frac{4}{5}x=0$$

Factor out x.

$$x\left(-\frac{7}{16}x+\frac{4}{5}\right)=0$$

To find equation solutions, solve x=0 and -\frac{7x}{16}+\frac{4}{5}=0.

$$x=0 x=\frac{64}{35}$$

Steps Using the Quadratic Formula

Calculate \frac{2}{5} to the power of 2 and get \frac{4}{25}.

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

Expand \left(\frac{3}{4}x\right)^{2}.

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

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{4}{25}+\frac{9}{16}x^{2}=\left(x-\frac{2}{5}\right)^{2}$$

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

$$\frac{4}{25}+\frac{9}{16}x^{2}=x^{2}-\frac{4}{5}x+\frac{4}{25}$$

Subtract x^{2} from both sides.

$$\frac{4}{25}+\frac{9}{16}x^{2}-x^{2}=-\frac{4}{5}x+\frac{4}{25}$$

Combine \frac{9}{16}x^{2} and -x^{2} to get -\frac{7}{16}x^{2}.

$$\frac{4}{25}-\frac{7}{16}x^{2}=-\frac{4}{5}x+\frac{4}{25}$$

Add \frac{4}{5}x to both sides.

$$\frac{4}{25}-\frac{7}{16}x^{2}+\frac{4}{5}x=\frac{4}{25}$$

Subtract \frac{4}{25} from both sides.

$$\frac{4}{25}-\frac{7}{16}x^{2}+\frac{4}{5}x-\frac{4}{25}=0$$

Subtract \frac{4}{25} from \frac{4}{25} to get 0.

$$-\frac{7}{16}x^{2}+\frac{4}{5}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{7}{16} for a, \frac{4}{5} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{4}{5}±\sqrt{\left(\frac{4}{5}\right)^{2}}}{2\left(-\frac{7}{16}\right)}$$

Take the square root of \left(\frac{4}{5}\right)^{2}.

$$x=\frac{-\frac{4}{5}±\frac{4}{5}}{2\left(-\frac{7}{16}\right)}$$

Multiply 2 times -\frac{7}{16}.

$$x=\frac{-\frac{4}{5}±\frac{4}{5}}{-\frac{7}{8}}$$

Now solve the equation x=\frac{-\frac{4}{5}±\frac{4}{5}}{-\frac{7}{8}} when ± is plus. Add -\frac{4}{5} to \frac{4}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{-\frac{7}{8}}$$

Divide 0 by -\frac{7}{8} by multiplying 0 by the reciprocal of -\frac{7}{8}.

$$x=0$$

Now solve the equation x=\frac{-\frac{4}{5}±\frac{4}{5}}{-\frac{7}{8}} when ± is minus. Subtract \frac{4}{5} from -\frac{4}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{8}{5}}{-\frac{7}{8}}$$

Divide -\frac{8}{5} by -\frac{7}{8} by multiplying -\frac{8}{5} by the reciprocal of -\frac{7}{8}.

$$x=\frac{64}{35}$$

The equation is now solved.

$$x=0 x=\frac{64}{35}$$

Steps for Completing the Square

Calculate \frac{2}{5} to the power of 2 and get \frac{4}{25}.

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

Expand \left(\frac{3}{4}x\right)^{2}.

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

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{4}{25}+\frac{9}{16}x^{2}=\left(x-\frac{2}{5}\right)^{2}$$

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

$$\frac{4}{25}+\frac{9}{16}x^{2}=x^{2}-\frac{4}{5}x+\frac{4}{25}$$

Subtract x^{2} from both sides.

$$\frac{4}{25}+\frac{9}{16}x^{2}-x^{2}=-\frac{4}{5}x+\frac{4}{25}$$

Combine \frac{9}{16}x^{2} and -x^{2} to get -\frac{7}{16}x^{2}.

$$\frac{4}{25}-\frac{7}{16}x^{2}=-\frac{4}{5}x+\frac{4}{25}$$

Add \frac{4}{5}x to both sides.

$$\frac{4}{25}-\frac{7}{16}x^{2}+\frac{4}{5}x=\frac{4}{25}$$

Subtract \frac{4}{25} from both sides.

$$-\frac{7}{16}x^{2}+\frac{4}{5}x=\frac{4}{25}-\frac{4}{25}$$

Subtract \frac{4}{25} from \frac{4}{25} to get 0.

$$-\frac{7}{16}x^{2}+\frac{4}{5}x=0$$

Divide both sides of the equation by -\frac{7}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{7}{16}x^{2}+\frac{4}{5}x}{-\frac{7}{16}}=\frac{0}{-\frac{7}{16}}$$

Dividing by -\frac{7}{16} undoes the multiplication by -\frac{7}{16}.

$$x^{2}+\frac{\frac{4}{5}}{-\frac{7}{16}}x=\frac{0}{-\frac{7}{16}}$$

Divide \frac{4}{5} by -\frac{7}{16} by multiplying \frac{4}{5} by the reciprocal of -\frac{7}{16}.

$$x^{2}-\frac{64}{35}x=\frac{0}{-\frac{7}{16}}$$

Divide 0 by -\frac{7}{16} by multiplying 0 by the reciprocal of -\frac{7}{16}.

$$x^{2}-\frac{64}{35}x=0$$

Divide -\frac{64}{35}, the coefficient of the x term, by 2 to get -\frac{32}{35}. Then add the square of -\frac{32}{35} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{64}{35}x+\left(-\frac{32}{35}\right)^{2}=\left(-\frac{32}{35}\right)^{2}$$

Square -\frac{32}{35} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{64}{35}x+\frac{1024}{1225}=\frac{1024}{1225}$$

Factor x^{2}-\frac{64}{35}x+\frac{1024}{1225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{32}{35}\right)^{2}=\frac{1024}{1225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{32}{35}\right)^{2}}=\sqrt{\frac{1024}{1225}}$$

Simplify.

$$x-\frac{32}{35}=\frac{32}{35} x-\frac{32}{35}=-\frac{32}{35}$$

Add \frac{32}{35} to both sides of the equation.

$$x=\frac{64}{35} x=0$$