Solve x^2+left(5-xright)^2=9+{left(8/5x-4right)}^2

$${ x }^{ 2 } + { \left(5-x \right) }^{ 2 } = 9+ { \left( \frac{ 8 }{ 5 } x-4 \right) }^{ 2 }$$

Solve for x

$$x=5$$

$$x=0$$

Steps Using Factoring

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

$$x^{2}+25-10x+x^{2}=9+\left(\frac{8}{5}x-4\right)^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+25-10x=9+\left(\frac{8}{5}x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{8}{5}x-4\right)^{2}.

$$2x^{2}+25-10x=9+\frac{64}{25}x^{2}-\frac{64}{5}x+16$$

Add 9 and 16 to get 25.

$$2x^{2}+25-10x=25+\frac{64}{25}x^{2}-\frac{64}{5}x$$

Subtract 25 from both sides.

$$2x^{2}+25-10x-25=\frac{64}{25}x^{2}-\frac{64}{5}x$$

Subtract 25 from 25 to get 0.

$$2x^{2}-10x=\frac{64}{25}x^{2}-\frac{64}{5}x$$

Subtract \frac{64}{25}x^{2} from both sides.

$$2x^{2}-10x-\frac{64}{25}x^{2}=-\frac{64}{5}x$$

Combine 2x^{2} and -\frac{64}{25}x^{2} to get -\frac{14}{25}x^{2}.

$$-\frac{14}{25}x^{2}-10x=-\frac{64}{5}x$$

Add \frac{64}{5}x to both sides.

$$-\frac{14}{25}x^{2}-10x+\frac{64}{5}x=0$$

Combine -10x and \frac{64}{5}x to get \frac{14}{5}x.

$$-\frac{14}{25}x^{2}+\frac{14}{5}x=0$$

Factor out x.

$$x\left(-\frac{14}{25}x+\frac{14}{5}\right)=0$$

To find equation solutions, solve x=0 and -\frac{14x}{25}+\frac{14}{5}=0.

$$x=0 x=5$$

Steps Using the Quadratic Formula

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

$$x^{2}+25-10x+x^{2}=9+\left(\frac{8}{5}x-4\right)^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+25-10x=9+\left(\frac{8}{5}x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{8}{5}x-4\right)^{2}.

$$2x^{2}+25-10x=9+\frac{64}{25}x^{2}-\frac{64}{5}x+16$$

Add 9 and 16 to get 25.

$$2x^{2}+25-10x=25+\frac{64}{25}x^{2}-\frac{64}{5}x$$

Subtract 25 from both sides.

$$2x^{2}+25-10x-25=\frac{64}{25}x^{2}-\frac{64}{5}x$$

Subtract 25 from 25 to get 0.

$$2x^{2}-10x=\frac{64}{25}x^{2}-\frac{64}{5}x$$

Subtract \frac{64}{25}x^{2} from both sides.

$$2x^{2}-10x-\frac{64}{25}x^{2}=-\frac{64}{5}x$$

Combine 2x^{2} and -\frac{64}{25}x^{2} to get -\frac{14}{25}x^{2}.

$$-\frac{14}{25}x^{2}-10x=-\frac{64}{5}x$$

Add \frac{64}{5}x to both sides.

$$-\frac{14}{25}x^{2}-10x+\frac{64}{5}x=0$$

Combine -10x and \frac{64}{5}x to get \frac{14}{5}x.

$$-\frac{14}{25}x^{2}+\frac{14}{5}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{14}{25} for a, \frac{14}{5} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{14}{5}±\sqrt{\left(\frac{14}{5}\right)^{2}}}{2\left(-\frac{14}{25}\right)}$$

Take the square root of \left(\frac{14}{5}\right)^{2}.

$$x=\frac{-\frac{14}{5}±\frac{14}{5}}{2\left(-\frac{14}{25}\right)}$$

Multiply 2 times -\frac{14}{25}.

$$x=\frac{-\frac{14}{5}±\frac{14}{5}}{-\frac{28}{25}}$$

Now solve the equation x=\frac{-\frac{14}{5}±\frac{14}{5}}{-\frac{28}{25}} when ± is plus. Add -\frac{14}{5} to \frac{14}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{-\frac{28}{25}}$$

Divide 0 by -\frac{28}{25} by multiplying 0 by the reciprocal of -\frac{28}{25}.

$$x=0$$

Now solve the equation x=\frac{-\frac{14}{5}±\frac{14}{5}}{-\frac{28}{25}} when ± is minus. Subtract \frac{14}{5} from -\frac{14}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{28}{5}}{-\frac{28}{25}}$$

Divide -\frac{28}{5} by -\frac{28}{25} by multiplying -\frac{28}{5} by the reciprocal of -\frac{28}{25}.

$$x=5$$

The equation is now solved.

$$x=0 x=5$$

Steps for Completing the Square

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

$$x^{2}+25-10x+x^{2}=9+\left(\frac{8}{5}x-4\right)^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+25-10x=9+\left(\frac{8}{5}x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{8}{5}x-4\right)^{2}.

$$2x^{2}+25-10x=9+\frac{64}{25}x^{2}-\frac{64}{5}x+16$$

Add 9 and 16 to get 25.

$$2x^{2}+25-10x=25+\frac{64}{25}x^{2}-\frac{64}{5}x$$

Subtract \frac{64}{25}x^{2} from both sides.

$$2x^{2}+25-10x-\frac{64}{25}x^{2}=25-\frac{64}{5}x$$

Combine 2x^{2} and -\frac{64}{25}x^{2} to get -\frac{14}{25}x^{2}.

$$-\frac{14}{25}x^{2}+25-10x=25-\frac{64}{5}x$$

Add \frac{64}{5}x to both sides.

$$-\frac{14}{25}x^{2}+25-10x+\frac{64}{5}x=25$$

Combine -10x and \frac{64}{5}x to get \frac{14}{5}x.

$$-\frac{14}{25}x^{2}+25+\frac{14}{5}x=25$$

Subtract 25 from both sides.

$$-\frac{14}{25}x^{2}+\frac{14}{5}x=25-25$$

Subtract 25 from 25 to get 0.

$$-\frac{14}{25}x^{2}+\frac{14}{5}x=0$$

Divide both sides of the equation by -\frac{14}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{14}{25}x^{2}+\frac{14}{5}x}{-\frac{14}{25}}=\frac{0}{-\frac{14}{25}}$$

Dividing by -\frac{14}{25} undoes the multiplication by -\frac{14}{25}.

$$x^{2}+\frac{\frac{14}{5}}{-\frac{14}{25}}x=\frac{0}{-\frac{14}{25}}$$

Divide \frac{14}{5} by -\frac{14}{25} by multiplying \frac{14}{5} by the reciprocal of -\frac{14}{25}.

$$x^{2}-5x=\frac{0}{-\frac{14}{25}}$$

Divide 0 by -\frac{14}{25} by multiplying 0 by the reciprocal of -\frac{14}{25}.

$$x^{2}-5x=0$$

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=5 x=0$$