Solve x^2+(5-x)^2=(5-2x)^2

$$x ^ { 2 } + ( 5 - x ) ^ { 2 } = ( 5 - 2 x ) ^ { 2 }$$

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}.

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

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

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

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

$$2x^{2}+25-10x=25-20x+4x^{2}$$

Subtract 25 from both sides.

$$2x^{2}+25-10x-25=-20x+4x^{2}$$

Subtract 25 from 25 to get 0.

$$2x^{2}-10x=-20x+4x^{2}$$

Add 20x to both sides.

$$2x^{2}-10x+20x=4x^{2}$$

Combine -10x and 20x to get 10x.

$$2x^{2}+10x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$2x^{2}+10x-4x^{2}=0$$

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

$$-2x^{2}+10x=0$$

Factor out x.

$$x\left(-2x+10\right)=0$$

To find equation solutions, solve x=0 and -2x+10=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}.

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

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

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

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

$$2x^{2}+25-10x=25-20x+4x^{2}$$

Subtract 25 from both sides.

$$2x^{2}+25-10x-25=-20x+4x^{2}$$

Subtract 25 from 25 to get 0.

$$2x^{2}-10x=-20x+4x^{2}$$

Add 20x to both sides.

$$2x^{2}-10x+20x=4x^{2}$$

Combine -10x and 20x to get 10x.

$$2x^{2}+10x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$2x^{2}+10x-4x^{2}=0$$

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

$$-2x^{2}+10x=0$$

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

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

Take the square root of 10^{2}.

$$x=\frac{-10±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-10±10}{-4}$$

Now solve the equation x=\frac{-10±10}{-4} when ± is plus. Add -10 to 10.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-10±10}{-4} when ± is minus. Subtract 10 from -10.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

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}.

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

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

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

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

$$2x^{2}+25-10x=25-20x+4x^{2}$$

Add 20x to both sides.

$$2x^{2}+25-10x+20x=25+4x^{2}$$

Combine -10x and 20x to get 10x.

$$2x^{2}+25+10x=25+4x^{2}$$

Subtract 4x^{2} from both sides.

$$2x^{2}+25+10x-4x^{2}=25$$

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

$$-2x^{2}+25+10x=25$$

Subtract 25 from both sides.

$$-2x^{2}+10x=25-25$$

Subtract 25 from 25 to get 0.

$$-2x^{2}+10x=0$$

Divide both sides by -2.

$$\frac{-2x^{2}+10x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{10}{-2}x=\frac{0}{-2}$$

Divide 10 by -2.

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

Divide 0 by -2.

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

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.

$$x=5 x=0$$