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

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

Solve for x

$$x=6$$

$$x=5$$

Steps Using Factoring

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

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

Subtract x from both sides.

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

Combine -10x and -x to get -11x.

$$x^{2}-11x+25=-5$$

Add 5 to both sides.

$$x^{2}-11x+25+5=0$$

Add 25 and 5 to get 30.

$$x^{2}-11x+30=0$$

To solve the equation, factor x^{2}-11x+30 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-11 ab=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$a=-6 b=-5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-6\right)\left(x-5\right)$$

To find equation solutions, solve x-6=0 and x-5=0.

$$x=6 x=5$$

Steps Using Factoring By Grouping

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

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

Subtract x from both sides.

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

Combine -10x and -x to get -11x.

$$x^{2}-11x+25=-5$$

Add 5 to both sides.

$$x^{2}-11x+25+5=0$$

Add 25 and 5 to get 30.

$$x^{2}-11x+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 30=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$a=-6 b=-5$$

Rewrite x^{2}-11x+30 as \left(x^{2}-6x\right)+\left(-5x+30\right).

$$\left(x^{2}-6x\right)+\left(-5x+30\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-6\right)-5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-5\right)$$

To find equation solutions, solve x-6=0 and x-5=0.

$$x=6 x=5$$

Steps Using the Quadratic Formula

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

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

Subtract x from both sides.

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

Combine -10x and -x to get -11x.

$$x^{2}-11x+25=-5$$

Add 5 to both sides.

$$x^{2}-11x+25+5=0$$

Add 25 and 5 to get 30.

$$x^{2}-11x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 30}}{2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 30}}{2}$$

Multiply -4 times 30.

$$x=\frac{-\left(-11\right)±\sqrt{121-120}}{2}$$

Add 121 to -120.

$$x=\frac{-\left(-11\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-11\right)±1}{2}$$

The opposite of -11 is 11.

$$x=\frac{11±1}{2}$$

Now solve the equation x=\frac{11±1}{2} when ± is plus. Add 11 to 1.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{11±1}{2} when ± is minus. Subtract 1 from 11.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

The equation is now solved.

$$x=6 x=5$$

Steps for Completing the Square

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

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

Subtract x from both sides.

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

Combine -10x and -x to get -11x.

$$x^{2}-11x+25=-5$$

Subtract 25 from both sides.

$$x^{2}-11x=-5-25$$

Subtract 25 from -5 to get -30.

$$x^{2}-11x=-30$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-30+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-30+\frac{121}{4}$$

Add -30 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{1}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{1}{2} x-\frac{11}{2}=-\frac{1}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=6 x=5$$