Solve (x-5)^2=1

$$( x - 5 ) ^ { 2 } = 1$$

Solve for x

$$x=6$$

$$x=4$$

Steps Using Factoring

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

$$x^{2}-10x+25=1$$

Subtract 1 from both sides.

$$x^{2}-10x+25-1=0$$

Subtract 1 from 25 to get 24.

$$x^{2}-10x+24=0$$

To solve the equation, factor x^{2}-10x+24 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.

$$a+b=-10 ab=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

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

$$\left(x-6\right)\left(x-4\right)$$

To find equation solutions, solve x-6=0 and x-4=0.

$$x=6 x=4$$

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}.

$$x^{2}-10x+25=1$$

Subtract 1 from both sides.

$$x^{2}-10x+25-1=0$$

Subtract 1 from 25 to get 24.

$$x^{2}-10x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite x^{2}-10x+24 as \left(x^{2}-6x\right)+\left(-4x+24\right).

$$\left(x^{2}-6x\right)+\left(-4x+24\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-6\right)-4\left(x-6\right)$$

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

$$\left(x-6\right)\left(x-4\right)$$

To find equation solutions, solve x-6=0 and x-4=0.

$$x=6 x=4$$

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}.

$$x^{2}-10x+25=1$$

Subtract 1 from both sides.

$$x^{2}-10x+25-1=0$$

Subtract 1 from 25 to get 24.

$$x^{2}-10x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 24}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 24}}{2}$$

Multiply -4 times 24.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2}$$

Add 100 to -96.

$$x=\frac{-\left(-10\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-\left(-10\right)±2}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2}{2}$$

Now solve the equation x=\frac{10±2}{2} when ± is plus. Add 10 to 2.

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

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{10±2}{2} when ± is minus. Subtract 2 from 10.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

The equation is now solved.

$$x=6 x=4$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-5=1 x-5=-1$$

Add 5 to both sides of the equation.

$$x=6 x=4$$