Solve (x-8)^2=25

$$( x - 8 ) ^ { 2 } = 25$$

Solve for x

$$x=13$$

$$x=3$$

Steps Using Factoring

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

$$x^{2}-16x+64=25$$

Subtract 25 from both sides.

$$x^{2}-16x+64-25=0$$

Subtract 25 from 64 to get 39.

$$x^{2}-16x+39=0$$

To solve the equation, factor x^{2}-16x+39 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.

$$a+b=-16 ab=39$$

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 39.

$$-1,-39 -3,-13$$

Calculate the sum for each pair.

$$-1-39=-40 -3-13=-16$$

The solution is the pair that gives sum -16.

$$a=-13 b=-3$$

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

$$\left(x-13\right)\left(x-3\right)$$

To find equation solutions, solve x-13=0 and x-3=0.

$$x=13 x=3$$

Steps Using Factoring By Grouping

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

$$x^{2}-16x+64=25$$

Subtract 25 from both sides.

$$x^{2}-16x+64-25=0$$

Subtract 25 from 64 to get 39.

$$x^{2}-16x+39=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+39. To find a and b, set up a system to be solved.

$$a+b=-16 ab=1\times 39=39$$

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 39.

$$-1,-39 -3,-13$$

Calculate the sum for each pair.

$$-1-39=-40 -3-13=-16$$

The solution is the pair that gives sum -16.

$$a=-13 b=-3$$

Rewrite x^{2}-16x+39 as \left(x^{2}-13x\right)+\left(-3x+39\right).

$$\left(x^{2}-13x\right)+\left(-3x+39\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-13\right)-3\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(x-3\right)$$

To find equation solutions, solve x-13=0 and x-3=0.

$$x=13 x=3$$

Steps Using the Quadratic Formula

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

$$x^{2}-16x+64=25$$

Subtract 25 from both sides.

$$x^{2}-16x+64-25=0$$

Subtract 25 from 64 to get 39.

$$x^{2}-16x+39=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -16 for b, and 39 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 39}}{2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 39}}{2}$$

Multiply -4 times 39.

$$x=\frac{-\left(-16\right)±\sqrt{256-156}}{2}$$

Add 256 to -156.

$$x=\frac{-\left(-16\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-\left(-16\right)±10}{2}$$

The opposite of -16 is 16.

$$x=\frac{16±10}{2}$$

Now solve the equation x=\frac{16±10}{2} when ± is plus. Add 16 to 10.

$$x=\frac{26}{2}$$

Divide 26 by 2.

$$x=13$$

Now solve the equation x=\frac{16±10}{2} when ± is minus. Subtract 10 from 16.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=13 x=3$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-8=5 x-8=-5$$

Add 8 to both sides of the equation.

$$x=13 x=3$$