Solve (x-5)^2-9=0

$$( x - 5 ) ^ { 2 } - 9 = 0$$

Solve for x

$$x=8$$

$$x=2$$

Steps Using Factoring

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

$$x^{2}-10x+25-9=0$$

Subtract 9 from 25 to get 16.

$$x^{2}-10x+16=0$$

To solve the equation, factor x^{2}-10x+16 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.

$$a+b=-10 ab=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -10.

$$a=-8 b=-2$$

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

$$\left(x-8\right)\left(x-2\right)$$

To find equation solutions, solve x-8=0 and x-2=0.

$$x=8 x=2$$

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}.

$$x^{2}-10x+25-9=0$$

Subtract 9 from 25 to get 16.

$$x^{2}-10x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 16=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -10.

$$a=-8 b=-2$$

Rewrite x^{2}-10x+16 as \left(x^{2}-8x\right)+\left(-2x+16\right).

$$\left(x^{2}-8x\right)+\left(-2x+16\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-8\right)-2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-2\right)$$

To find equation solutions, solve x-8=0 and x-2=0.

$$x=8 x=2$$

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}.

$$x^{2}-10x+25-9=0$$

Subtract 9 from 25 to get 16.

$$x^{2}-10x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 16}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 16}}{2}$$

Multiply -4 times 16.

$$x=\frac{-\left(-10\right)±\sqrt{100-64}}{2}$$

Add 100 to -64.

$$x=\frac{-\left(-10\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-\left(-10\right)±6}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±6}{2}$$

Now solve the equation x=\frac{10±6}{2} when ± is plus. Add 10 to 6.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{10±6}{2} when ± is minus. Subtract 6 from 10.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=8 x=2$$

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}.

$$x^{2}-10x+25-9=0$$

Subtract 9 from 25 to get 16.

$$x^{2}-10x+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-10x=-16$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-16+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-16+25$$

Add -16 to 25.

$$x^{2}-10x+25=9$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-5=3 x-5=-3$$

Add 5 to both sides of the equation.

$$x=8 x=2$$