Solve (x-4)^2-9=0

$$( x - 4 ) ^ { 2 } - 9 = 0$$

Solve for x

$$x=7$$

$$x=1$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-9=0$$

Subtract 9 from 16 to get 7.

$$x^{2}-8x+7=0$$

To solve the equation, factor x^{2}-8x+7 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.

$$a+b=-8 ab=7$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-7 b=-1$$

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

$$\left(x-7\right)\left(x-1\right)$$

To find equation solutions, solve x-7=0 and x-1=0.

$$x=7 x=1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-9=0$$

Subtract 9 from 16 to get 7.

$$x^{2}-8x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 7=7$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-7 b=-1$$

Rewrite x^{2}-8x+7 as \left(x^{2}-7x\right)+\left(-x+7\right).

$$\left(x^{2}-7x\right)+\left(-x+7\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(x-7\right)-\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x-1\right)$$

To find equation solutions, solve x-7=0 and x-1=0.

$$x=7 x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-9=0$$

Subtract 9 from 16 to get 7.

$$x^{2}-8x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 7}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 7}}{2}$$

Multiply -4 times 7.

$$x=\frac{-\left(-8\right)±\sqrt{64-28}}{2}$$

Add 64 to -28.

$$x=\frac{-\left(-8\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-\left(-8\right)±6}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±6}{2}$$

Now solve the equation x=\frac{8±6}{2} when ± is plus. Add 8 to 6.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{8±6}{2} when ± is minus. Subtract 6 from 8.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=7 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-9=0$$

Subtract 9 from 16 to get 7.

$$x^{2}-8x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-8x=-7$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-7+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-7+16$$

Add -7 to 16.

$$x^{2}-8x+16=9$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-4=3 x-4=-3$$

Add 4 to both sides of the equation.

$$x=7 x=1$$