Solve (X-9)^2=36

$$( X - 9 ) ^ { 2 } = 36$$

Solve for X

$$X=15$$

$$X=3$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(X-9\right)^{2}.

$$X^{2}-18X+81=36$$

Subtract 36 from both sides.

$$X^{2}-18X+81-36=0$$

Subtract 36 from 81 to get 45.

$$X^{2}-18X+45=0$$

To solve the equation, factor X^{2}-18X+45 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.

$$a+b=-18 ab=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -18.

$$a=-15 b=-3$$

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

$$\left(X-15\right)\left(X-3\right)$$

To find equation solutions, solve X-15=0 and X-3=0.

$$X=15 X=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(X-9\right)^{2}.

$$X^{2}-18X+81=36$$

Subtract 36 from both sides.

$$X^{2}-18X+81-36=0$$

Subtract 36 from 81 to get 45.

$$X^{2}-18X+45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as X^{2}+aX+bX+45. To find a and b, set up a system to be solved.

$$a+b=-18 ab=1\times 45=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -18.

$$a=-15 b=-3$$

Rewrite X^{2}-18X+45 as \left(X^{2}-15X\right)+\left(-3X+45\right).

$$\left(X^{2}-15X\right)+\left(-3X+45\right)$$

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

$$X\left(X-15\right)-3\left(X-15\right)$$

Factor out common term X-15 by using distributive property.

$$\left(X-15\right)\left(X-3\right)$$

To find equation solutions, solve X-15=0 and X-3=0.

$$X=15 X=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(X-9\right)^{2}.

$$X^{2}-18X+81=36$$

Subtract 36 from both sides.

$$X^{2}-18X+81-36=0$$

Subtract 36 from 81 to get 45.

$$X^{2}-18X+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$X=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 45}}{2}$$

Square -18.

$$X=\frac{-\left(-18\right)±\sqrt{324-4\times 45}}{2}$$

Multiply -4 times 45.

$$X=\frac{-\left(-18\right)±\sqrt{324-180}}{2}$$

Add 324 to -180.

$$X=\frac{-\left(-18\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$X=\frac{-\left(-18\right)±12}{2}$$

The opposite of -18 is 18.

$$X=\frac{18±12}{2}$$

Now solve the equation X=\frac{18±12}{2} when ± is plus. Add 18 to 12.

$$X=\frac{30}{2}$$

Divide 30 by 2.

$$X=15$$

Now solve the equation X=\frac{18±12}{2} when ± is minus. Subtract 12 from 18.

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

Divide 6 by 2.

$$X=3$$

The equation is now solved.

$$X=15 X=3$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(X-9\right)^{2}}=\sqrt{36}$$

Simplify.

$$X-9=6 X-9=-6$$

Add 9 to both sides of the equation.

$$X=15 X=3$$