Solve (4-x)(4-x)=9

$$(4-x)(4-x)=9$$

Solve for x

$$x=1$$

$$x=7$$

Steps Using the Quadratic Formula

Multiply 4-x and 4-x to get \left(4-x\right)^{2}.

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

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

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

Subtract 9 from both sides.

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

Subtract 9 from 16 to get 7.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

Multiply 4-x and 4-x to get \left(4-x\right)^{2}.

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

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

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

Subtract 16 from both sides.

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

Subtract 16 from 9 to get -7.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

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.

$$x=7 x=1$$