Solve 3(x-3)=left(x-3right)^2

$$3(x-3)= { \left(x-3 \right) }^{ 2 }$$

Solve for x

$$x=6$$

$$x=3$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 3 by x-3.

$$3x-9=\left(x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$3x-9=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$3x-9-x^{2}=-6x+9$$

Add 6x to both sides.

$$3x-9-x^{2}+6x=9$$

Combine 3x and 6x to get 9x.

$$9x-9-x^{2}=9$$

Subtract 9 from both sides.

$$9x-9-x^{2}-9=0$$

Subtract 9 from -9 to get -18.

$$9x-18-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+9x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-18\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 9.

$$a=6 b=3$$

Rewrite -x^{2}+9x-18 as \left(-x^{2}+6x\right)+\left(3x-18\right).

$$\left(-x^{2}+6x\right)+\left(3x-18\right)$$

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

$$-x\left(x-6\right)+3\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+3\right)$$

To find equation solutions, solve x-6=0 and -x+3=0.

$$x=6 x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x-3.

$$3x-9=\left(x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$3x-9=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$3x-9-x^{2}=-6x+9$$

Add 6x to both sides.

$$3x-9-x^{2}+6x=9$$

Combine 3x and 6x to get 9x.

$$9x-9-x^{2}=9$$

Subtract 9 from both sides.

$$9x-9-x^{2}-9=0$$

Subtract 9 from -9 to get -18.

$$9x-18-x^{2}=0$$

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.

$$-x^{2}+9x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-9±\sqrt{81+4\left(-18\right)}}{2\left(-1\right)}$$

Multiply 4 times -18.

$$x=\frac{-9±\sqrt{81-72}}{2\left(-1\right)}$$

Add 81 to -72.

$$x=\frac{-9±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$x=\frac{-9±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-9±3}{-2}$$

Now solve the equation x=\frac{-9±3}{-2} when ± is plus. Add -9 to 3.

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

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-9±3}{-2} when ± is minus. Subtract 3 from -9.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=3 x=6$$

Steps for Completing the Square

Use the distributive property to multiply 3 by x-3.

$$3x-9=\left(x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$3x-9=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$3x-9-x^{2}=-6x+9$$

Add 6x to both sides.

$$3x-9-x^{2}+6x=9$$

Combine 3x and 6x to get 9x.

$$9x-9-x^{2}=9$$

Add 9 to both sides.

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

Add 9 and 9 to get 18.

$$9x-x^{2}=18$$

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.

$$-x^{2}+9x=18$$

Divide both sides by -1.

$$\frac{-x^{2}+9x}{-1}=\frac{18}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{9}{-1}x=\frac{18}{-1}$$

Divide 9 by -1.

$$x^{2}-9x=\frac{18}{-1}$$

Divide 18 by -1.

$$x^{2}-9x=-18$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-18+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-18+\frac{81}{4}$$

Add -18 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{9}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{9}{2}=\frac{3}{2} x-\frac{9}{2}=-\frac{3}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=6 x=3$$