Solve 4(x+3)^2-12(x+3)+9=0

$$4 ( x + 3 ) ^ { 2 } - 12 ( x + 3 ) + 9 = 0$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Factoring By Grouping

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

$$4\left(x^{2}+6x+9\right)-12\left(x+3\right)+9=0$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36-12\left(x+3\right)+9=0$$

Use the distributive property to multiply -12 by x+3.

$$4x^{2}+24x+36-12x-36+9=0$$

Combine 24x and -12x to get 12x.

$$4x^{2}+12x+36-36+9=0$$

Subtract 36 from 36 to get 0.

$$4x^{2}+12x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=12 ab=4\times 9=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 12.

$$a=6 b=6$$

Rewrite 4x^{2}+12x+9 as \left(4x^{2}+6x\right)+\left(6x+9\right).

$$\left(4x^{2}+6x\right)+\left(6x+9\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(2x+3\right)+3\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(2x+3\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve 2x+3=0.

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

Steps Using the Quadratic Formula

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

$$4\left(x^{2}+6x+9\right)-12\left(x+3\right)+9=0$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36-12\left(x+3\right)+9=0$$

Use the distributive property to multiply -12 by x+3.

$$4x^{2}+24x+36-12x-36+9=0$$

Combine 24x and -12x to get 12x.

$$4x^{2}+12x+36-36+9=0$$

Subtract 36 from 36 to get 0.

$$4x^{2}+12x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 12 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\times 9}}{2\times 4}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 4\times 9}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-12±\sqrt{144-16\times 9}}{2\times 4}$$

Multiply -16 times 9.

$$x=\frac{-12±\sqrt{144-144}}{2\times 4}$$

Add 144 to -144.

$$x=\frac{-12±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{12}{2\times 4}$$

Multiply 2 times 4.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

Steps for Completing the Square

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

$$4\left(x^{2}+6x+9\right)-12\left(x+3\right)+9=0$$

Use the distributive property to multiply 4 by x^{2}+6x+9.

$$4x^{2}+24x+36-12\left(x+3\right)+9=0$$

Use the distributive property to multiply -12 by x+3.

$$4x^{2}+24x+36-12x-36+9=0$$

Combine 24x and -12x to get 12x.

$$4x^{2}+12x+36-36+9=0$$

Subtract 36 from 36 to get 0.

$$4x^{2}+12x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+12x=-9$$

Divide both sides by 4.

$$\frac{4x^{2}+12x}{4}=-\frac{9}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{12}{4}x=-\frac{9}{4}$$

Divide 12 by 4.

$$x^{2}+3x=-\frac{9}{4}$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-\frac{9}{4}+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=\frac{-9+9}{4}$$

Add -\frac{9}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=0$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{3}{2}=0 x+\frac{3}{2}=0$$

Subtract \frac{3}{2} from both sides of the equation.

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

The equation is now solved. Solutions are the same.

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