Solve 4u^2+9=-12u

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

Solve for u

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

Steps Using Factoring By Grouping

Add 12u to both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4u^{2}+au+bu+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 4u^{2}+12u+9 as \left(4u^{2}+6u\right)+\left(6u+9\right).

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

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

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

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

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

Rewrite as a binomial square.

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

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

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

Steps Using the Quadratic Formula

Add 12u to both sides.

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

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.

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

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}.

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

Square 12.

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

Multiply -4 times 4.

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

Multiply -16 times 9.

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

Add 144 to -144.

$$u=\frac{-12±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

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

Multiply 2 times 4.

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

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

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

Steps for Completing the Square

Add 12u to both sides.

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

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

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide 12 by 4.

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

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.

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

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

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

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.

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

Factor u^{2}+3u+\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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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

The equation is now solved. Solutions are the same.

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