Solve 4a^2+36a+81/a^2+9=13

$$\frac { 4 a ^ { 2 } + 36 a + 81 } { a ^ { 2 } + 9 } = 13$$

Solve for a

$$a=2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by a^{2}+9.

$$4a^{2}+36a+81=13\left(a^{2}+9\right)$$

Use the distributive property to multiply 13 by a^{2}+9.

$$4a^{2}+36a+81=13a^{2}+117$$

Subtract 13a^{2} from both sides.

$$4a^{2}+36a+81-13a^{2}=117$$

Combine 4a^{2} and -13a^{2} to get -9a^{2}.

$$-9a^{2}+36a+81=117$$

Subtract 117 from both sides.

$$-9a^{2}+36a+81-117=0$$

Subtract 117 from 81 to get -36.

$$-9a^{2}+36a-36=0$$

Divide both sides by 9.

$$-a^{2}+4a-4=0$$

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

$$a+b=4 ab=-\left(-4\right)=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite -a^{2}+4a-4 as \left(-a^{2}+2a\right)+\left(2a-4\right).

$$\left(-a^{2}+2a\right)+\left(2a-4\right)$$

Factor out -a in the first and 2 in the second group.

$$-a\left(a-2\right)+2\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(-a+2\right)$$

To find equation solutions, solve a-2=0 and -a+2=0.

$$a=2 a=2$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by a^{2}+9.

$$4a^{2}+36a+81=13\left(a^{2}+9\right)$$

Use the distributive property to multiply 13 by a^{2}+9.

$$4a^{2}+36a+81=13a^{2}+117$$

Subtract 13a^{2} from both sides.

$$4a^{2}+36a+81-13a^{2}=117$$

Combine 4a^{2} and -13a^{2} to get -9a^{2}.

$$-9a^{2}+36a+81=117$$

Subtract 117 from both sides.

$$-9a^{2}+36a+81-117=0$$

Subtract 117 from 81 to get -36.

$$-9a^{2}+36a-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 36 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-36±\sqrt{36^{2}-4\left(-9\right)\left(-36\right)}}{2\left(-9\right)}$$

Square 36.

$$a=\frac{-36±\sqrt{1296-4\left(-9\right)\left(-36\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$a=\frac{-36±\sqrt{1296+36\left(-36\right)}}{2\left(-9\right)}$$

Multiply 36 times -36.

$$a=\frac{-36±\sqrt{1296-1296}}{2\left(-9\right)}$$

Add 1296 to -1296.

$$a=\frac{-36±\sqrt{0}}{2\left(-9\right)}$$

Take the square root of 0.

$$a=-\frac{36}{2\left(-9\right)}$$

Multiply 2 times -9.

$$a=-\frac{36}{-18}$$

Divide -36 by -18.

$$a=2$$

Steps for Completing the Square

Multiply both sides of the equation by a^{2}+9.

$$4a^{2}+36a+81=13\left(a^{2}+9\right)$$

Use the distributive property to multiply 13 by a^{2}+9.

$$4a^{2}+36a+81=13a^{2}+117$$

Subtract 13a^{2} from both sides.

$$4a^{2}+36a+81-13a^{2}=117$$

Combine 4a^{2} and -13a^{2} to get -9a^{2}.

$$-9a^{2}+36a+81=117$$

Subtract 81 from both sides.

$$-9a^{2}+36a=117-81$$

Subtract 81 from 117 to get 36.

$$-9a^{2}+36a=36$$

Divide both sides by -9.

$$\frac{-9a^{2}+36a}{-9}=\frac{36}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$a^{2}+\frac{36}{-9}a=\frac{36}{-9}$$

Divide 36 by -9.

$$a^{2}-4a=\frac{36}{-9}$$

Divide 36 by -9.

$$a^{2}-4a=-4$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-4a+\left(-2\right)^{2}=-4+\left(-2\right)^{2}$$

Square -2.

$$a^{2}-4a+4=-4+4$$

Add -4 to 4.

$$a^{2}-4a+4=0$$

Factor a^{2}-4a+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}.

$$\left(a-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$a-2=0 a-2=0$$

Add 2 to both sides of the equation.

$$a=2 a=2$$

The equation is now solved. Solutions are the same.

$$a=2$$