Tamara J. answered • 04/08/13

Math Tutoring - Algebra and Calculus (all levels)

I'm thinking you meant to put an equals sign in the middle rather than another division sign, which would make the problem look like the following:

(4a^{2} + 4a + 1)/(4 - 9a^{2}) = (3a - 2)/(4a^{2} + 2a)

If this is the case, then let's first factor each polynomial separately as follows:

4a^{2} + 4a + 1 = (2a + 1)(2a + 1)

4 - 9a^{2} = (2)^{2} - (3a)^{2} = (2 - 3a)(2 + 3a)

3a - 2 = -1(-3a + 2) = -1(2 - 3a)

4a^{2} + 2a = 2a(2a + 1)

Substitute the factored forms of these polynomials back into the original equation. In doing so, we arrive at the following:

((2a + 1)(2a + 1))/((2 - 3a)(2 + 3a)) = (-1(2 - 3a))/(2a(2a + 1))

Looking at this equation, the numerator on the left hand side and the denominator on the right hand side have a term in common that being (2a + 1), which means that these two terms cancel each other out. Likewise, the denominator on the left hand side and the numerator on the right hand side also share a common term that being (2 - 3a) and they, in turn, also cancel each other out.

(**(2a + 1)**(2a + 1))/(**(2 - 3a)**(2 + 3a)) = (-1**(2 - 3a)**)/(2a**(2a + 1)**)

With this, we are left with the following:

(2a + 1)/(2 + 3a) = (-1)/(2a)

Cross multiply:

(2a)(2a + 1) = (-1)(2 + 3a)

4a^{2} + 2a = -2 - 3a

Add 2 and 3a to both sides of the equation to set the equation equal to 0:

4a^{2} + 5a + 2 = 0

Since this equation can't be factored into 2 binomials, use the quadratic equation to solve for a (let 'x' represent the variable 'a' in this case since a is the leading coefficient in the standard form of quadratic equations ==> ax^{2} + bx + c = 0):

x = (-b ± √(b^{2} - 4ac)) / 2a

x = (-5 ± √((5)^{2} - 4(4)(2))) / 2(4)

x = (-5 ± √(25 - 32)) / 8

x = (-5 ± √(-7)) / 8

x = (-5 ± √(-1)·√(7) / 8

x = (-5 ± i√7)/8