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

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1 Answer

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:

     (4a2 + 4a + 1)/(4 - 9a2) = (3a - 2)/(4a2 + 2a)

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

     4a2 + 4a + 1 = (2a + 1)(2a + 1)

     4 - 9a2 = (2)2 - (3a)2 = (2 - 3a)(2 + 3a)

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

     4a2 + 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)

     4a2 + 2a = -2 - 3a

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

     4a2 + 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 ==> ax2 + bx + c = 0):

     x = (-b ± √(b2 - 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