how do you solve: "find the area of a square with sides (3x)/(4x^2-9x)?" Thank you for your help.
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s^2 = A
[3x / (4x^2 - 9x)]^2 = A
9x^2 / (16x^4 - 36x^3 - 36x^3 + 81x^2)
9x^2 / (16x^4 - 72x^3 + 81x^2)
9x^2 / x^2(16x^2 - 72x + 81)
9 / (16x^2 - 72x + 81) = A
[3x / (4x^2 - 9x)]^2 = A
9x^2 / (16x^4 - 36x^3 - 36x^3 + 81x^2)
9x^2 / (16x^4 - 72x^3 + 81x^2)
9x^2 / x^2(16x^2 - 72x + 81)
9 / (16x^2 - 72x + 81) = A
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(3x)^2/(4x^2-9x)^2
9/(4x-9)^2
9/(4x-9)^2