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 (x^2-4x-12)/(x^2-25)≥0

 May 28, 2018
edited by Bobbly  May 30, 2018
edited by Bobbly  May 30, 2018
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Hey bobbly!

 

 \(\frac5{x+3}+\frac{3}{x-2}=4\\ \frac{5\cdot(x-2)+3\cdot(x+3)}{(x-2)(x+3)}=4\\ \frac{5x-10+3x+9}{x^2+x-6}=4\\ 5x-10+3x+9=4x^2+4x-24\\ 4x^2-4x-23=0\)

 

We then use the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \), to solve for x. 

 

\(x_1=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:4\left(-23\right)}}{2\cdot \:4}=\frac{1+2\sqrt{6}}{2}\)

\(x_2=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:4\left(-23\right)}}{2\cdot \:4}=\frac{1-2\sqrt{6}}{2}\)

 

I hope this helped,

 

Gavin

 May 28, 2018

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