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Solve \(\dfrac{x^2 + 2x}{x^2 - 4x} < 0\)

 May 24, 2020
 #1
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Solve

\(\dfrac{x^2 + 2x}{x^2 - 4x} < 0\)

 

zero in numerator:

\(\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \\ x(x + 2) &=& 0 \\ \hline x= 0 &\text{or}& x=-2 \\ \hline \end{array} \)

 

zero in denominator:

\(\begin{array}{|rcll|} \hline x^2 - 4x &=& 0 \\ x(x -4) &=& 0 \\ \hline x= 0 &\text{or}& x=4 \\ \hline \end{array}\)

 

\(\begin{array}{|r|c|c|c|c|c|c|c|} \hline & (-\infty,-2) & -2 & (-2,0) & 0 & (0,4) & 4 & (4,+\infty) \\ \hline x^2 + 2x & +&0&-&0&+&+&+\\ x^2 - 4x & +&+&+&0&-&0&+\\ \hline \text{fraction} & + & 0 & - & / & - & / & + \\ \hline \end{array}\)

 

\(-2 < x < 0\) and \(0 < x < 4\)

 

laugh

 May 25, 2020
edited by heureka  May 25, 2020

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