Question: Find the solution set of the inequality $$\frac{x^2-1}{(x+2)(x+3)}>2$$
From the answer given in the previous problem I got this:
First $x\neq -2,-3$. solving the equation I get $-(2\sqrt{3}+5)<x<(2\sqrt{3}-5)$.
Is this ok?
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$$\frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right) } >2\\ \frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right) } -2>0\\ \frac { -{ x }^{ 2 }-10x-13 }{ \left( x+2 \right) \left( x+3 \right) } >0\\ \frac { { x }^{ 2 }+10x+13 }{ \left( x+2 \right) \left( x+3 \right) } <0\\ \frac { \left( x+2\sqrt { 3 } +5 \right) \left( x+5-2\sqrt { 3 } \right) \left( x+2 \right) \left( x+3 \right) }{ { \left( x+2 \right) }^{ 2 }{ \left( x+3 \right) }^{ 2 } } <0\\ \left( x+2\sqrt { 3 } +5 \right) \left( x+5-2\sqrt { 3 } \right) \left( x+2 \right) \left( x+3 \right) <0\\ \\ \\ \\ $$can you take from here? (your answer is wrong by the way)
haqnatural
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2The inequality should be flipped between steps 3 and 4? – Tai Jul 26 '17 at 17:02
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the inequality flipped, i think! – user467365 Jul 26 '17 at 17:03
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Your fourth step is wrong. When you multiply an inequality by a negative number, the direction of the inequality is reversed. – N. F. Taussig Jul 27 '17 at 09:28
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@N.F.Taussig,thank you,it was typo,now i fixed – haqnatural Jul 27 '17 at 09:45