Solving the inequality x2+4x+3&gt;0

Question

Solving the inequality x2+4x+3&amp;gt;0

Jesse Gates · Accepted Answer

You have to consider several scenarios.In general, when is ab&amp;gt;0? This happens when- a, b are both positive, or- a, b are both negativeSo, consider: you have (x+1)(x+3)&amp;gt;0. Then,- x+1,x+3 are both positive, or- x+1,x+3 are both negativeIn your working, you assume only the former, and get that either x≻1 or x≻3. Of course, if both are positive, we need x≻1, so that gives one solution.But what if both are negative? Then, when you solve (x+1)(x+3)&amp;gt;0, you are dividing by one of those factors and therefore must reverse the inequality sign. Hence,(x+1)(x+3)&amp;gt;0⇒x+1&amp;lt;0  or  x+3&amp;lt;0when both are negative. Therefore, x&amp;lt;−1or x&amp;lt;−3. Of course, again, we see that both factors are only negative if x&amp;lt;−3.Thus, the solution set is the set of x where x≻1 or x&amp;lt;−3.

Declan Cameron · Accepted Answer

Option:(x+2)2&amp;gt;1;⇒|x+2∣&amp;gt;1, i.e.x+2&amp;gt;1, or x+2&amp;lt;−1.

Solving the inequality \(\displaystyle{x}^{{{2}}}+{4}{x}+{3}{>}{0}\)

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