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How would I solve the following inequality.

$x^2+10x \gt-24$

How would I solve it and put it in a number line?

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1  
Follow the procedure in the examples here. –  David Mitra Jan 20 '13 at 19:30
    
Here's the number line. Note the unfilled dots, because we have $>$ and not $\ge$. –  George V. Williams Jan 20 '13 at 19:37

4 Answers 4

up vote 3 down vote accepted

Hint: solve for

$x^2+10x+ 24 = 0$, find the roots, and determine when the factors are positive:

Solve: $$x^2 + 10 x + 24 = (x+4)(x+6) > 0$$

When is $(x+4)(x + 6)$ positive?:

$\quad$When both factors are positive, or when both factors are negative.

$$ (x+4)(x+6) > 0 \implies \begin{cases}(x+4) > 0, & (x+6) > 0 \longrightarrow x>-4\\ \\(x+4) < 0, & (x+6) < 0 \longrightarrow x<-6 \end{cases} $$

Your task is to plot the intervals on which $x$ satisfies the inequality.


Edit: if you want to confirm the solution "graph", compare to:

enter image description here

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Yes it makes sense thanks for the help. –  Fernando Martinez Jan 20 '13 at 20:08

$$x^2 + 10x > -24 \implies x^2 + 10x + 24 > 0 \implies (x+4)(x+6) > 0$$ Recall that if $ab > 0$, then either $a>0 \,\, \& \,\, b > 0$ or $a < 0 \,\, \& \,\, b < 0$. Hence, $$(x+4)(x+6) > 0 \implies \begin{cases}(x+4) > 0; & (x+6) > 0 \implies x>-4\\ (x+4) < 0; & (x+6) < 0 \implies x<-6 \end{cases}$$ Hence, we get that $$x \in (-\infty,-6) \cup (-4, \infty)$$

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thanks for the help. –  Fernando Martinez Jan 20 '13 at 20:10

Solve the quadratic $x^2+10x+24=0$ and see what happens between the two roots.

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$x^2+10x>−24$ iff $x^2-10x+25 > 1$ iff $(x-5)^2 > 1$. (Note: "iff" means "if and only if".)

Because of the magical property of $1$ being its own square root (funny how this happens, eh?) this is true iff $|x-5| > 1$ which is true iff $x<4$ or $x > 6$.

By completing the square like this, you are implicitly solving the equation. If you start with the inequality $x^2-2bx > c$ ($b=5$ and $c=-24$ in your case) this becomes $(x-b)^2 > c + b^2$. If $c + b^2 < 0$, all $x$ satisfy this; otherwise this can be rewritten as $|x-b| > \sqrt{c+b^2}$ for which the solutions are $x < b-\sqrt{c+b^2}$ and $x > b+\sqrt{c+b^2}$

If you start with the inequality $x^2-2bx < c$ this becomes $(x-b)^2 < c + b^2$. If $c + b^2 < 0$, no $x$ satisfies this; otherwise this can be rewritten as $|x-b| < \sqrt{c+b^2}$ for which the solutions are $b-\sqrt{c+b^2} < x < b+\sqrt{c+b^2}$

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