How do you solve $x^{2} + x > 0$ $x^2+x>0$ using a sign chart?

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How do you solve $x^{2} + x > 0$ $x^2+x>0$ using a sign chart?

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Answer 1 · 2016-12-31T13:52:53

The answer is $x \in] - \infty, - 1 [\cup] 0, + \infty [$ $x in ] -oo,-1 [ uu ] 0, +oo[$

Explanation:

Let's factorise the inequality

$x^{2} + x = x (x + 1) > 0$ $x^2+x=x(x+1)>0$

Let $f (x) = x (x + 1)$ $f(x)=x(x+1)$

We can now establish the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a$ $color(white)(aaaaa)$ $- \infty$ $-oo$ $a a a a a$ $color(white)(aaaaa)$ $- 1$ $-1$ $a a a a a$ $color(white)(aaaaa)$ $0$ $0$ $a a a a a$ $color(white)(aaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 1$ $x+1$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a a a a a a$ $color(white)(aaaaaaaaaa)$ $-$ $-$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $+$ $+$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) > 0$ $f(x)>0$ , when $x \in] - \infty, - 1 [\cup] 0, + \infty [$ $x in ] -oo,-1 [ uu ] 0, +oo[$

graph{x(x+1) [-3.08, 3.078, -1.54, 1.54]}

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How do you solve $x^{2} + x > 0$ $x^2+x>0$ using a sign chart?

1 Answer

Explanation:

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