How do you solve ${(q + 4)}^{2} > 10 q + 31$ $(q+4)^2>10q+31$ using a sign chart?

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How do you solve ${(q + 4)}^{2} > 10 q + 31$ $(q+4)^2>10q+31$ using a sign chart?

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Answer 1 · 2017-08-07T18:02:02

The solution is $q \in (- \infty, - 3) \cup (5, + \infty)$ $q in (-oo, -3) uu (5, +oo)$

Explanation:

Let's rearrange and factorise the inequality

${(q + 4)}^{2} > 10 q + 31$ $(q+4)^2>10q+31$

$q^{2} + 8 q + 16 > 10 q + 31$ $q^2+8q+16>10q+31$

$q^{2} + 8 q + 16 - 10 q - 31 > 0$ $q^2+8q+16-10q-31>0$

$q^{2} - 2 q - 15 > 0$ $q^2-2q-15>0$

$(q + 3) (q - 5) > 0$ $(q+3)(q-5)>0$

Let $f (q) = (q + 3) (q - 5)$ $f(q)=(q+3)(q-5)$

We can build the sign chart

$a a a a$ $color(white)(aaaa)$ $q$ $q$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 3$ $-3$ $a a a a$ $color(white)(aaaa)$ $5$ $5$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $q + 3$ $q+3$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $q - 5$ $q-5$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (q)$ $f(q)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (q) > 0$ $f(q)>0$ when $q \in (- \infty, - 3) \cup (5, + \infty)$ $q in (-oo, -3) uu (5, +oo)$

graph{(x+4)^2-10x-31 [-27.53, 23.8, -17.2, 8.47]}

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How do you solve ${(q + 4)}^{2} > 10 q + 31$ $(q+4)^2>10q+31$ using a sign chart?

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Explanation:

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How do you solve (q+4)^2>10q+31(q+4)2>10q+31(q+4)^2>10q+31 using a sign chart?

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Explanation:

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How do you solve ${(q + 4)}^{2} > 10 q + 31$ $(q+4)^2>10q+31$ using a sign chart?