Question

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

Answer 1

The solution is `q in (-oo, -3) uu (5, +oo)`

Explanation:

Let's rearrange and factorise the inequality

`(q+4)^2>10q+31`

`q^2+8q+16>10q+31`

`q^2+8q+16-10q-31>0`

`q^2-2q-15>0`

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

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

We can build the sign chart

`color(white)(aaaa)` `q` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `5` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `q+3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `q-5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(q)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(q)>0` when `q in (-oo, -3) uu (5, +oo)`

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