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

1 Answer
Aug 7, 2017

Answer:

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]}