Question

The height of a triangle is 4 in. greater than twice its base. The area of the triangle is no more than 168 in.2. Which inequality can be used to find the possible lengths, x, of the base of the triangle?

Answer 1

so you can writing h=4+2b
and you know area of a triangle is A=h*b/2 so substituted in place of h this value
h=4+2b will get

A=(4+2b)*b/2 = (4b+2b^2)/2 = 2(2b+b^2)/2 simplify it by 2 and will get

A=2b+b^2
- and you know that A<168 in^2 so than you can writing

2b+b^2 < 168

b^2 +2b - 168 < 0

b^2 +2b -168 =0
D=4+672=676
b_1,2= (-2+/- sqrt(676))/2 = (-2+/- 26)/2 = 12 and -24



- result from this table that among values -24 and 12 of b , but without values -24 and 12 ,area A will be least than 168 in^2

hope so much that this will be understandably sure easy