x=base
(2x+4)=height
Area of a triangle=(base x height)/2
We can suggest this inequality:
x(2x+4)/2<168
x(2x+4)<336
2x²+4x<336
2x²+4x-336<0
x²+2x-168<0
1) we change the inequality sign of inequality by equality sign.
x²+2x-168=0
x=[-2⁺₋√(4+672)] / 2
x=(-2⁺₋√676)/2
we have two solutions:
x₁=(-2-√676)/2=-14
x₂=(-2⁺√676)/2=12
2) we make intervals with these values.
(-∞, -14 )
(-14,12)
(12, +∞)
3) we check it out if these intervals work
(-∞,-14), For instead; if x=-15⇒ (-15)²+2(-15)-168=27>0. this solution interval don´t work.
(-14,12), For example; if x=1 ⇒(-1)²+2(1)-168=-168<0 this solution interval works.
but the solutions have to be always positive numbers, therefore, the solution is: (0,12).
(12,+∞) , For example, if x=13 ⇒ (13)²+2(13)-168=27>0, this solution interval don´t work.
Answer: the inequality is : x(2x+4)/2<168, and the interval solution is (0,12) for the values of the base.
