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

Answer:

The actual answer choice is B

Explanation:

Took the quiz on edg and got it right.

Answer 2

The height (h) of a triangle is 4 in. greater than twice its base (x): h = 2x + 4
The area of the triangle is no more than 168 in.2: A ≤ 168 in²

The area of the triangle is A = x*h/2
Hence x*h/2 ≤ 168 in²

Substitute h in the formula for the area:
h = 2x + 4 = 2(x + 2)
x*h/2 ≤ 168 in²

x * 2(x + 2)/2 ≤ 168
x * (x + 2) ≤ 168
x² + 2x ≤ 168
x² + 2x - 168 ≤ 0

Using the formula for quadratic equation:
$x_{1,2} = \frac{-b+/- \sqrt{ b^{2}-4ac } }{2a} = \frac{-2+/- \sqrt{ 2^{2}-4*1*(-168) } }{2*1}= \frac{-2+/- \sqrt{ 4+672 } }{2}= \\ \\ \frac{-2+/- \sqrt{ 676 } }{2}=\frac{-2+/- 26 }{2} \\ \\ x_1 = \frac{-2+26}{2} = \frac{24}{2} =12 \\ \\ x_1 = \frac{-2-26}{2} = \frac{-28}{2} =-14$