Question

Carmen is using the quadratic equation (x + 15)(x) = 100 where x represents the width of a picture frame. Which statement about the solutions x = 5 and x = –20 is true? A. The solutions x = 5 and x = –20 are reasonable. B. The solution x = 5 should be kept, but x = –20 is unreasonable. C. The solution x = –20 should be kept, but x = 5 is unreasonable. D. The solutions x = 5 and x = –20 are unreasonable.

Answer 1

We write the equation in the standard form which is:

ax^2 + bx + c = 0

(x + 15)(x) = 100
x^2 + 15x - 100 = 0
(x + 20) (x-5) = 0

The solutions are

x = -20 and x = 5

Therefore, the correct answer is option B. The solution x = 5 should be kept, but x = –20 is unreasonable.

Answer 2

(x+15) x = 100
x^2 + 15x - 100 = 0
x^2 + 20x - 5x - 100 = 0
x (x + 20) - 5 (x + 20) = 0
(x - 5) (x + 20) = 0

x = 5 or x = -20
 x = 5 is the answer since -20 is not reasonable.