Answer:
The width of the rectangle is k+4 inches
Step-by-step explanation:
you have [tex]k^{2}+19k+60[/tex] which can be factorized to (k+4)(k+15)
if the length of the rectangle is k-5 that would mean that we can write [tex]k^{2}+19k+60[/tex] as (k-5)n which we know is false, so the only one that applies is k+4
The true statement is:
"The width of the rectangle is k + 4"
Remember that for a rectangle of length L and width W, the area is given by:
A = L*W
So we want to factorize the area equation, which is a quadratic equation, into a product of two terms.
A = k^2 + 19k + 60
The two zeros are given by Bhaskara's formula:
[tex]k = \frac{-19 \pm \sqrt{19^2 - 4*1*60} }{2} \\\\k = \frac{-19 \pm 11 }{2}[/tex]
So we have two zeros, these are:
k = (-19 - 11)/2 = -15
k = (-19 +11)/2 = -4
So we can factorize the area as:
A = (k - (-15))*(k - (-4)) = (k + 15)*(k + 4).
From this, the only statement that can be true is:
"The width of the rectangle is k + 4"
If you want to learn more about quadratic equations, you can read:
https://brainly.com/question/1480401
