width --- x
length = 2x+7
x(2x+7) = 99
2x^2 + 7x - 99 = 0
Solve for x using your favourite method, reject the negative answer
hint: the solution is a rational number
The length of a rectangle is 7 yd more than double the width, and the area of the rectangle is 99 yd^2. find the dimensions of the rectangle.
3 answers
L= 2w+7 (equation for length width would just be w)
A= 99 yd^2 (what you are given)
A= LW (Equation for area)
99= W(2W+7) (substitution)
99= 2W^2+7W (distributed)
2W^2+7w-99 (factor out)
(2W-11)(W+9)
W= 11/2, -9 (rectangle can't have - dimensions so 11/2)
L=2(11/2)+7 = 18
L=18 and W= 11/2
A= 99 yd^2 (what you are given)
A= LW (Equation for area)
99= W(2W+7) (substitution)
99= 2W^2+7W (distributed)
2W^2+7w-99 (factor out)
(2W-11)(W+9)
W= 11/2, -9 (rectangle can't have - dimensions so 11/2)
L=2(11/2)+7 = 18
L=18 and W= 11/2
Width = W.
Length = 2W + 7
W*(2W+7) = 99.
2w^2 + 7w - 99 = 0,
Use Quadratic formula to find W.
W = (-7 +- sqrt(49 + 792))/4 = 5.5, and -9 yds. Use the positive value.
Length = 2W + 7
W*(2W+7) = 99.
2w^2 + 7w - 99 = 0,
Use Quadratic formula to find W.
W = (-7 +- sqrt(49 + 792))/4 = 5.5, and -9 yds. Use the positive value.
2 answers