SOLUTION: I just solved a problem on my algebra test, and I am not sure how I did it... In regards to a rectangle with sides X and 2x+5, and area being 168
X(2x+5)=168
Solve for X
W
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-> SOLUTION: I just solved a problem on my algebra test, and I am not sure how I did it... In regards to a rectangle with sides X and 2x+5, and area being 168
X(2x+5)=168
Solve for X
W
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Question 366606: I just solved a problem on my algebra test, and I am not sure how I did it... In regards to a rectangle with sides X and 2x+5, and area being 168
X(2x+5)=168
Solve for X
While trying to work on it, I was factoring 168, and realized the factors 8,21 fit for X and 2x+5... it was an intuitive leap. I would like to know how to actually solve this properly Found 2 solutions by Fombitz, ewatrrr:Answer by Fombitz(32388) (Show Source):
Hi,
x(2x+5)=168
2x^2 + 5x - 168 = 0
Yes, factoring works:
Coming up with the product of -8*21 = 168 and using that
(2x +21)(x -8) = 0
(2x + 21)=0
x = -(21/10) = -10.5
(x -8) = 0
x = 8
The quadratic equations "always" works as well:
x = (-5 + 37)/4 = 8
x = (-5 -37)/4 = -42/4 = -21/2 = -10.5