SOLUTION: Help! I don't know how to solve this! one rectangle is (2x + 3) units wide and (7x - 5) units long. A second rectangle has a width of (3x - 9) units and a length of (8x

SOLUTION: Help! I don't know how to solve this! one rectangle is (2x + 3) units wide and (7x - 5) units long. A second rectangle has a width of (3x - 9) units and a length of (8x - 2) uni

Question 363981: Help! I don't know how to solve this!
one rectangle is (2x + 3) units wide and (7x - 5) units long. A second rectangle has a width of (3x - 9) units and a length of (8x - 2) units. Which has the greater area? How much greater?
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
Rectangle 1: (2x+3)(7x-5)=14x^2+11x-15
Rectangle 2: (3x-9)(8x-2)=24x^2-78x+18
.
14x^2+11x-15=0
x=-1.5, x=.714... Between x=-1.5 and x=.714 y is negative.
.
24x^2-78x+18=0
x=.25, x=3 Between x=.25 and x=3 y is negative.
.
24x^2-78x+18-(14x^2+11x-15)=10x^2-89x+33
10x^2-89x+33=0
x=.3877..., x=8.512...
For both rectangles have a negative area.
For Rectangle 1 has a larger area than rectangle 2
For and rectangle 2 has a larger area than rectangle 1.
.
Ed

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