SOLUTION: Hi, I have a question. It is a word problem I am to state as a quadratic equation. I can use one of three methods: factoring, completing the square, or the quadratic formula.

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Question 711733: Hi,
I have a question. It is a word problem I am to state as a quadratic equation. I can use one of three methods: factoring, completing the square, or the quadratic formula.
The problem is:
The length of a rectangle is 2 1/2 times its width. Its area is 90 square units. What are its dimensions? (Hint: length x width = area)
Let w = width in units
(2 1/2)w = 5w/2 = length in units
Thanks,
Amber
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +w+ = the width
+2.5w+ = length
+A+=+w%2A%28+2.5w+%29+
+A+=+2.5w%5E2+
---------------
given:
+A+=+90+ square units
+2.5w%5E2+=+90+
----------------
Divide both sides by +2.5+
+w%5E2+=+36+
+w%5E2+-+36+=+0+
You can apply the quadratic formula
+w+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+0+
+c+=+-36+
+w+=+%28-0+%2B-+sqrt%28+0%5E2-4%2A1%2A%28-36%29+%29%29%2F%282%2A1%29+
+w+=+%28+sqrt%28+144+%29%29+%2F+2+
and
+w+=+%28+-sqrt%28+144+%29%29+%2F+2+
------------------------
+w+=+12%2F2+
+w+=+6+
and also
+w+=+-12%2F2+
+w+=+-6+
You can disregard the negative solution since
you don't have negative lengths, so
+w+=+6+
+2.5w+=+15+
The width is 6 units
The length is 15 units