SOLUTION: I really need your help with this problem "The length of a rectangle is 2 cm less than twice the width. The area of the rectangle is 180 cm^2. Find the length and width of the tria

Question 31040: I really need your help with this problem "The length of a rectangle is 2 cm less than twice the width. The area of the rectangle is 180 cm^2. Find the length and width of the triangle"
-thank you!
Answer by mbarugel(146) (Show Source): You can put this solution on YOUR website!
Hello!
There appears to be a typo in your question. You wrote "Find the length and width of the triangle", but it looks like you meant "rectangle" instead of triangle. I'll asume you meant that.
As in all word problems, the main difficulty is in "translating" the given data into algebraic terms.
Let's call X to the rectangle length, and Y to its width. We're told that:
- "The length of a rectangle is 2 cm less than twice the width". Twice the width is 2Y. Two less than twice the width is 2Y - 2. So we have:
- "The area of the rectangle is 180 cm^2". The formula for the area of a rectangle is length times width (XY). So we get the equation
What we have here is a system of equations:

Replacing the 1st equation into the second one, we get:

So we're left with a quadratic equation, which can be solved with the standard procedure:

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=361 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 10, -9. Here's your graph:

The solutions are 10 and -9; but -9 clearly doesn't make sense (width can't be negative). So we conclude that the width of this rectangle is 10. Finally, we knew that . Plugging Y = 10, we get:

So the length is 18 cm and the width is 10 cm

I hope this helps!
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