SOLUTION: Good evening tutor can you tell me if I have this problem correct? THe length of the retangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm^2, fin

Question 105084: Good evening tutor can you tell me if I have this problem correct?
THe length of the retangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm^2, find the width of the rectangle to the nearest hundredth.
Could you tell me if the answer is 3.41 cm? Thanks for your help.
Found 2 solutions by bucky, Earlsdon:
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Your answer is correct.
.
The method is:
.
Length = 5*w + 2
.
Width = w
.
Area = L * w = 65 sq cm
.
Multiply L * w and set it equal to the area to get the equation:
.
(L * w) = 65
.
Substitute 5*w + 2 for L to change the equation to:
.
(5w + 2)*w = 65
.
Multiply out the left side:
.

.
Subtract 65 from both sides:
.

.
Apply the quadratic formula to solve for w. When you do you get:
.

.
The term in the radical simplifies to:
.

.
Substituting this for the radical results in:
.

.
If you let the +- sign be negative you get a negative answer for w. That doesn't make sense
so we just use the + sign to get:
.
and this rounds to 3.41 cm.
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This agrees with your answer, so you are correct.
.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let L = the length and W = the width.
From the problem description, you can write:
L = 5W+2 "The length (L) of the rectangle is (=) 2 more (+2) than 5 times its width (5W)"
A = 65 sq.cm. "The area of the rectangle is 65 sq.cm."
The formula for the area of a rectangle is:
A = L*W Make the appropriate substitutions.
Simplify and solve for W.
Subtract 65 from both sides.
Use the quadratic formula to solve for W.

Discard the negative solution.

The width is 3.41 cm
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