SOLUTION: please help me solve this problem the area of a rectangle is 22cm2 the width is w the length is 3w+5 I got as far as: w(3w+5)=22cm2 can you help me solve this?

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Question 23365: please help me solve this problem
the area of a rectangle is 22cm2
the width is w
the length is 3w+5
I got as far as:
w(3w+5)=22cm2
can you help me solve this?
Found 2 solutions by stanbon, rapaljer:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
w(3w+5)=22
3w^2+5w-22=0
Using the ac method to factor, get:
3w^2-6w+11w-22=0
3w(w-2)+11(w-2)=0
(w-2)(3w+11)=0
w=2 or w=-11/3
If w=2 then l=3w+5 = 11
If w=-11/3; it can't because width cannot be negative.
So, the only answer is w=2 and l=11

Cheers,
Stan H.

Answer by rapaljer(4671) (Show Source):
You can put this solution on YOUR website!
w(3w+5)=22

This is going to be quadratic, so multiply out the parentheses, and set it equal to zero:
3w^2 + 5w -22 = 0
(3x ______)(w ______)=0

Factoring this will probably be the hard part of the problem. Here we go:

Two numbers whose product is 22 would probably be 2 * 11. Try the 11 in the first slot and the 2 in the last slot, and the middle term should come out to -6w +11w= 5w:

(3w+11)(w-2) = 0

Two solutions:
3w+11= 0
3w=-11
w=-11/3 Reject! The width cannot be negative!

w-2=0
w=2 cm. This one works!

Now, find the length:
L = 3w+5
L =3*2+5 = 6+5 =11 cm.

Check: Area = LW = 11*2 = 22 cm^2.

R^2 at SCC