Word Problem: rectangle is twice as long as it is wide....

[npaggs]

A rectangle is twice as long as it is wide. If the length and width are both increased by 5cm, the resulting rectangle has an area of 50cm^2. Find the dimensions of the original.

The equation I figured out was (2x + 5)(x + 5) = 50
FOIL:

2x^2 + 10x +25 = 50
2x^2 + 10x - 25 = 0

Now I plugged them into the Quad equation just to make sure I wouldn't make any mistakes and got
-5 ± 5 sqrt 3 / 2

Would that be the correct answer after plugging it in?
And if so, how do I find the original dimensions given these two x values?

 

[stapel]

The equation I figured out was (2x + 5)(x + 5) = 50
FOIL: 2x^2 + 10x +25 = 50

Are you sure?

. . . . .(2x + 5)(x + 5) = (2x)(x) + (5)(x) + (2x)(5) + (5)(5)

I don't think the linear term should be "10x".... :wink:

Eliz.

 

[npaggs]

Aye, thank you.

So my new values of x would be -15 ± 5 sqrt17 / 4
Even worse.

Now how would I use these to figure out the original dimensions? ;s

 

[camgrover]

Ok, so what you got is 2x^2+15x-25

so quadratic equation:
x= (-b+-srt(b^2-4ac))/2a
x= (-(-15)+-srt(15^2-4(2)(-25))/2(2)
x= (15+-srt(345))/4
Now, because you only care about positive values, because a rectangle cant have a negative distance:
x= (15+srt(345))/4
x is about equal to 8.3954
Unless of course I suck..

 

[Denis]

So my new values of x would be -15 ± 5 sqrt17 / 4
Even worse.
Now how would I use these to figure out the original dimensions? ;s

CORRECT; you need brackets! : [-15 + 5sqrt(17)] / 4 : the "-" not required, since value needs to be positive.

You can show original dimensions like this: x by 2x, where x = [-15 + 5sqrt(17)] / 4,
since you were not asked to go any further.