SOLUTION: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm what are the dimensions (length and width) of the rectangle. to start i get

Algebra ->  Pythagorean-theorem -> SOLUTION: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm what are the dimensions (length and width) of the rectangle. to start i get       Log On


   

Question 46607: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm what are the dimensions (length and width) of the rectangle.
to start i get
x^2 + (x+1)^2 = 4^2
x^2 + x^2 + 2x + 1 = 16
2x^2 + 2x - 15 = 0
x^2 + x - 7.5 = 0
This is where I can't go futher and I need to turn it in tonight.
Any help please and thank you.
Answer by adamchapman(301) About Me  (Show Source):
You can put this solution on YOUR website!
You have got to: x%5E2+%2B+x+-+7.5+=+0+
Now use the quadratic solver x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
where a=1, b=1 and c=-7.5
So:
x+=+%287.5+%2B-+sqrt%28+56.25%2B30+%29%29%2F2+
x+=+%287.5+%2B-+sqrt%28+86.25+%29%29%2F2+
x+=+%287.5+%2B-+9.287%29%2F2+
so the two existing solutions are x=8.394 or x=-0.984 (to 3 decimal places)
I hope this helps,
Adam.
P.S. please check out my website, it may be helpful to you:
http://www.geocities.com/quibowibbler