SOLUTION: I'm having difficulties with this word problem. Any help that you can give me would be wonderful! Thank you in advance!! The length of a rectangle is 1cm longer than its width.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I'm having difficulties with this word problem. Any help that you can give me would be wonderful! Thank you in advance!! The length of a rectangle is 1cm longer than its width.      Log On


   

Question 87295: I'm having difficulties with this word problem. Any help that you can give me would be wonderful! Thank you in advance!!
The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length and width) of the rectangle?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
X^2+(X+1)^2=4^2
X^2+X^2+2X+1=16
2X^2+2X+1-16=0
2X^2+2X-15=0
USING THE QUADRATIC EQUATION x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ THUS:
X=(-2+-SQRT[2^2-4*2*-15])/2*2
X=(-2+-SQRT[4+12])/4
X=(-2+-SQRT124)/4
X=(-2+11.14)/4
X=9.14/4
X=2.285 ANSWER.