SOLUTION: 6. 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 (the length and width) of rectangle?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 6. 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 (the length and width) of rectangle?       Log On


   

Question 46343This question is from textbook beginning algebra
: 6. 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 (the length and width) of
rectangle?
Is this right so far. I don't understand after this.
x2 +(x+1)2=4sq
x2 +x2 + 1x +1 = 16
2x2 + 1x - = 0 I don't understand how to get the next part? This question is from textbook beginning algebra

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
you have the right idea but made a few Yistakes
x~2+(x+1)~2=4~2 or x~2+x~2+2x+1=16 or 2x~2+2x+1-16=0 or 2x~2+2x-15=0 or
x~2+x-7.5 now using the quadradic equation -- the solutions are x=2.283882181415 & x=-3.283882181415. Thus the solution is the width of the rectangle is 2.28cm & the length is 2.28+1 or 3.28cm.
Proof 2.283882181415*2.283882181415+3.2823882181415*3.283888211415=4*4 or 5.21612+10.78388=16 or 16=16