SOLUTION: Hello, my question is this: 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 r

Question 91120: Hello, my question is this: 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?
I started with a sketch of the problem and came up with the equation
x^2+(x+1)^2=4^2
x^2+x^2+2x+1=16
2x^2+2x-15=0
Now I don't know where to go from here Help me please!
Found 2 solutions by vertciel, jim_thompson5910:
Answer by vertciel(183) (Show Source): You can put this solution on YOUR website!
I am not sure if I was the person who helped you with this problem earlier; this problem seems familiar.
2x^2 + 2x - 15 = 0
You can't factor to solve for the x roots so you'll have the use the quadratic formula.

There you are.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=2, b=2, and c=-15

Square 2 to get 4

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

So these expressions approximate to

or

So our solutions are:
or

However, since a negative length doesn't make sense, our only solution is

So the width is about 2.284 cm and the length is 3.284 cm
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