Radical expressions?

Question

The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 5cm, what are the length and width of the rectangle?
I came up with 5^2=1+x^2+x^2 so far.

Answer 1

width = w
lenght = w+1

length x lenght + width x width = diag x diag
=> (w+1)(w+1) + w^2 = 25
=> w^2+2w+1+w^2 = 25
=> 2(w^2)+2w-24=0
=> w^2+w-12=0
=> (w-3)(w+4)=0
=> w=3 or w=-4
here: w=3 since w>0

Answer 2

i split it into a triangle so i have the hyp. = 5 and one side equal to x and the other equal to x+1. the pathagarium therom (spelling?) is a^2+b^2=c^2, c being the long side or 5 in our case. so i squared that. and got 25, then i figured out two numbers, one that is one greater to the other that when squared and added together equal 25, and those numbers are 3 and 4, so the width of the rectangle is 3 and the length 4

Answer 3

l = w + 1

l^2 + w^2 = d^2

(w + 1)^2 + w^2 = 5^2
((w + 1)(w + 1)) + w^2 = 25
(w^2 + w + w + 1) + w^2 = 25
w^2 + 2w + 1 + w^2 = 25
2w^2 + 2w - 24 = 0
2(w^2 + w - 12) = 0
2(w + 4)(w - 3)

w = -4 or 3
w = 3

l = 3 + 1 = 4

Length = 4cm
Width = 3cm

Answer 4

l - 1 = w

5^2 = d^2 = w^2 + l^2 = (l-1)^2 + l^2 = l^2 - 2*l +1 + l^2 = 2*l^2 - 2*l +1

so 2*l^2 - 2*l -24 = 0, or
l^2 - l -12 =0

Plugging into the quadratic equation gives l = -3, 4. Obviously discard the negative solution, so l=4, w=3.

Answer 5

You have a very good idea what to do, but you forgot to put parenthesis around the 1+x on the right side.

So your equation should be:
5^2 = (1+x)^2 + x^2

Than you do all your powers, set the equation equal to zero and solve for x.

Nice job trying the problem on your own! :)

Answer 6

5^2=(x+1)^2 +x^2
25 = x^2 +2x+1 +x^2
2x^2 +2x -24=0
x^2 +x -12=0
(x+4)(x-3) =0
x=3 or x=-4 (not possible)
Length=4
width =3