Answer:
5
Explanation:
Pythagoras : a^2 + b^2 ≈ c^2
so, x^2 + (2x+2)^2 = (2x+3)^2
=> x^2 + 4x^2 + 8x + 4 = 4x^2 + 12x + 9
=> x^2 = 4x + 5
=> x^2 - 4x - 5 = 0
now solve the quadratic equating. a little bit of creative thinking and trying with the options of a (x-a)×(x+b)=0 approach (multiplying and adding of factors need to give us -5 but only -4x in the result, so we try a=-5 but then b=1 as "correcting" factor), and yes
(x-5)×(x+1) = x^2 - 4x - 5 = 0
this is only true, if at least one of the multiplication factors are 0.
so, x-5=0 or x+1=0. or both, but that is not possible.
x-5=0, when x=5.
x+1=0, when x=-1
-1 as the length of a side does not make any sense.
so, the only valid solution is x=5