It's going to be kind of crazy, but you need to use Pythagorean's Theorem for this. That will look like this: [tex](2 \sqrt{x} +1)^2=( \sqrt{x} +1)^2+(2 \sqrt{x} )^2[/tex]. FOIL out the left side to get [tex]4x+4 \sqrt{x} +1[/tex]. FOIL out the first of the 2 expressions on the right to get [tex]x+2 \sqrt{x} +1[/tex], and the second of the 2 to get 4x. Our equation now looks like this: [tex]4x+4 \sqrt{x} +1=x+2 \sqrt{x} +1+4x[/tex]. Combine like terms to get an equation that still has square roots in it that we have to deal with: [tex]4x-x-4x=-2 \sqrt{x} [/tex] and [tex]x=2 \sqrt{x} [/tex]. We will square both sides to get rid of the square root sign. [tex] x^{2} =2x[/tex]. This is a polynomial now that can be factored to solve for x. Bring the 2x over by subtraction and set the polynomial equal to 0. [tex] x^{2} -2x=0[/tex]. Factor out an x, leaving us with x(x-2)=0. That means that x = 0 or x - 2 = 0 and x = 2. Of course if we are solving for the length of a side we know it can't have a side length of 0, so it must have a side length that is a multiple of 2. x = 2
