First observe that if [tex]a+b>0[/tex],
[tex](a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}[/tex]
Let [tex]a=0[/tex] and [tex]b=x[/tex]. It follows that
[tex]a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}[/tex]
Now let [tex]b=1[/tex], so [tex]a^2+a=4x[/tex]. Solving for [tex]a[/tex],
[tex]a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2[/tex]
which means
[tex]a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}[/tex]
Now solve for [tex]x[/tex].
[tex]x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2[/tex]
(note that we assume [tex]2x-1\ge0[/tex])
[tex]4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}[/tex]
(we omit [tex]x=0[/tex] since [tex]2\cdot0-1=-1\ge0[/tex] is not true)
