Answer:
5
Step-by-step explanation:
→ Let x = √20 + √20 .....
x
→ Square both sides
x² = 20 + √20 +√20 + √20
→ Replace √20 +√20 + √20 with x
x² = 20 + x
→ Move everything to the left hand side
x² - 20 - x = 0
→ Factorise
( x + 4 ) ( x - 5 )
→ Solve
x = -4 , 5
→ Discard negative result
x = 5
Note that
[tex]\left(\sqrt{x+a} + b\right)^2 = \left(\sqrt{x+a}\right)^2 + 2b \sqrt{x+a} + b^2 = x + 2b \sqrt{x+a} + a+b^2[/tex]
Taking square roots on both sides, we have
[tex]\sqrt{x+a} + b = \sqrt{x + 2b \sqrt{x+a} + a+b^2}[/tex]
Now suppose [tex]a+b^2=0[/tex] and [tex]2b=1[/tex]. Then [tex]b=\frac12[/tex] and [tex]a=-\frac14[/tex], so we can simply write
[tex]\sqrt{x-\dfrac14} + \dfrac12 = \sqrt{x + \sqrt{x - \dfrac14}}[/tex]
This means
[tex]\sqrt{x-\dfrac14} = -\dfrac12 + \sqrt{x + \sqrt{x - \dfrac14}}[/tex]
and substituting this into the root expression on the right gives
[tex]\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x + \sqrt{x - \dfrac14}}}[/tex]
and again,
[tex]\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac14}}}[/tex]
and again,
[tex]\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac14}}}}[/tex]
and so on. After infinitely many iterations, the right side will converge to an infinitely nested root,
[tex]\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{\cdots}}}}}[/tex]
We get the target expression when [tex]x=\frac{41}2[/tex], since [tex]\frac{41}2-\frac12=\frac{40}2=20[/tex], which indicates the infinitely nested root converges to
[tex]\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{\cdots}}}} = \sqrt{\dfrac{41}2 - \dfrac14} + \dfrac12 \\\\ ~~~~~~~~ = \sqrt{\dfrac{81}4} + \dfrac12 \\\\ ~~~~~~~~ = \dfrac92 + \dfrac12 = \dfrac{10}2 = \boxed{5}[/tex]