Question

Please help.. (This is meant to be infinite, so I can't use a calculator)

Answer 1

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

Answer 2

Note that

$\left(\sqrt{x+a} + b\right)^2 = \left(\sqrt{x+a}\right)^2 + 2b \sqrt{x+a} + b^2 = x + 2b \sqrt{x+a} + a+b^2$

Taking square roots on both sides, we have

$\sqrt{x+a} + b = \sqrt{x + 2b \sqrt{x+a} + a+b^2}$

Now suppose $a+b^2=0$ and $2b=1$. Then $b=\frac12$ and $a=-\frac14$, so we can simply write

$\sqrt{x-\dfrac14} + \dfrac12 = \sqrt{x + \sqrt{x - \dfrac14}}$

This means

$\sqrt{x-\dfrac14} = -\dfrac12 + \sqrt{x + \sqrt{x - \dfrac14}}$

and substituting this into the root expression on the right gives

$\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x + \sqrt{x - \dfrac14}}}$

and again,

$\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac14}}}$

and again,

$\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac14}}}}$

and so on. After infinitely many iterations, the right side will converge to an infinitely nested root,

$\sqrt{x - \dfrac14} + \dfrac12 = \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{x - \dfrac12 + \sqrt{\cdots}}}}}$

We get the target expression when $x=\frac{41}2$, since $\frac{41}2-\frac12=\frac{40}2=20$, which indicates the infinitely nested root converges to

$\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{\cdots}}}} = \sqrt{\dfrac{41}2 - \dfrac14} + \dfrac12 \\\\ ~~~~~~~~ = \sqrt{\dfrac{81}4} + \dfrac12 \\\\ ~~~~~~~~ = \dfrac92 + \dfrac12 = \dfrac{10}2 = \boxed{5}$