Question

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

Answer 1

Note that

`\left(√(x+a) + b\right)^2 = \left(√(x+a)\right)^2 + 2b √(x+a) + b^2 = x + 2b √(x+a) + a+b^2`

Taking square roots on both sides, we have

`√(x+a) + b = \sqrt{x + 2b √(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

`√(x-\frac14) + \frac12 = \sqrt{x + √(x - \frac14)}`

This means

`√(x-\frac14) = -\frac12 + \sqrt{x + √(x - \frac14)}`

and substituting this into the root expression on the right gives

`√(x - \frac14) + \frac12 = \sqrt{x - \frac12 + \sqrt{x + √(x - \frac14)}}`

and again,

`√(x - \frac14) + \frac12 = \sqrt{x - \frac12 + \sqrt{x - \frac12 + √(x - \frac14)}}`

and again,

`√(x - \frac14) + \frac12 = \sqrt{x - \frac12 + \sqrt{x - \frac12 + \sqrt{x - \frac12 + √(x - \frac14)}}}`

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

`√(x - \frac14) + \frac12 = \sqrt{x - \frac12 + \sqrt{x - \frac12 + \sqrt{x - \frac12 + \sqrt{x - \frac12 + √(\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+√(\cdots)}}} = \sqrt{\frac{41}2 - \frac14} + \frac12 \\\\ ~~~~~~~~ = \sqrt{\frac{81}4} + \frac12 \\\\ ~~~~~~~~ = \frac92 + \frac12 = \frac{10}2 = \boxed{5}`

Answer 2

5

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