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

Please help.. (This is meant to be infinite, so I can't use a calculator)-example-1
User Clyde Lobo
by
8.2k points

2 Answers

5 votes

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}

User Florian Straub
by
7.5k points
6 votes

Answer:

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

User AlexMelw
by
7.4k points

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