Square root problem :D?

Question

Let x= sqrt(56+sqrt(56+sqrt(56+sqrt(56+sqrt(56+...

Let y= sqrt(56-sqrt(56-sqrt(56-sqrt(56-sqrt(56-...

what is x+y?

Answer 1

The first one converges on 8.

The second one converges on 7.

Thus x + y converges on 8 + 7.

Answer:

x + y = 15

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x = √( 56 + √( 56 + √( 56 + ...

Square both sides, which removes the outside square root:

x² = 56 + √( 56 + √( 56 + ...

Move 56 over:

x² - 56 = √( 56 + √( 56 + ...

The right hand side is the same as the original x:

x² - 56 = x

Find the roots of this equation:

x² - x - 56 = 0

( x - 8 )·( x + 7 ) = 0

x = 8 or x = -7

Since the above continually adds smaller and smaller values to the root of 56, the answer is positive.

x = 8

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The same approach works with y:

y = √( 56 - √( 56 - √( 56 - ...

y = √( 56 - y )

y² = 56 - y

y² - 56 = -y

y² + y - 56 = 0

( y + 8 )·( y - 7 ) = 0

y = -8 or y = 7

The above continually takes away smaller and smaller values from the root of 56, which are never bigger than it, the answer is 7.

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x = 8

y = 7

x + y = 15

Answer 2

Note since x= sqrt(56+sqrt(56+sqrt(56+sqrt(56+sqrt(56+...

then x+y=0  

Answer 3

x = sqrt(56 + x)

Or x^2 = 56 + x

Or x^2 - x - 56 = 0-------------(1)

y = sqrt(56 - y)

Or y^2 = 56 - y

Or y^2 + y - 56 = 0--------------(2)

Hint: From (1), solve for x. From (2), solve for y.

Then find x + y

You will get two values for x. Take the one for which 56 + x > 0

You will get two values for y. Take the one for which 56- y > 0 or y < 56

Note: Contact me if you need further help.

Answer 4

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