Bunuel wrote:

tonebeeze wrote:

Bunuel -

So the approach to problems such as this one is to work your way from the most inner root, out toward the main root? Always keeping track of whether the underlying roots are (1) negative (2) are larger than the main root.

Consider another example: For which of the following values of x \(\sqrt{1-\sqrt{4-\sqrt{x}}}\) is NOT defined as a real number?

A. 16

B. 12

C.10

D. 9

E. 4

First see whether \(4-\sqrt{x}\) could be negative for some value of \(x\) so you should test max value of \(x\): \(4-\sqrt{x_{max}}=4-\sqrt{16}=0\). As it's not negative then see whether \(1-\sqrt{4-\sqrt{x}}\) can be negative for some value of \(x\), so you should test min value of \(x\) to maximize \(4-\sqrt{x}\): \(1-\sqrt{4-\sqrt{x_{min}}}=1-\sqrt{4-\sqrt{4}}=1-1.41=-0.41<0\).

Answer E.

Hope it's clear.

Bunuel i got back again to this question

i have a question

why in the example above you kept 1 \(1-\sqrt{4-\sqrt{x_{min}}}=1-\sqrt{4-\sqrt{4}}=1-1.41=-0.41<0\). and how you got this sqr of 4 is 2 so I see two 4s under two radical signs hence \(\sqrt{4-4}\) = 2-2=0

And in the example below you removed 1 btw square root of 1 is 1 ?

We have \(\sqrt{1-\sqrt{2-\sqrt{x}}}\). For \({x=5}\) expression becomes:\(\sqrt{1-\sqrt{2-\sqrt{5}}}\) and \(2-\sqrt{5}<0\), thus square root from this expression is not a real number.

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