Crazy hard NEED HELP

[sage]

Solve for x in terms of a, for all real numbers.
x=radical[a-radical(a+x)]

 

[soroban]

Hello, sage!

An ugly problem . . . Where did it come from?

\(\displaystyle \text{Solve for }x\text{ in terms of }a\text{, for all real numbers: }\:x \:=\:\sqrt{a-\sqrt{a+x}}\)

Square both sides: .\(\displaystyle x^2 \:=\:a - \sqrt{a-x} \quad\Rightarrow\quad a - x^2 \:=\:\sqrt{a-x}\)

Square both sides: .\(\displaystyle a^2 - 2ax^2 + x^4 \:=\:a-x \quad\Rightarrow\quad x^4 - 2ax^2 + x + a^2-a \:=\:0\)

Now just solve this quartic equation . . .

 

[Deleted member 4993]

I hate to admit brilliance of hockey-fan - but it is......

 

[lookagain]

\(\displaystyle x = \sqrt{a - \sqrt{a + x}}\)

x = [-1 +- SQRT(4a + 1)] / 2
or
x = [1 +- SQRT(4a - 3)] / 2

There are restrictions on what x and a can equal.


For instance, to keep real values for x here, it is necessary that \(\displaystyle a \ge 0,\) but maybe not sufficient.
Why?

Suppose the radicand (a + x) is nonnegative. Then, it would be \(\displaystyle \sqrt{(neg. \ \#) - (at \ least \ 0)},\)
which is not real.


And because x is equal to the square root of a quantity (the basic idea of the given equation),
then x has to be nonnegative.


Additionally, the solution x = [-1 - SQRT(4a + 1)]/2 would have to be thrown out,
because it is necessarily negative.


Let's see. More restrictions may exist independently for x, for a, and/or for x in terms of a.


Edit:


And the radicand \(\displaystyle a - \sqrt{a + x}\) must be greater than or equal to 0.


Solving that gives \(\displaystyle x \le a^2 - a,\) but because \(\displaystyle x \ge 0 \ \ and \ \ a \ge 0, \)

then \(\displaystyle a = 0 \ \ or \ \ a \ge 1.\)