# Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

#### anemone

##### MHB POTW Director
Staff member
Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

Find the exact value of the solution to the equation
$\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x$

I wanted so much to share with you all this question because it's an easy yet the tricky one, imho.

#### chisigma

##### Well-known member
Re: Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

Find the exact value of the solution to the equation
$\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x$

I wanted so much to share with you all this question because it's an easy yet the tricky one, imho.
If You consider the first order difference equation...

$\displaystyle a_{n+1}= \sqrt{2-(-1)^{n}\ a_{n}}\ ;\ a_{0}=x$ (1)

... Your problem is the search of solutions of (1) with periodicity 4.In order to symplify the procedure You can start searching solutions of periodicity 2 that, of course, are also solutions with periodicity 4. The solutions with periodicity 2 are solutions of the equation...

$\displaystyle \sqrt{2+\sqrt{2-x}}=x$ (2)

... i.e. a non negative solution of the fourth order equation...

$x^{4} -4\ x^{2} + x + 2= (x-1)\ (x+2)\ (x^{2}-x-1)=0$ (3)

... that are $\displaystyle x_{1}=1$ and $\displaystyle x_{2}=\frac{1 + \sqrt{5}}{2}$. A simple test shows that $x_{2}$ is 'good solution'. Other solutions with periodicity 4 are to be found...

Kind regards

$\chi$ $\sigma$

Last edited:
• anemone and sbhatnagar

#### sbhatnagar

##### Active member
Re: Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

$$\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x$$

Write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}=x$$

Again write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}}}}}=x$$

Repeat this procedure till infinity.

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\cdots}}}}}}}}}}}}=x$$

This means

$$\sqrt{2+\sqrt{2-x}}=x$$

Take squares of both sides

$$2+\sqrt{2-x}=x^2 \quad \quad \cdots (1)$$

Multiply both sides by $2-\sqrt{2-x}$:

$$2-\sqrt{2-x}=\frac{2+x}{x^2} \quad \quad \cdots (2)$$

$$4=x^2+ \frac{2+x}{x^2}$$

$$x^4-4x^2+x+2=0$$

Factorize:

$$(x-1)(x-2)(2x-\sqrt{5}-1)(2x+\sqrt{5}-1)=0$$

This gives:

$$x=\frac{1}{2}(\sqrt{5}+1)$$

which is the desired solution.

• anemone and chisigma

#### anemone

##### MHB POTW Director
Staff member
Re: Find the exact value of the solution to the equation √ 2+(√ 2-(√ 2+√( 2-x))))=x.

If You consider the first order difference equation...

$\displaystyle a_{n+1}= \sqrt{2-(-1)^{n}\ a_{n}}\ ;\ a_{0}=x$ (1)

... Your problem is the search of solutions of (1) with periodicity 4.In order to symplify the procedure You can start searching solutions of periodicity 2 that, of course, are also solutions with periodicity 4. The solutions with periodicity 2 are solutions of the equation...

$\displaystyle \sqrt{2+\sqrt{2-x}}=x$ (2)

... i.e. a non negative solution of the fourth order equation...

$x^{4} -4\ x^{2} + x + 2= (x-1)\ (x+2)\ (x^{2}-x-1)=0$ (3)

... that are $\displaystyle x_{1}=1$ and $\displaystyle x_{2}=\frac{1 + \sqrt{5}}{2}$. A simple test shows that $x_{2}$ is 'good solution'. Other solutions with periodicity 4 are to be found...

Kind regards

$\chi$ $\sigma$
Hi chisigma, you're always capable of solving all kinds of maths-related questions using some sophisticated approachs, huh?
You're awesome!

$$\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x$$

Write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}=x$$

Again write this as:

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}}}}}}}}}=x$$

Repeat this procedure till infinity.

$$\sqrt{2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-\cdots}}}}}}}}}}}}=x$$

This means

$$\sqrt{2+\sqrt{2-x}}=x$$
I see it this way:

Since $f^{-1} (f(x))=x$, and we're given $\sqrt {2 +\sqrt {2-\sqrt{2+\sqrt{ 2-x}}}}=x$,

then if I solve the following for x,
$\sqrt{2+\sqrt{ 2-x}}=x$
that means I solve the original equation for x too. (By using your way or substitution method.)