1. ## Solve the equation

$\displaystyle \frac 1{\left(1+\sqrt{1+\sqrt{x}}\right)^4}+\frac 1{\left(1-\sqrt{1+\sqrt{x}}\right)^4}+\frac 2{\left(1+\sqrt{1+\sqrt{x}}\right)^3}+\frac 2{\left(1-\sqrt{1+\sqrt{x}}\right)^3}=0$

2. Originally Posted by james_bond
$\displaystyle \frac 1{\left(1+\sqrt{1+\sqrt{x}}\right)^4}+\frac 1{\left(1-\sqrt{1+\sqrt{x}}\right)^4}+\frac 2{\left(1+\sqrt{1+\sqrt{x}}\right)^3}+\frac 2{\left(1-\sqrt{1+\sqrt{x}}\right)^3}=0$
Put $\displaystyle y=1+\sqrt{x}$, then your equation becomes:

$\displaystyle \frac 1{\left(1+\sqrt{y}\right)^4}+\frac 1{\left(1-\sqrt{y}\right)^4}+\frac 2{\left(1+\sqrt{y}\right)^3}+\frac 2{\left(1-\sqrt{y}\right)^3}=0$

$\displaystyle \frac {\left(1-\sqrt{y}\right)^4} {(1-y)^4} + \frac {\left(1+\sqrt{y}\right)^4}{(1-y)^4}+ \frac {2\left(1-\sqrt{y}\right)^3}{(1-y)^3}+ \frac {2\left(1+\sqrt{y}\right)^3}{(1-y)^3}=0$

and since $\displaystyle y \ge 0$:

$\displaystyle \left(1-\sqrt{y}\right)^4 + \left(1+\sqrt{y}\right)^4+ 2\left(1-\sqrt{y}\right)^3(1-y)+ 2\left(1+\sqrt{y}\right)^3(1-y)=0$

Now expand the powers and simplify and see what you have.

RonL

3. I posted a carefully crafted reply to this before it was moved.

What happened to my post?

I started with $\displaystyle a=\sqrt{1+\sqrt{x}}$

4. Originally Posted by CaptainBlack
$\displaystyle \left(1-\sqrt{y}\right)^4 + \left(1+\sqrt{y}\right)^4+ 2\left(1-\sqrt{y}\right)^3(1+y)+ 2\left(1+\sqrt{y}\right)^3(1+y)=0$

Now expand the powers and simplify and see what you have.
I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).

5. Originally Posted by james_bond
I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
Let me check this out for you...

$\displaystyle 14y^2+28y+6=0$

$\displaystyle y^2+2y+\frac{3}{7}=0$

$\displaystyle y=\frac{-2 \pm \sqrt{4-4\frac{3}{7}}}{2}$

$\displaystyle y=\frac{-2 \pm \sqrt{\frac{28-12}{7}}}{2}$

$\displaystyle y=\frac{-2 \pm \sqrt{\frac{16}{7}}}{2}$

$\displaystyle y=\frac{-2 \pm 4\sqrt{\frac{1}{7}}}{2}$

$\displaystyle y=-1 \pm 2\sqrt{\frac{1}{7}}$

A solution of $\displaystyle y=1.6$ is impossible if
$\displaystyle 14y^2+28y+6=0$ is the quadratic equation. Notice that any positive number makes the equation greater than zero, and never less than or equal to zero.

6. Originally Posted by colby2152

A solution of $\displaystyle y=1.6$ is impossible if
$\displaystyle 14y^2+28y+6=0$ is the quadratic equation. Notice that any positive number makes the equation greater than zero, and never less than or equal to zero.
Yes but why would $\displaystyle y$ be $\displaystyle 1.6$
Both roots are negative. Ohh so $\displaystyle \sqrt{x}$ would be negative - contradiction - no solutions.

7. Originally Posted by james_bond
Yes but why would $\displaystyle y$ be $\displaystyle 1.6$
Both roots are negative. Ohh so $\displaystyle \sqrt{x}$ would be negative - contradiction - no solutions.
The discriminant would not be negative. I used the quadratic formula in my solution. There are two solutions, and both are negative real numbers. (SIDE NOTE: I suppose this is a contradiction since to be negative implies real?)

8. Originally Posted by james_bond
I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
Now I think you should get: $\displaystyle 10y^2-20y-6=0$

(which if I've got this right gives a root of $\displaystyle \approx 1.6$, which agrees with
the graphical sdolution)

(I must admit that I used "a tutors" substitution rather than my own as
it is easier to simplify - and then convert back as if I had made my substitution)

RonL

9. Originally Posted by james_bond
I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
Sorry, but note the deliberate (or was it just a transcription) error, now
corrected.

RonL

10. Originally Posted by a tutor
I posted a carefully crafted reply to this before it was moved.

What happened to my post?

I started with $\displaystyle a=\sqrt{1+\sqrt{x}}$
Oppsss.. that will be me. Over enthusiatic moderating (we had a glut of
multiple posts and other moderation issues yesterday)

RonL