# Find a positive REAL value

1. May 23, 2008

### Nusc

1. The problem statement, all variables and given/known data

I need to find a positive REAL value x such that

$$\frac{2-(4+x^2-x\sqrt (4+x^2) \sqrt (\frac{2+x^2+x\sqrt 4+x^2 }{8+2x^2})}{4\sqrt (4+x^2) } == 0$$

x cannot be complex or zero! So far, I don't even think its possible.

Any suggestions?

2. Relevant equations

3. The attempt at a solution

Last edited: May 23, 2008
2. May 23, 2008

### Dick

The formatting on that is not very clear. But why do you need that? Is this the solution to some more basic problem?

3. May 23, 2008

### Nusc

Is that not good enough?

4. May 23, 2008

### Dick

It is now, sort of. Wasn't before. I'm still curious why. If I'm reading it right, then x=0 doesn't work anyway, correct?

Last edited: May 23, 2008
5. May 24, 2008

### Nusc

You're right in this case.

6. May 24, 2008

### Defennder

Well I could reduce it to $$x^3 -3x^2 -4=0$$

Since I don't know how to solve for exact solutions, I used http://www.1728.com/cubic.htm
and it appears that x= 3.3553013976081196 would be an approximate answer.

I noticed there is a double equal sign ('==') in the expression. Does that mean anything?

7. May 24, 2008

### epenguin

Really checked through your transcription? - I mean they usually don't give questions with messy answers. In real life yes, but not so often in textbooks.

Is
$$\frac{2-(4+x^2-x\sqrt (4+x^2) \sqrt (\frac{2+x^2+x\sqrt (4+x^2) }{8+2x^2})}{4\sqrt (4+x^2) } == 0$$

a bit nearer by any chance?

8. May 24, 2008

### Nusc

That's strange, I got mine to cancel out which is what I expected.

So nevermind.

9. May 24, 2008

### Nusc

Actually I was wrong, how did you simplify it to a cubic?

10. May 24, 2008

### Nusc

If you plot that function in mathematica, from say 0-5 you can find values of x that would yield an approximate answer to 0.

11. May 24, 2008

### DavidWhitbeck

The cubic has an exact solution, you can look it up or you can just use Newton's method to find an approximate solution.

12. May 24, 2008

### Nusc

The function I'm trying to approximate is not a cubic, I'm just wondering how the other guy got it. The function of interest is at the top. What's better in this case Bisection or Newton?

13. May 24, 2008

### DavidWhitbeck

Have you tried solving for x? You haven't even simplified your expression! Throw the denominator away, and notice that there are two square root factors on one of the terms that cancel each other (up to a factor of 1/2). I could tell you what they are, but I would prefer to figure out what I'm talking about.

Once you do that it's easy to isolate the square root term and then square both sides of the equation. You will have either a cubic or a quartic.

14. May 24, 2008

### Defennder

I'm not entirely sure I interpreted your expression correctly. For one thing, is it supposed to be $\sqrt{4+x^2}$ or $\sqrt{4} + x^2$? It appears it should be the former. In that case, it's best written as

$$\frac{2-(4+x^2-x\sqrt{4+x^2} \sqrt{\frac{2+x^2+x\sqrt{4+x^2} }{2(4+x^2)}}}{4\sqrt{4+x^2}} = 0$$

Again, what does the double equal sign '==' mean?

Assuming my interpretation of your expression is correct, this time it reduces to $2x^4+5x^2+4=0$.

That gives $$x^2=-\frac{5}{4} \pm \frac{\sqrt{7}}{4}i$$

Finding the square roots of these would hence give you 4 solutions. But then again you mentioned that you thought that x cannot be complex. Is that something stated by the question?

15. May 24, 2008

### rootX

Use some numerical method:
Newton doesn't look good here
Bisection ..
or some good one

16. May 24, 2008

### Vid

From a programming prospective a single equal sign is an assignment operator that gives the value of the right side to the left, and a double equal sign is a comparison operator that returns a value of true if the two sides are indeed equal.

17. May 24, 2008

### jimvoit

Ballance the parentheses in the numerator so that the expression is unambiguous.

18. May 24, 2008

### Nusc

Below is my MAPLE procedure but it gives me values that are very large...

restart;
bisection := proc (f1, c, d, TOL, N0)
local f, a, b, p, i, j, k;
f := f1;
p := evalf((1/2)*c+(1/2)*d);
a := c;
b := d;
i := 1;
j := 0;
k := 1;
while j = 0 and k <= N0 do
if evalf(f(a)*f(p)) < 0 then b := evalf(p)
else a := evalf(p) end if;
print(k);
print("Value of P:");
print(p);
if abs(f(p)) < TOL or N0 <= i then printf("Procedure completed successfully. Approximate value of P:"), P;
j := 1 end if;
k := k+1;
i := i+1;
p := evalf((1/2)*a+(1/2)*b)
end do
end proc

What can be done to improve the code?

19. May 24, 2008

### jimvoit

The double equal sign is used in Mathematica to indicate an equation. I plotted the left side of the equation and found it asymtotic to the x axis, which indicates that there is no solution. That's no help with the algebra I know, and I may have transcribed the equation incorrectly..But that's that for what it's worth.

20. May 25, 2008

### Nusc

It's not asymptotic to the x-axis... Try -1000..1000