Find the set that contains the real solution to an equation

[diredragon]

What did you get after simplifying, but before squaring?

After the initial setting i simplified to
$-(x+ \sqrt{3})(\sqrt{x} - \sqrt{x+ \sqrt{3}}) + (x- \sqrt{3})(\sqrt{x} + \sqrt{x- \sqrt{3}}) = \sqrt{3x}$

Did you get after rationalizing $$(x+\sqrt{3})^{3/2}+(x-\sqrt{3})^{3/2}=\sqrt{27x}$$?
After squaring once and simplifying, you get an upper limit for a real root.

I don't know how you got to this expression by rationalising.

The initial substitution leads to:

$\frac{y+1}{\sqrt{y} + \sqrt{y+1}} + \frac{y-1}{\sqrt{y} - \sqrt{y-1}} = \sqrt{y}$

Then, multiplying through gives:

$y \sqrt{y} - \sqrt{y -1} - \sqrt{y+1} = -\sqrt{y} \sqrt{y+1} \sqrt{y-1}$

And then it's easy enough! The upper limit involves $\sqrt{5}$. I hope that's not giving too much away.

I don't see how is the substitutes initial expression equivalent to this one. How do you know it is?

 

[Samy_A]

After the initial setting i simplified to
$-(x+ \sqrt{3})(\sqrt{x} - \sqrt{x+ \sqrt{3}}) + (x- \sqrt{3})(\sqrt{x} + \sqrt{x- \sqrt{3}}) = \sqrt{3x}$

I don't know how you got to this expression by rationalising.

In your expression, group the terms in $\sqrt{x+ \sqrt{3}}$ and $\sqrt{x- \sqrt{3}}$, and you will get the same expression as ehild and I got.
For example, you will get $(x+ \sqrt{3})( \sqrt{x+ \sqrt{3}})$, and that is nothing more than $(x+\sqrt{3})^{3/2}$.

I don't see how is the substitutes initial expression equivalent to this one. How do you know it is?

He multiplied and simplified. A lot of terms happen to cancel out.

 

[diredragon]

In your expression, group the terms in $\sqrt{x+ \sqrt{3}}$ and $\sqrt{x- \sqrt{3}}$, and you will get the same expression as ehild and I got.
For example, you will get $(x+ \sqrt{3})( \sqrt{x+ \sqrt{3}})$, and that is nothing more than $(x+\sqrt{3})^{3/2}$.He multiplied and simplified. A lot of terms happen to cancel out.

I see. I now get $(x- \sqrt{3})^3 + (x + \sqrt{3})^3 + 2(x^2 - 3)^{3/2} = 27x$
I tried the binomial expansion, doesn't get me anywhere, I am left with $2x^3$ term. How to solve the $0$ of this function?

 

[Samy_A]

I see. I now get $(x- \sqrt{3})^3 + (x + \sqrt{3})^3 + 2(x^2 - 3)^{3/2} = 27x$
I tried the binomial expansion, doesn't get me anywhere, I am left with $2x^3$ term. How to solve the $0$ of this function?

So you are trying it the "hard" way, by computing it.
Well, when you do the binomial expansions, you are left with the square root on one side and $9x-2x³$ on the other side. Square this again, do all the expansions, and you will be left with a rather easy equation for x.

 

[PeroK]

Alternatively:

$y \sqrt{y} - \sqrt{y -1} - \sqrt{y+1} = -\sqrt{y} \sqrt{y+1} \sqrt{y-1}$

$y \sqrt{y} - \sqrt{y -1} = \sqrt{y+1} (1 - \sqrt{y} \sqrt{y-1})$

Leads to an easier finish.

 

[diredragon]

I get $x^2= 3$, so i choose a) as my final answer :).

 

[PeroK]

I get $x^2= 3$, so i choose a) as my final answer :).

$x = \sqrt{3}$ is not a solution.

 

[diredragon]

$x = \sqrt{3}$ is not a solution.

I made an algebraic mistake at the end. I now gey $x^2 = 4$ no mistakes i think. So is this the solution?

 

[Samy_A]

I made an algebraic mistake at the end. I now gey $x^2 = 4$ no mistakes i think. So is this the solution?

Yes, I got that too.

But what is the final answer? :)

 

[PeroK]

I made an algebraic mistake at the end. I now gey $x^2 = 4$ no mistakes i think. So is this the solution?

I didn't realize you were all intent on solving the equation!

 

[diredragon]

Yes, I got that too.

But what is the final answer? :)

2 is found in interval $[\sqrt{3}, 2\sqrt{3})$ so that should be correct

 

[Samy_A]

2 is found in interval $[\sqrt{3}, 2\sqrt{3})$ so that should be correct

Indeed.

There are ways to solve this exercise without actually doing the whole computation.

@PeroK ? :)

 

[PeroK]

Indeed.

There are ways to resolve this exercise without actually doing the whole computation.

@PeroK ? :)

I assumed a solution would be hard to find, so I concentrated on identifying the possible interval. But, it would have been a much better problem just to ask you to solve the equation and forget the multiple choice. I think the extra effort was worth it to get a solution!

In fact, if you replace $\sqrt{3}$ by any positive $\alpha$ in the original equation, then you get $x = \frac{2\alpha}{\sqrt{3}}$.