Find the set that contains the real solution to an equation

[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 $x = \frac{2\alpha}{\sqrt{3}}$.