Solving an equation

[Luciano Campos]

Hy everyone, I need help to solve this equation for alpha:

[math]\alpha = 2 \sqrt{\alpha\beta}+\frac{1}{2}[/math]
I know the solution is

[math]\sqrt{\alpha} = \sqrt{\beta} + \sqrt{\beta +1/2}[/math]
But I just don't know how to get it.

I'd really appreciate if anyone could help.

Best,

 

[Dr.Peterson]

Hy everyone, I need help to solve this equation for alpha:

[math]\alpha = 2 \sqrt{\alpha\beta}+\frac{1}{2}[/math]
I know the solution is

[math]\sqrt{\alpha} = \sqrt{\beta} + \sqrt{\beta +1/2}[/math]

Technically that's not the solution, since you were told to solve for [imath]\alpha[/imath], not for [imath]\sqrt\alpha[/imath].

But I would arrange the equation as a quadratic in [imath]\sqrt\alpha[/imath] and use the quadratic formula. Then you need to decide whether both solutions are meaningful.

 

[Luciano Campos]

Thanks, Dr Peterson. However, I still cannot find the solution. If you could give me more details on the math, that'll be great.

The final solution needs to be:

[math]\alpha=\left(\sqrt{\beta}+\sqrt{\beta+1/2}\right)^2[/math]
Best,

 

[JeffM]

The first thing I would is to clear fractions, isolate the radical, and square.

[math]\alpha = 2\sqrt{\alpha \beta } + \dfrac{1}{2} \implies 2 \alpha = 4 \sqrt {\alpha * \beta} + 1 \implies \\ 2 \alpha - 1 = 4 \sqrt { \alpha \beta } \implies 4 \alpha^2 - 4 \alpha + 1 = 16 \alpha \beta. [/math]
Now you have a quadratic so what next?

 

[Luciano Campos]

Thanks Jeff, that looks nice. And then what?

 

[blamocur]

Thanks Jeff, that looks nice. And then what?

And then you solve the quadratic equation for [imath]\alpha[/imath]. Can you search the web for "quadratic equation"?

 

[Dr.Peterson]

Thanks, Dr Peterson. However, I still cannot find the solution. If you could give me more details on the math, that'll be great.

The final solution needs to be:

[math]\alpha=\left(\sqrt{\beta}+\sqrt{\beta+1/2}\right)^2[/math]
Best,

What I expected you to do is to make some attempt and show your work, so we could see where you need help:

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In any case, please note that my suggestion is different from @JeffM's, and takes a lot less work; if you had a further question about it, I was going to suggest a substitution, like [imath]x=\sqrt{\alpha}[/imath], to make it look more familiar. But since you were able to do the formatting so well, I figured you might not need that immediately.

 

[Steven G]

As Dr Peterson stated early on, you are already given a quadratic equation. So put all your energy into writing that equation in standard form for a quadratic and then just use the quadratic formula.

 

[Luciano Campos]

That's great! Thank you all!

 

[JeffM]

You have been given two ways to solve this problem.

One is to “see“ that it can be transformed:

[math]\alpha = 2\sqrt{ \alpha \beta } + \dfrac{1}{2} \implies \\ (\sqrt { \alpha })^2 = \sqrt{4 \beta} * \sqrt{ \alpha } + \dfrac{1}{2}[/math]
which is a simple quadratic in [imath]\sqrt { \alpha }[/imath].

The other way is to eliminate the radical and end up with a somewhat messy quadratic in [imath]\alpha[/imath].

In either case, you end up with a quadratic.

Solve it, and you are 80% of the way to completion.

 

[Luciano Campos]

Thanks Jeff! Appreciate your help.