Solving an equation, help needed.

**skalik** · July 21st 2011, 10:44 PM

It seems pretty simple, but it's keeping me busy for some time:

x^2-a-√(x-a)=0

Find roots of X in a dependency of A.

Thanks a lot.

**pickslides** · July 21st 2011, 10:47 PM

Here's a kicker

$\displaystyle x^2-a-\sqrt{x-a}=0$

$\displaystyle x^2-a= \sqrt{x-a}$

Now square both sides

$\displaystyle (x^2-a)^2= x-a$

Your turn, expand the left hand side, what do you get?

**skalik** · July 22nd 2011, 12:01 AM

Been there:

x^4-2x^2a+a^2=x-a

It's not that simple as it seems.

**pickslides** · July 22nd 2011, 12:13 AM

True, I jumped pretty quick on that, if the RHS side was independant of x you could sub $u = x^2$ and make it a quadratic.

Other than that, have you seen the quartic formula?

Quartic function - Wikipedia, the free encyclopedia

**skalik** · July 22nd 2011, 12:26 AM

I'm not sure Quartic is in place here, although i've gave it a shot yesterday. Maybe there is a need in a more complexed substitution? or maybe breaking the x^2-a into something like: (x-√a)(x+√a)?

**Duke** · July 22nd 2011, 01:24 AM

x^2-a-√(x-a)=0 - Wolfram|Alpha. The solutions are ridiculos so make sure you have correct equation.

**skalik** · July 22nd 2011, 01:35 AM

Apparently Quartic was the solution, "x^4-2x^2a+a^2=x-a" what so called a depressed quartic equation and the solution is the exact Ferrari method shown in wiki's link from above... THANKS a lot.

**chisigma** · July 22nd 2011, 04:52 AM

In cases like this is better to spend a minute in thinking before to connect to Wiki or Wolfram Alpha...

The relation...

$x^{2}-a - \sqrt{x-a}=0$ (1)

... define implicity an $x= f(a)$ or equivalently an $a=f^{-1}(x)$ . Clearly finding $f(a)$ requires the [impratical...] solution of a quartic and finding $f^{-1}(x)$ requires the [easy enough...] solution of a quadratic, so that the second alternative is preferable. Ther result is...

$a= \frac{2\ x^{2} -1 \pm \sqrt{1+4\ x -4\ x^{2}}}{2}$ (2)

Kind regards

$\chi$ $\sigma$

**Duke** · July 22nd 2011, 05:28 AM

In cases like this, it is better spend a miunte READING THE QUESTION before answering. The task was to find x=f(a).

**chisigma** · July 22nd 2011, 06:04 AM

In cases like this it is better a little of knowledge of elementary theory!

...

The first branch of the function...

$a=f^{-1}(x)= \frac{2\ x^{2}-1 + \sqrt{1 + 4\ x -4\ x^{2}}}{2}$ (1)

... has the properies that $a(0)=0$ so that exists a neighborhood of $x=0$ where its inverse $x(a)$ exists and can also be effectively computed. The same is for the other branch...

$a=f^{-1}(x)=\frac{2\ x^{2}-1 - \sqrt{1 + 4\ x -4\ x^{2}}}{2}$ (2)

... for which is $a(0)=-1$ ...

Kind regards

$\chi$ $\sigma$

**Duke** · July 22nd 2011, 07:27 AM

Can you please explain what you are doing there?

**mr fantastic** · July 22nd 2011, 02:16 PM

Originally Posted by skalik

It seems pretty simple, but it's keeping me busy for some time:

x^2-a-√(x-a)=0

Find roots of X in a dependency of A.

Thanks a lot.

Is the the whole question? Or has it come from another question?

**skalik** · July 22nd 2011, 10:20 PM

It was just a stand alone equation, given in a spirit of a riddle between colleagues at work.
Thanks for the effort.

**skalik** · July 22nd 2011, 10:33 PM

Originally Posted by chisigma

In cases like this is better to spend a minute in thinking before to connect to Wiki or Wolfram Alpha...

The relation...

$x^{2}-a - \sqrt{x-a}=0$ (1)

... define implicity an $x= f(a)$ or equivalently an $a=f^{-1}(x)$ . Clearly finding $f(a)$ requires the [impratical...] solution of a quartic and finding $f^{-1}(x)$ requires the [easy enough...] solution of a quadratic, so that the second alternative is preferable. Ther result is...

$a= \frac{2\ x^{2} -1 \pm \sqrt{1+4\ x -4\ x^{2}}}{2}$ (2)

Kind regards

$\chi$ $\sigma$

Wasn't satisfied from this "Quartic" solution myself, had no need to break the equation down as mentioned in Wiki (i've got the main idea). After couple of minutes was on my way to perform the same, didn't finished but the idea above was clear enough, shame i saw your post just now

Thanks.

**mr fantastic** · July 23rd 2011, 02:45 AM

Originally Posted by skalik

It was just a stand alone equation, given in a spirit of a riddle between colleagues at work.
Thanks for the effort.

Perhaps you should ask your colleague to tell you his/her proposed solution and then you can share it here.

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