# Math Help - Solving an equation, help needed.

1. ## Solving an equation, help needed.

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.

2. ## Re: Solving an equation, help needed.

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?

3. ## Re: Solving an equation, help needed.

Been there:

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

It's not that simple as it seems.

4. ## Re: Solving an equation, help needed.

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

5. ## Re: Solving an equation, help needed.

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)?

6. ## Re: Solving an equation, help needed.

x&#94;2-a-&radic;&#40;x-a&#41;&#61;0 - Wolfram|Alpha. The solutions are ridiculos so make sure you have correct equation.

7. ## Re: Solving an equation, help needed.

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.

8. ## Re: Solving an equation, help needed.

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

$\chi$ $\sigma$

9. ## Re: Solving an equation, help needed.

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

10. ## Re: Solving an equation, help needed.

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

$\chi$ $\sigma$

11. ## Re: Solving an equation, help needed.

Can you please explain what you are doing there?

12. ## Re: Solving an equation, help needed.

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?

13. ## Re: Solving an equation, help needed.

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

14. ## Re: Solving an equation, help needed.

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.

15. ## Re: Solving an equation, help needed.

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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