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So I have an equation of a graph, $x^4=x^2-\frac{x} {2}+\frac{1}{16}$. I need to find the number of real roots of said equation using an algebraic method. The thing is, I have no clue how to do this as there is no way I know to get $f(x)$ here. I know I can use Sturm's theorem to find the number of real roots but obviously I can't use it without $f(x)$. How can I find the number of real roots of this equation?

BlueCP
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    Not following. Equations have solutions, not roots. Presumably you meant "find the roots of the quartic $x^4-x^2+\frac 12x-\frac 1{16}$", in which case you have your $f(x)$. – lulu Sep 09 '18 at 14:49
  • See my comment below. – BlueCP Sep 09 '18 at 15:27

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To find f(x), simply subtract $x^4$ from both sides. On the other hand, to get it in standard form let $$x^4-x^2+\frac{x}{2}-\frac{1}{16}=0$$ I think the equation can also be expressed be better for solving. Let me know if it works.

$$\frac{1}{16}(2x-1)^2(4x^2 +4x-1)=0$$

poetasis
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  • Perhaps I've explained myself badly. The equation you've given that's equal to 0 is the actual equation of the graph, not the values of $x$ at $y=0$. Just as to find f(x) of the graph $x+y=0$, you would rearrange to get $y=-x$, which is equivalent to $f(x)=-x$. In that equation 0 is not the substitution of $f(x)$. Same with my equation. Except in my equation, there is no $y$. And consequently, as far as I can see, no way to get $f(x)$. – BlueCP Sep 09 '18 at 15:23
  • @RomanJanik: with the factored form you know that for $x$ to be a root one of the factors has to be $0$, so you can solve for that for each factor. You should be able to see that $x=\frac 12$ is a double root and that the other factor contributes two roots. – Ross Millikan Sep 09 '18 at 15:38
  • If you look at the $(2x-1)^2$ factor in the alternate form, your can see that $\frac{1}{2}$ is two of the roots. You can get the other two roots from $(4x^2 +4x-1)$ via the quadratic formula. – poetasis Sep 09 '18 at 15:47
  • Okay, I see it now. Thank you all for your help. – BlueCP Sep 09 '18 at 15:53
  • Can someone tell me why this answer was downvoted? It doesn't have to be 'that' person. I'm just wondering what I did wrong. Also, @Roman Janik do you think my answer is correct or no? – poetasis Sep 11 '18 at 18:55
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    @poetasis your answer is correct, forgot to accept it though. – BlueCP Sep 18 '18 at 21:26