Problem finding roots of a depressed quartic using the Ferrari Method.

Question

I attempted to follow the Ferrari method for a depressed quartic as described at wikipedia for the function $x^4+5/8x^2-5/8x-51/256$

The roots are $-1/4, 3/4, -1/4-i$, and $-1/4+i$. However, when I follow the method, I get a $1/4, -3/4, 1/4-i$, and $1/4+i$, or each root times $-1$.

I followed the process with other depressed quartics, and sometimes they were correct, and sometimes they were, again, the negative of the correct value.

For the quartic in question, I calculated the resolvent cubic of $x^3+5/8x^2-5/8x-51/256$. Which has one real root, m = 1/8. When I use this m, I get the negative of each of the roots.

I actually went through the derivation, and it seems correct. Does anyone have any insight as to why it is sometimes incorrect?

Edit: Here is the process as laid out on Wikipedia:

$$y^4+\frac{5}{8}y^2-\frac{5}{8}y-\frac{51}{256} = 0$$

Add $\frac{p^2}{4}$ to each side: $(y^2 + 5/16)^2 = 5/8y + 9/64 $

Add $2y^2m + pm+m^2$ to each side: $(y^2 + 5/16 + m)^2 = 2my^2 + 5/8y + m^2 + 5/8m + 9/64$

Set discriminant to zero and solve for m: $8m^3+5m^2+\frac{19}{8}m-\frac{25}{64}=0$

Use real root: $m=\frac{1}{8}$

Final equation is: $\left(y^2+\frac{5}{16}+\frac{1}{8}+\frac{1}{2}y+\frac{5}{8}\right)\left(y^2+\frac{5}{16}+\frac{1}{8}-\frac{1}{2}y-\frac{5}{8}\right)$

or: $\left(y^2+\frac{27}{16}+\frac{1}{2}y\right)\left(y^2- \frac{3}{16}-\frac{1}{2}y\right)= 0$

The roots of the right side are $-\frac{1}{4}$ and $\frac{3}{4}$

I'll simplify the wikipedia final equations for comparison.

Answer 1

I solve equations by Ferrari's method in the following stile.

We need to solve $$256x^4+160x^2-160x-51=0$$ or $$(16x^2+k)^2-((32k-160)x^2+160x+x^2+51)=0.$$

Now, we'll find a value of $k$, for which

$(32k-160)x^2+160x+x^2+51$ is a square.

For which we need $$80^2-(x^2+51)(32k-160)=0$$ or $$k^3-5k^2+51k-455=0$$ and since $7$ is a divisor of $455$, we get a multiplier $k-7$, which gives $$(k-7)(k^2+2k+65)=0$$ or $k=7$ only.

Thus, we need to solve $$(16x^2+7)^2-(64x^2+160x+100)=0$$ or $$(16x^2+7+8x+10)(16x^2+7-8x-10)=0$$ or $$(16x^2+8x+17)(16x^2-8x-3)=0$$ and the rest is smooth.

Answer 2

The roots of

$$y^2-\frac y2-\frac3{16}=0$$ are indeed $\color{red}-\dfrac14$ and $\color{red}+\dfrac34$.

Seems that you need to review your quadratic solver :)