# Thread: solving a quatric equation

1. ## solving a quatric equation

I just need to solve a quatric equation like $ax^4 + bx^3 +cx^2+dx+e=0$
I need the 4 roots in terms a,b,c,d,e.

Thanks

2. Originally Posted by bjkrishna
I just need to solve a quatric equation like $ax^4 + bx^3 +cx^2+dx+e=0$
I need the 4 roots in terms a,b,c,d,e.

Thanks
Wrong forum.

Use synthetic division.

3. all i can say is that its not going to be just solving it man...

its a pretty long process.

Its not difficult but it does take time. anyway this is what you need to do:

First you need to read up on depressed quartic equations. Once you have make your equation into a depressed quartic.

After youve done that you need to get familiar with Ferrari's solution. It has quite a few outcomes depending on the initial conditions so it might be worth taking a note of that when you show the different methods of solving the polynomial.

4. Originally Posted by bjkrishna
I just need to solve a quatric equation like $ax^4 + bx^3 +cx^2+dx+e=0$
I need the 4 roots in terms a,b,c,d,e.

Thanks
It turns out that solving a quartic boils down to solving a cubic.

We have $f(x)=ax^4+bx^3+cx^2+dx+e$. Let the roots of $f(x)$ be $\alpha_1, \; \alpha_2, \; \alpha_3, \; \alpha_4$.

Under the substitution $x=y-\tfrac a4$ the quartic becomes $g(y) = y^4+py^2+qy+r$, where $p=\tfrac18(-3a^2+8b), \; q=\tfrac18(a^3-4ab+8c), \; r=\tfrac{1}{256}(-3a^4+16a^2b-64ac+256d)$.

The roots of $g(y)$ are $\alpha_1+\tfrac a4, \; \alpha_2+\tfrac a4, \; \alpha_3+\tfrac a4, \; \alpha_4+\tfrac a4$. Let's call these $\beta_1, \; \beta_2, \; \beta_3, \; \beta_4$ respectively.

Now define $h(x) = x^3-2px^2+(p^2-4r)x+q^2$. Direct verification shows the roots of $h(x)$ are

$\theta_1 = (\beta_1+\beta_2)(\beta_3+\beta_4)$
$\theta_2 = (\beta_1+\beta_3)(\beta_2+\beta_4)$
$\theta_3 = (\beta_1+\beta_4)(\beta_2+\beta_3)$

So we have $\theta_1 = (\beta_1+\beta_2)(\beta_3+\beta_4)$ from above and from $g(x)$ we get $(\beta_1+\beta_2)+(\beta_3+\beta_4)=0$ since the sum of the roots will give the coefficient of the cubic term, and in this case it's zero.

This gives us $\beta_1+\beta_2 = \sqrt{-\theta_1}$ and $\beta_3+\beta_4 = -\sqrt{-\theta_1}$.

Similarly
$\beta_1+\beta_3 = \sqrt{-\theta_2}$ and $\beta_2+\beta_4 = -\sqrt{-\theta_2}$
$\beta_1+\beta_4 = \sqrt{-\theta_3}$ and $\beta_2+\beta_3 = -\sqrt{-\theta_3}$.

Also an easy computation shows $\sqrt{-\theta_1}\sqrt{-\theta_2}\sqrt{-\theta_3} = -q$, so the choice of two of the square roots determines the third. Using some algebra we get

$2\beta_1 = \sqrt{-\theta_1}+\sqrt{-\theta_2}+\sqrt{-\theta_3}$
$2\beta_2 = \sqrt{-\theta_1}-\sqrt{-\theta_2}-\sqrt{-\theta_3}$
$2\beta_3 = -\sqrt{-\theta_1}+\sqrt{-\theta_2}-\sqrt{-\theta_3}$
$2\beta_4 = -\sqrt{-\theta_1}-\sqrt{-\theta_2}+\sqrt{-\theta_3}$

From above, knowing the $\beta_i$'s lets us know the $\alpha_i$'s. Thus we've found the roots of $f(x)$.

And here's the final result after making the substitutions back to $a,b,c,d,e$: quartic formula.