I just need to solve a quatric equation like $\displaystyle ax^4 + bx^3 +cx^2+dx+e=0 $

I need the 4 roots in terms a,b,c,d,e.

Can any one please help me solving the quatric equation.

Thanks

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- Apr 25th 2010, 01:05 PMbjkrishnasolving a quatric equation
I just need to solve a quatric equation like $\displaystyle ax^4 + bx^3 +cx^2+dx+e=0 $

I need the 4 roots in terms a,b,c,d,e.

Can any one please help me solving the quatric equation.

Thanks - Apr 25th 2010, 01:13 PMdwsmith
- Apr 25th 2010, 01:20 PMschmeling
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. - Apr 25th 2010, 01:27 PMdwsmith
- Apr 25th 2010, 03:05 PMchiph588@
It turns out that solving a quartic boils down to solving a cubic.

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

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

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

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

$\displaystyle \theta_1 = (\beta_1+\beta_2)(\beta_3+\beta_4) $

$\displaystyle \theta_2 = (\beta_1+\beta_3)(\beta_2+\beta_4) $

$\displaystyle \theta_3 = (\beta_1+\beta_4)(\beta_2+\beta_3) $

So we have $\displaystyle \theta_1 = (\beta_1+\beta_2)(\beta_3+\beta_4) $ from above and from $\displaystyle g(x) $ we get $\displaystyle (\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 $\displaystyle \beta_1+\beta_2 = \sqrt{-\theta_1} $ and $\displaystyle \beta_3+\beta_4 = -\sqrt{-\theta_1} $.

Similarly

$\displaystyle \beta_1+\beta_3 = \sqrt{-\theta_2} $ and $\displaystyle \beta_2+\beta_4 = -\sqrt{-\theta_2} $

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

Also an easy computation shows $\displaystyle \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

$\displaystyle 2\beta_1 = \sqrt{-\theta_1}+\sqrt{-\theta_2}+\sqrt{-\theta_3} $

$\displaystyle 2\beta_2 = \sqrt{-\theta_1}-\sqrt{-\theta_2}-\sqrt{-\theta_3} $

$\displaystyle 2\beta_3 = -\sqrt{-\theta_1}+\sqrt{-\theta_2}-\sqrt{-\theta_3} $

$\displaystyle 2\beta_4 = -\sqrt{-\theta_1}-\sqrt{-\theta_2}+\sqrt{-\theta_3} $

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

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