There exists exact solutions to the straight line, the quadratic and cubic polynomials.
I have found an unique exact solution to the Quartic
for real coefficients.
The straight line is in the form y = ax + b
The quadratic is in the form y = ax² + bx + c
The Cubic is in the form y = ax³ + bx² + cx + d
The Quartic is in the form y = ax⁴ + bx³ + cx² + dx + e

  Quartic Equation

The general Quartic is given by



Ax4 + Bx3 + Cx2 + Dx + E = 0


Dividing by A we therefore solve for

 x4 + ax3 + bx2 + cx + d =0 

where a, b, c and d are all members of the real numbers. 

We therefore restrict our solution to a,b,c,d all real

Reducing the fourth order polynomial




Given x4 + ax3 + bx2 + cx + d =0     eq1

Let  x = x1 + m
         
Then x2 = x21 + 2mx1 + m2
          
and  x3 = x31 + 3mx21 + 3x1m2 + m3
       
and x4 = x41 + 4mx31 + 6m2x21 + 4x1m3 + m4
            

Substitute the above into eq1 and we get


x41 + x31(4m+a) + x21(6m2+3am+b) + x1(4m3+3am2+2bm+ c) + ( m4+am3+bm2+cm+d)=0     eq2

We eliminate the third order term by

Letting 4m + a = 0 and thus m = -a/4 and m2=a2/16 and m3=-a3/64


Eq2 then reduces to x41 + ex21 + fx1 + g = 0


where  e = (6m2 + 3am + b) = 6a2 /16 - 3a2 /4 + b = -3a2 /8 + b  eq3


where f = 4m3 + 3am2 + 2bm + c = -a3 /16 + 3a3 /16 -2ab/4 +c 

Thus f = c - ab/2 + a3 /8   eq4


and g = m4 + am3 + bm2 + cm + d 

     
Thus g = a4 /256 - a4 /64 + ba2 /16 - ac/4 + d 


resulting in g = d - ac/4 + ba2 /16 - 3a4 /256   eq5

Let  x1 = x2  /q
   

Thus x42  + eq2x22  + fx2q3 + gq4 = 0 


Now we consider two cases, the first for e>0 and the second 
for e<0.


Case one: e>0




Let q2 = 1/e   eq6



Substitute eq6 and we get

x42 + x22  + hx2 - K = 0  eq7

SQRT is "square root of"
                  
where h = fq3 = f / SQRT( e3)       eq8      
            

and  K = - g /e2               eq9

We therefore have reduced our original equation to the form as
in eq7 with h and K both real. If we can get an exact solution
for this equation then it will be straight forward to get x by
working backwards.

End of case one



Case two: e<0



Let q2 =-1/e , so q = 1/sqrt(|e|)  where |e| is the possitive value of e 

So if e = -2 then |e|=2 and if e=2 then |e|=2

sqrt means square root of , and i = sqrt(-1)

and |e| = positive value of e

and the equation reduces to x42 - x22 + hx2 - K =0

where K = -gq4 = -g/e2

and h = fq3 = f/sqrt(|e3|)  


End case 2


So we now have depending on which case a quadratic term which is
either unity positive or unity negative and we have both h and K 
real numbers.



Roots of the fourth order polynomial

We shall assume that as in the case of the cubic equation, that
the root will consists of two independent terms.We also assume
that each term will be a root of the fourth order.

We have reduced the problem to the following equation


x42  + x22  + hx2 - K = 0  with h and K both real for case 1

And to x42 - x22 + hx2 - K = 0 with h and K both real for case 2
 
Now we just rewrite these equations so that it reads easier.

So x2 becomes x, h becomes K1 and -K becomes K2


We will consider the two cases separately

We will however see in the end that the method we use for the
two cases is exactly the same and this should not come as a
surprise as the equations look much alike. The results should
therefore also have common features.


Case one


Let us then examine x4 + x2 + K1 x + K2 = 0   eq1 for case 1

                      

Let us assume a root x = z1/41  - iz1/42      eq2
 
          
where i =sqrt(-1)

Therefore x2  = z1/21 - 2i(z1z2)1/4  - z1/22       eq3    


and x4  = z1 - 4iz3/41z1/42  - 6(z1z2)1/2 +4iz1/41z3/42 + z2     eq4
    

We know that for the equation to be 0, that the imaginary and
 the real part of the equation have to be 0.

We substitute eq's 2,3 and 4 into eq1 and write the real and
 imaginary parts separately as two independent equations.

       
Real part     : z1  - 6(z1z2) 1/2 + z2 + z1/21 - z1/22 + K1z1/41  + K2 = 0    eq5
              

For the imaginary part

- K1 z1/42< - 2(z1z2)1/4  - 4z3/41z1/42  + 4z1/41z3/42 = 0  eq6


  
divide eq6 with z1/42   because it is a factor of all the terms
     


then -K 1 - 2z1/41  -  4z3/41  + 4z1/4 1z1/22  = 0  eq7
             
multiply eq7 with z1/41
          
   
then   -4z1/21z1/22  =  -K1  z1/41  - 2z1/22   - 4z1       eq8   
    
 
divide eq7 with z1/41 
  

then -K1 /z1/41  - 2  -4z1/21  +  4z1/22   = 0
   

and   -K1 /4z1/41  - 1/2 = z1/21  -  z1/22   = T   eq9
   


Out of eq5 we get


(z1/21 - z1/22)2 - 4(z1z2) 1/2  + (z1/21 - z1/22) + K1z1/41  + K2 = 0  eq10

Substitute eq8 into eq10 and also substituting T we get


 
T2 -2z1/21  -4z1  + T + K2 = 0     eq11


Substitute eq9 into eq11 and we get


K21 / 16z1/21 + K1 / 4z1/41 + 1/4 - 2z1/21 - 4z1 - K1 / 4z1/41 - 1/2 + K2 = 0  eq12



          
multiply eq12 with z1/21
        


then  K21 /16  - 0.25z1/21 - 2z1 -4z3/21  + K2 z1/21 = 0    eq13
    

 
Let  z1 = v2    eq16 and substitute into eq13


   
then -4v3 -2v2 + v(K2 - .25) + K21 /16 = 0  eq14
             

Thus  v3 + v2/2 + v(1/16 - K2/4) - K21 /64 = 0  eq15


We thus have a equation in v that we can solve.

This means that we can get v from eq15 and z1  from eq16.
               

And  from eq9 we can then get z2
               

and from eq2 we can get x.


We know that complex roots always have a conjugate partner and
we can therefore write the other root as

     
x = z1/41  + iz1/42
   

What we have done here is to demonstrate that there is an exact 
method by which we can determine the roots of the fourth order
polynomial for case 1.

We will now do an example to demonstrate




Example

Let's look at x4 + x2 + 2x - 24 = 0

Through inspection we know that  x = 2 

and that K1 = 2 and K2 = -24
        

If we take the factor x-2 out of the equation we are left with



x3 + 2x2 + 5x + 12 = 0 which have a real root at

x = -2,2029812583

The quadratic term that's left then delivers

x = 0,10149062915 +- i 2,3317082922 as complex roots

We have done all the above to check our method that we are going to use.

We now use eq15 to evaluate v.


 Substituting we get


v3 + 0,5v2 + v(1/16 +24/4) - 4/64 = 0

  
Therefore v3 + 0,5v2 + 6,0625v - 0,0625 = 0

We get a real root at v = 0,010300347807

  
Therefore z1 = v2 = 0,000106097
 

  
and  z1/41  = 0,101490625
     

            
We also know   z1/21 - z1/22 =( -2 / 4z1/41)  -1/2
              
                           
            
                      
       
Therefore z1/4 2   = 2,331708333
        

The complex root therefore is

x = 0,101490625 +- i 2,331708333

which is the root we got through inspection.



Now we examine case 2


                  
Let us then examine x4 - x2 + K1 x + K2 = 0   eq1 for case 2
                       
 



                 
Let us assume a root x = z1/41  - iz1/42      eq2
                  
    
We now go through the exact same steps as in case one and we will end up with


  
v3 - v2/2   + v (1/16 - K2/4)  - K21/64 = 0  eq15
                    

We thus have an equation in v that we can solve and this equation 
nearly looks the same as the one for case one..

So we can again solve for x


We have seen therefore that the path we follow for case 2 is 
exactly the same as for case one and this should not come as a 
surprise.

We have therefore shown that an exact solution does exist for 
the quartic with real coefficients.



End