Solutions Of Algebraic Equations

Quadratic Equation : $ax^2 +bx +c=0$

1. $x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If $a,b,c$ are real and if  $D=b^2-4ac$ is the $discriminant$, then the roots are 

     (i) real and unequal if $D>0$

     (ii) real and equal if $D=0$

      (iii) complex conjugate if $D<0$

2. If $x_1,x_2$ are roots , then $x_1+x_2=-b/a$ and $x_1x_2=c/a$

Cubic Equation : $x^3+a_1x^2 +a_2x+a_3=0$

Let       $Q =\cfrac{3a_2-a_1^2}{9} , R=\cfrac{9a_1a_2 -27a_3 -2a_1^3}{54},$

              $S=\sqrt[3]{R+\sqrt{Q^3+R^3}} , T=\sqrt[3]{R -\sqrt{Q^3+R^3}}$

1. Solutions :                $\begin{cases} x_1 = S+T-\cfrac{1}{3}a_1 \\ x_2 = -\cfrac{1}{2}(S+T) -\cfrac{1}{3}a_1 +\cfrac{1}{2}i\sqrt{3}(S-T) \\ x_3 =-\cfrac{1}{2}(S+T)-\cfrac{1}{3}a_1 -\cfrac{1}{3}i\sqrt{3}(S-T)   \end{cases}$

If $a_1,a_2,a_3$ are real and if $D=Q^3+R^2$ is the $discriminant$, then 

     (i) one root real and two complex conjugate if $D>0$

     (ii) all roots are real and at least two are equal if $D=0$

     (iii) all roots are real and unequal if $D<0$

If $D<0$, computation is simplified by use of trigonometry .

2. Solutions if $D<0 :   \begin{cases}x_1 = 2\sqrt{-Q}\cos(\cfrac{1}{3}\theta)-\cfrac{1}{3}a_1 \\ x_2 =2\sqrt{-Q}\cos(\cfrac{1}{3}\theta +120^\circ)-\cfrac{1}{3}a_1 \\ x_3=2\sqrt{-Q}\cos(\cfrac{1}{3}\theta +240^\circ)-\cfrac{1}{3}a_1  \end{cases}  $ where $\cos \theta =R/\sqrt{-Q^3}$

3. $x_1+x_2+x_3 =-a_1 ,x_1x_2 +x_2x_3+x_3x_1 =a_2, $  $x_1x_2x_3 =-a_3$

where $x_1,x_2,x_3$ are the three roots .

Quartic Equation : $x^4+a_1x^3+a_2x^2+a_3x+a_4=0$

Let $y_1$ be a root of the cubic equation 

1. $y^3-a_2y^2 +(a_1a_3-4a_4)y+(4a_2a_4-a_3^2-a_1^2a_4)=0$

2. Solutions : The 4 roots of $z^2+\cfrac{1}{2}\{a_1\pm\sqrt{a_1^2 -4a_2+4y_1}\}z+\cfrac{1}{2}\{y_1\mp\sqrt{y_1^2-4a_4}\}=0$

     If all roots of 1 are real , computation is simplified by using that particular real root which produces all real coefficients in the quadratic equation 2 

3.  $\begin{cases}x_1+x_2+x_3+x_4=-a_1 \\ x_1x_2 +x_2x_3 +x_3x_4 +x_4x_1 +x_1x_3 +x_2x_4 =a_2 \\ x_1x_2x_3 +x_2x_3x_4 +x_1x_2x_4 +x_1x_3x_4 =-a_3 \\ x_1x_2x_3x_4=a_4  \end{cases}$ 

where $x_1,x_2,x_3,x_4$ are the four roots .