Solutions of Algebraic Equations

Quadric Equation: \( ax^2 + bx + c = 0 \)

Solutions (roots):

\( x_{1,2} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)

If \( D = b^2 – 4ac \) is the discriminant , then the roots are

1. real and unique if D>0

2. real and equal if D=0

3. complex conjugate if D<0

Cubic Equation: \( x^3 + a_1x^2 + a_2x + a_3 = 0 \)

Let

\begin{aligned}
Q &= \frac{3a_2 – a_1^2}{9} \\
R &= \frac{9a_1a_2 – 27a_3 – 2a_1^3}{54} \\
S &= \sqrt[\Large3]{R + \sqrt{Q^3 + R^2}} \\
T &= \sqrt[\Large3]{R – \sqrt{Q^3 + R^2}}
\end{aligned}

Then solutions (roots) of the cubic equation are:

\begin{aligned}
x_1 &= S + T – \frac{1}{3}a_1 \\
x_2 &= -\frac{1}{2} (S + T) – \frac{1}{3}a_1 + \frac{1}{2}\,i\,\sqrt{3}(S-T) \\
x_3 &= -\frac{1}{2} (S + T) – \frac{1}{3}a_1 – \frac{1}{2}\,i\,\sqrt{3}(S-T)
\end{aligned}

If \( D = Q^3 + R^2 \) is the discriminant of the cubic equation, then:

1. one root is real and two complex conjugate if D>0

2. all roots are real and at last two are equal if D=0

3. all roots are real and unequal if D<0

Quartic Equation: \( x^4 + a_1x^3 + a_2x^2 + a_3x + a_4 = 0 \)

Let y1 be a real root of the cubic equation

\( y^3 – a_2y^2 + (a_1a_3-4a_4)y+(4a_2a_4 – a_3^2 – a_1^2a_4) = 0 \)

Then solutions of the quartic equation are the 4 roots of

\( z^2 + \frac{1}{2}\left(a_1 \pm \sqrt{a_1^2 – 4a_2+4y_1}\right)z +\frac{1}{2}\left(y_1 \pm \sqrt{y_1^2 – 4a_4}\right)= 0 \)

 

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