Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org

Math formulas: Solutions of algebraic equations

0 formulas included in custom cheat sheet

Quadric Equation: ax2+bx+c=0ax^2 + bx + c = 0

Solutions (roots):

x1,2=b±b24ac2a x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

If D=b24ac D = b^2 - 4ac is the discriminant , then the roots are

1. real and unique if D>0 D > 0

2. real and equal if D=0 D = 0

3. complex conjugate if D<0 D < 0

Cubic Equation: x3+a1x2+a2x+a3=0x^3 + a_1x^2 + a_2x + a_3 = 0

Let

Q=3a2a129R=9a1a227a32a1354S=R+Q3+R23T=RQ3+R23 \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:

x1=S+T13a1x2=12(S+T)13a1+12i3(ST)x3=12(S+T)13a112i3(ST) \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=Q3+R2D = Q^3 + R^2 is the discriminant of the cubic equation, then:

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

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

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

Quartic Equation:x4+a1x3+a2x2+a3x+a4=0x^4 + a_1x^3 + a_2x^2 + a_3x + a_4 = 0

Let y1y_1 be a real root of the cubic equation

y3a2y2+(a1a34a4)y+(4a2a4a32a12a4)=0 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

z2+12(a1±a124a2+4y1)z+12(y1±y124a4)=0 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

Was these formulas helpful?

Yes No