Solving Cubic Equations with Real Coefficients

The conjugate root theorem we've seen earlier can extend to cubic equations as well. If all coefficients of a cubic equation are real, then the roots are all real, or two of them are complex conjugates and the other is real.

From factor theorem, we know that cubic can be factored into lower degree expressions.

C=Q \cdot L=0 \\ C=L \cdot L \cdot L = 0

Where $C$ is a cubic expression, $Q$ is a quadratic expression and $L$ is a linear expression.

Example

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Form the cubic equations with roots $1+i$ and $3$ .

We are given two roots, one complex and one real, meaning that the remaining root must be the conjugate of the complex root.

z=1+i,z=1-i,z=3

Let's form a quadratic factor using the two complex roots.

\begin{align*} S &= (1+i)+(1-i) \\ &= 2 \\ P &= (1+i)\cdot(1-i) \\ &= 1-i+i-i^2 \\ &= 1-i+i-(-1) \\ &= 2 \end{align*}

\begin{align*} z^2-Sz+P&=0 \\ z^2 -(2)z+(2)&=0 \\ z^2-2z+2&=0 \end{align*}

With the remaining (real) root, we can derive a linear factor :

\begin{align*} z &= 3 \\ (z-3) &=0 \end{align*}

Expand the two factors to derive the cubic equation :

\begin{align*} L \cdot Q &= C &=0 \\ (z-3)(z^2-2z+2)&= z^3-2z^2+2z-3z^2+6z-6 &=0 \\ &= z^3-5z^2+8z-6 &=0 \end{align*}

Example

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Find $k \in \mathbb{R}$ in the equation $z^3-kz^2+24z+34=0$ given that $5-3i$ .

All coefficients are real, meaning that the conjugate root theorem can be applied. Another root is the conjugate of the given complex one.

z=5-3i,z=5+3i

Let's assume the remaining (real) root is $a$ .

z=5-3i,z=5+3i, z=a

Let's express the cubic as an expansion of it's factors.

Q \cdot L=C

First, derive $Q$ using the two complex roots :

\begin{align*} S &= (5+3i)+(5-3i) \\ &= 10 \\ P &= (5+3i) \cdot (5-3i) \\ &= 25-15i+15i-9i^2 \\ &= 25-9(-1) \\ &= 34 \end{align*}

\begin{align*} Q&=z^2-Sz+P&=0 \\ &=z^2 -(10)z+(34)&=0 \\ &=z^2-10z+34&=0 \end{align*}

Derive $L$ :

\begin{align*} L=(z-a) \end{align*}

Now let's rewrite the cubic expression in terms of its factors :

\begin{align*} L \cdot Q &=(z-a)(z^2-10z+34) &= C\\ &= z^3-10z^2+34z-az^2+10az-34a &=C\\ &= z^3+(-a-10)z^2+(10a+34)z-34a &=C \end{align*}

Finally, equate the two representations of the cubic equations together :

\begin{align*} &z^3+(-a-10)z^2+(10a+34)z-34a &= z^3-kz^2+24z+34=0 \end{align*}

Compare term coefficients :

\begin{align*} &1z^3+(-a-10)z^2+(10a+34)z-34a &= 1z^3-kz^2+24z+34=0 \end{align*}

\begin{align*} \text{ constants} &: -34a &= 34 \\ &: a &=-1 \end{align*}

\begin{align*} z^2 \text{ terms} &: (-a-10) &= -k\\ &: (-(-1)-10) &= -k \\ &: -9 &=-k \\ &: 9 &=k \end{align*}

Solving Cubic Equations with Real Coefficients Simplified Revision Notes for Leaving Cert Mathematics

Solving Cubic Equations with Real Coefficients

Form the cubic equations with roots $1+i$ and $3$ .

Find $k \in \mathbb{R}$ in the equation $z^3-kz^2+24z+34=0$ given that $5-3i$ .

500K+ Students Use These Powerful Tools to Master Solving Cubic Equations with Real Coefficients For their Leaving Cert Exams.

Other Revision Notes related to Solving Cubic Equations with Real Coefficients you should explore

Solving Cubic Equations with Real Coefficients Simplified Revision Notes for Leaving Cert Mathematics

Solving Cubic Equations with Real Coefficients

Form the cubic equations with roots 1+i1+i1+i and 333.

Find k∈Rk \in \mathbb{R}k∈R in the equation z3−kz2+24z+34=0z^3-kz^2+24z+34=0z3−kz2+24z+34=0 given that 5−3i5-3i5−3i.

500K+ Students Use These Powerful Tools to Master Solving Cubic Equations with Real Coefficients For their Leaving Cert Exams.

Other Revision Notes related to Solving Cubic Equations with Real Coefficients you should explore

Form the cubic equations with roots $1+i$ and $3$ .

Find $k \in \mathbb{R}$ in the equation $z^3-kz^2+24z+34=0$ given that $5-3i$ .