Solution of a Quadratic Equation in Complex Number System

Solution of a Quadratic Equation in Complex Number System

Let the given equation be $ax^2+bx+c=0$ where $a,b,c\in\mathbb{R}$ and $a\neq0$. The solution of this quadratic equation is given by the familiar formula

so the two roots of $ax^2+bx+c=0$ are $\dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $\dfrac{-b-\sqrt{b^2-4ac}}{2a}$.

The expression $D=b^2-4ac$ is called the discriminant. If $D<0$, the roots of the quadratic equation are complex (this is exactly the case the real number system could not handle, and exactly the case the Fundamental Theorem of Algebra guarantees still has a solution in $\mathbb{C}$).

Note. If $p+iq$ is a root of $ax^2+bx+c=0$ (with $a,b,c\in\mathbb{R}$, $a\neq0$), then $p-iq$ is automatically also a root. So when the coefficients are real, complex roots always occur in conjugate pairs.

Worked Example 1: solve $x^2+x+1=0$.

Comparing with $ax^2+bx+c=0$: $a=1,b=1,c=1$. The roots are $x=\dfrac{-1\pm\sqrt{1-4}}{2}=\dfrac{-1\pm\sqrt{-3}}{2}=\dfrac{-1\pm\sqrt3\,i}{2}$. So the roots are $-\dfrac12+\dfrac{\sqrt3}{2}i$ and $-\dfrac12-\dfrac{\sqrt3}{2}i$ — a conjugate pair, exactly as the note above predicts (and, not coincidentally, exactly the two complex cube roots of unity $\omega,\omega^2$ of Section 1.7).

Worked Example 2: solve $x^2-(2\sqrt3+3i)x+6\sqrt3\,i=0$.

Here the coefficients are complex, but the quadratic formula still applies, with $a=1,\ b=-(2\sqrt3+3i),\ c=6\sqrt3\,i$. Compute the discriminant: $b^2-4ac=[-(2\sqrt3+3i)]^2-4(1)(6\sqrt3\,i)=(12+12\sqrt3i-9)-24\sqrt3i=3-12\sqrt3i=3(1-4\sqrt3i)$. Finding $\sqrt{1-4\sqrt3i}$ needs the square-root-of-a-complex-number method of Section 1.3: let $a+ib=\sqrt{1-4\sqrt3i}$, so $a^2-b^2=1$ and $2ab=-4\sqrt3$, i.e. $ab=-2\sqrt3$. Then $(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2=1+4(12)=49$, so $a^2+b^2=7$. With $a^2-b^2=1$: $2a^2=8\Rightarrow a=\pm2$, and $2b^2=6\Rightarrow b=\pm\sqrt3$. Checking which sign combination satisfies $ab=-2\sqrt3$: only $a=2,b=-\sqrt3$ and $a=-2,b=\sqrt3$ work (not $a=2,b=\sqrt3$). So $\sqrt{1-4\sqrt3i}=\pm(2-\sqrt3i)$, and $\sqrt{3(1-4\sqrt3i)}=\pm\sqrt3(2-\sqrt3i)$. The roots are then $x=\dfrac{(2\sqrt3+3i)\pm\sqrt3(2-\sqrt3i)}{2}$; working through both signs gives the roots $x=2\sqrt3$ and $x=3i$.

Worked Example 3: find the value of $x^3-x^2+2x+10$ when $x=1+\sqrt3\,i$.

Since $x=1+\sqrt3i$, $x-1=\sqrt3i$; squaring both sides, $(x-1)^2=(\sqrt3i)^2=-3$, i.e. $x^2-2x+1=-3$, so $x^2-2x=-4\ \ldots$(I). Now rewrite the target expression using (I): $x^3-x^2+2x+10=x^3-(x^2-2x)+10=x^3-(-4)+10=x^3+14$. Compute $x^3=(1+\sqrt3i)^3=1+3\sqrt3i-9-3\sqrt3i=-8$ (expanding by the binomial theorem and using $i^2=-1,i^3=-i$). So the value is $-8+14=6$. This shortcut — find the quadratic that $x$ satisfies, and use it to reduce a higher power before substituting — avoids repeatedly computing large complex powers directly.

Worked Example 4: if $x=-5+2\sqrt{-4}$, find the value of $x^4+9x^3+35x^2-x+64$. …