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Wize University Linear Algebra Textbook > Complex Numbers

Quadratic Formula

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Quadratic Formula

When using the quadratic formula, you may have run into the issue of taking the square root of a negative number.

Wize Concept
Recall the quadratic formula for finding the roots of the real quadratic ax2+bx+c=0ax^2+bx+c=0:
x=−b±b2−4ac2a\boxed{\quad x=\dfrac{-b\pm\sqrt{\colorTwo{b^2-4ac}}}{2a} \quad}

b2−4ac\colorTwo{b^2-4ac} is called the discriminant. For any real quadratic:
  • b2−4ac≥0    ⟹  \colorTwo{b^2-4ac} \ge 0 \ \ \implies real solutions
  • b2−4ac<0    ⟹  \colorTwo{b^2-4ac} <0 \ \ \implies complex solutions (taking the square root of a negative number)
Wize Tip
Complex solutions always come in conjugate pairs!
If a+bia+bi is one root, then the other root is a−bia-bi.

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Example
Find the roots of the real quadratic p(x)=x2+2x+2p(x)=x^2+2x+2.
We want to find the values of xx such that x2+2x+2=0x^2+2x+2=0.
We have a=1,  b=2,  c=2a=1,\ \ b=2,\ \ c=2. Let's apply the quadratic formula.
x=−2±22−4(1)(2)2(1)=−2±−42=−2±−142=−2±i⋅22=−1±i\begin{aligned} x &= \dfrac{-2 \pm \sqrt{2^2-4(1)(2)}}{2(1)}\\[1em] &=\dfrac{-2 \pm \sqrt{-4}}{2}\\[1em] &=\dfrac{-2 \pm \colorOne{\sqrt{-1}}\sqrt{4}}{2}\\[1em] &=\dfrac{-2 \pm \colorOne{i}\cdot 2}{2}\\[1em] &= \boxed{-1 \pm i} \end{aligned}


Wize Concept
The quadratic formula works even if the quadratic has complex coefficients.
Note: you may need to find square roots of a complex number. Remember to use polar form to find the roots!

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Example: Roots of Real Quadratics

A real quadratic has a root at 2−3i2-3i . Find the other root along with the standard form of a quadratic with these roots.
Since we are given one root of a real quadratic and the root is complex, the other root is the complex conjugate: 2+3i\boxed{2+3i}.
We can find the real quadratic by multiplying the two factors that yield these roots:
(x−(2−3i))\Big(x- (2-3i) \Big) and (x−(2+3i))\Big(x- (2+3i) \Big) are the factors that, when set to be equal to zero, yield the desired roots.
  ⟹  p(x)=(x−(2−3i))(x−(2+3i))=x2−(2+3i)x−(2−3i)x+(2−3i)(2+3i)=x2−2x−3ix−2x+3ix+4+9=x2−4x+13\begin{aligned} \implies p(x) &= \Big(x- (2-3i) \Big)\Big(x- (2+3i) \Big)\\[0.5em] &= x^2 -(2+3i)x -(2-3i)x + (2-3i)(2+3i)\\[0.5em] &= x^2 -2x -\cancel{3ix} -2x+\cancel{3ix} + 4 + 9\\[0.5em] &= \boxed{x^2 -4x +13}\\[0.5em] \end{aligned}
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Example: Quadratic Formula for Complex Quadratics

Find the roots of the complex quadratic q(z)=z2−(1+3i)z+(−2+i)q(z) = z^2-(1+3i)z+(-2+i).
Note that a=1,  b=−(1+3i),  c=−2+ia=1,\ \ b=-(1+3i),\ \ c=-2+i.
Applying the quadratic formula:
z=−(−(1+3i))±(−(1+3i))2−4(1)(−2+i)2(1)=(1+3i)±(1+3i)2−(−8+4i)2=(1+3i)±(1+6i−9)+8−4i2=(1+3i)±2i2\begin{aligned} z &= \dfrac{-(-(1+3i)) \pm \sqrt{(-(1+3i))^2 - 4(1)(-2+i)}}{2(1)}\\[1.3em] &= \dfrac{(1+3i) \pm \sqrt{(1+3i)^2 - (-8+4i)}}{2}\\[1.3em] &= \dfrac{(1+3i) \pm \sqrt{(1+6i-9) +8-4i}}{2}\\[1.3em] &= \dfrac{(1+3i) \pm \colorTwo{\sqrt{2i}}}{2}\\[1.3em] \end{aligned}
At this step, we notice that we must take the square root of a complex number. Convert 2i\colorTwo{2i} to polar form seiϕse^{i\phi}:
s=0+22=2s = \sqrt{0+2^2} = 2
ϕ=π2(since 2i is on the positive y-axis)\phi = \dfrac{\pi}{2} \quad \text{(since $2i$ is on the positive $y$-axis)}
∴  2i=2eiπ/2\therefore\ \ 2i=2e^{i\pi/2}
The square roots are of the form sei(ϕ+2πk)/2\sqrt{s}e^{\small i(\phi+ 2\pi k)/2}, with values of k=0,1k=0, 1:
2i=2ei(π/2)/2or2ei(π/2 + 2π)/2=2eiπ/4or2ei5π/4=2(cos⁡π4+isin⁡π4)or2(cos⁡5π4+isin⁡5π4)=2(12+i12)or2(−12+i(−12))=1+ior−1−i = −(1+i)\begin{array}{rclcl} \sqrt{2i} &=& \sqrt{2} e^{i(\pi/2)/2} &\quad \text{or} \quad& \sqrt{2} e^{i(\pi/2\ +\ 2\pi)/2}\\[0.5em] &=& \sqrt{2} e^{i\pi/4} &\quad \text{or} \quad& \sqrt{2} e^{i5\pi /4}\\[0.5em] &=& \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) &\quad \text{or} \quad& \sqrt{2} (\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4})\\[0.5em] &=& \sqrt{2}\left( \frac{1}{\sqrt2} + i \frac{1}{\sqrt2} \right) &\quad \text{or} \quad& \sqrt{2} \left(-\frac{1}{\sqrt2} + i \left(-\frac{1}{\sqrt2}\right)\right)\\[1em] &=& \colorTwo{1+i} &\quad \text{or} \quad& -1-i \ =\ \colorTwo{-(1+i)} \end{array}
We can now continue finding the roots using the fact that ±2i=±(1+i)\pm\colorTwo{\sqrt{2i}} = \pm \colorTwo{(1+i)}:
z=(1+3i)±2i2=(1+3i)±(1+i)2=2+4i2or2i2=1+2iori\begin{array}{rcccl} z &=& \dfrac{(1+3i) \pm \colorTwo{\sqrt{2i}}}{2}\\[1.3em] &=& \dfrac{(1+3i) \pm \colorTwo{(1+i)}}{2}\\[1.3em] &=& \dfrac{2+4i}{2} &\text{or} \qquad& \dfrac{2i}{2} \\[1.7em] &=& \boxed{1+2i} &\text{or} \qquad& \boxed{i}\\[1.7em] \end{array}

Practice: Quadratic with Complex Coefficients

Find the roots of x2−4ix+6=0x^2 - 4ix + 6 = 0.