Complex Roots of Quadratics: Practice

The discriminant is $\Delta = (-4)^2 - 4(1)(13) = 16 - 52 = -36$, so $\sqrt{-36} = 6i$.

$x = \frac{4 \pm 6i}{2} = 2 \pm 3i$

Both terms of the numerator are divided by $2$, giving the conjugate pair $2 \pm 3i$.

Convert each radical to $i$-form before multiplying; the rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ does not apply when both radicands are negative.

$\sqrt{-8}\cdot\sqrt{-2} = \left(2i\sqrt{2}\right)\left(i\sqrt{2}\right) = 2\,i^2\,\cdot 2 = -4$

Multiplying the radicands first would give $\sqrt{16} = 4$, which is wrong: the $i^2 = -1$ supplies a sign the shortcut misses.

Use the sum and product of the conjugate pair. The sum is $(2 + 5i) + (2 - 5i) = 4$, and the product is

$(2 + 5i)(2 - 5i) = 4 + 25 = 29.$

A monic quadratic is $x^2 - (\text{sum})x + (\text{product})$, so the equation is $x^2 - 4x + 29$.

The relation product $= \tfrac{c}{a}$ holds even when the roots are complex, because the sum and product formulas were re-proved for the pair $p \pm qi$.

$r_1 r_2 = \frac{c}{a} = \frac{13}{1} = 13$

As a check, the roots are $3 \pm 2i$ and $(3 + 2i)(3 - 2i) = 9 + 4 = 13$.

Solve with the formula: $\Delta = 4 - 20 = -16$, so $x = \tfrac{2 \pm 4i}{2} = 1 \pm 2i$. Verify $1 + 2i$ by substitution, using $(1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i$.

$(-3 + 4i) - 2(1 + 2i) + 5 = -3 + 4i - 2 - 4i + 5 = 0$

The imaginary parts cancel exactly, confirming $1 + 2i$ is a root.

Non-real roots occur exactly when the discriminant is negative.

$\Delta = k^2 - 100 < 0 \quad\Longleftrightarrow\quad k^2 < 100 \quad\Longleftrightarrow\quad -10 < k < 10$

At $k = \pm 10$ the discriminant is zero (a repeated real root), and outside that band the roots are real and distinct.

With $a = 4$, $b = -4$, $c = 5$, the discriminant is $\Delta = 16 - 80 = -64$, so $\sqrt{-64} = 8i$ and $2a = 8$.

$x = \frac{4 \pm 8i}{8} = \frac{1}{2} \pm i$

Dividing both terms by $8$ gives real part $\tfrac{1}{2}$ and imaginary part $\pm 1$.

The sum of the roots is $-\tfrac{b}{a}$ whether the roots are real or complex, since $(p + qi) + (p - qi) = 2p = -\tfrac{b}{a}$.

$r_1 + r_2 = -\frac{b}{a} = -\frac{5}{3}$

The imaginary parts of a conjugate pair cancel, so the sum is always real.

The other root is the conjugate $5 - 2i$, and $c$ equals the product of the roots for a monic quadratic.

$c = (5 + 2i)(5 - 2i) = 25 + 4 = 29$

The conjugate product adds the squares of the real and imaginary parts, so $c = 29$.

Compute the sum and product of the pair. The sum is $i\sqrt{7} + (-i\sqrt{7}) = 0$, and the product is

$\left(i\sqrt{7}\right)\left(-i\sqrt{7}\right) = -i^2 \cdot 7 = 7.$

The monic quadratic is $x^2 - 0\,x + 7 = x^2 + 7$, and indeed $x^2 + 7 = 0$ gives $x^2 = -7$, so $x = \pm i\sqrt{7}$.

The discriminant is $\Delta = 4 - 12 = -8$, so $\sqrt{-8} = 2i\sqrt{2}$ and $2a = 6$.

$x = \frac{2 \pm 2i\sqrt{2}}{6} = \frac{1 \pm i\sqrt{2}}{3}$

The common factor of $2$ cancels from every term, leaving the conjugate pair $\tfrac{1 \pm i\sqrt{2}}{3}$.

For a monic quadratic, $c$ is the product of the roots, and the conjugate product adds the squares.

$c = (p + qi)(p - qi) = p^2 - q^2 i^2 = p^2 + q^2$

Meanwhile the sum gives $b = -2p$, not $2p$, so the only relation listed that must hold is $c = p^2 + q^2$.