Complex Roots of Quadratics: Practice

Non-real roots require a negative discriminant.

$\Delta = 16 - 4k^2 < 0 \quad\Longleftrightarrow\quad k^2 > 4 \quad\Longleftrightarrow\quad k < -2 \ \text{ or } \ k > 2$

Solving $k^2 > 4$ as a quadratic inequality gives the two outer intervals, so the roots are non-real exactly when $|k| > 2$.

The discriminant is $\Delta = 36 - 100 = -64$, so the roots are $x = \tfrac{6 \pm 8i}{2} = 3 \pm 4i$.

$|3 \pm 4i| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

A shortcut confirms it: the product of a conjugate pair equals the squared modulus, and here the product is $\tfrac{c}{a} = 25$, so the modulus is $\sqrt{25} = 5$.

Real coefficients make the other root the conjugate $2 + 3i$, so the sum of the roots is $4$. The sum also equals $-\tfrac{b}{a}$ with $a = 2$.

$4 = -\frac{b}{2} \quad\Longrightarrow\quad b = -8$

(The product is $4 + 9 = 13 = \tfrac{c}{2}$, so $c = 26$ and the equation is $2x^2 - 8x + 26 = 0$.)

The other root is the conjugate $1 - i\sqrt{3}$, and $c$ is the product of the roots.

$c = \left(1 + i\sqrt{3}\right)\left(1 - i\sqrt{3}\right) = 1 + 3 = 4$

The conjugate product is $p^2 + q^2$ with $p = 1$ and $q = \sqrt{3}$. (The sum gives $b = -2$, so the equation is $x^2 - 2x + 4 = 0$.)

First square: $(1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i$. Then substitute into the expression.

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

The result is $0$, which is exactly the statement that $1 + 2i$ is a root of $x^2 - 2x + 5 = 0$.

The product of the roots equals the constant term for a monic quadratic, and the conjugate product is $p^2 + q^2$.

$2^2 + q^2 = 40 \quad\Longrightarrow\quad q^2 = 36 \quad\Longrightarrow\quad q = 6$

So the roots are $2 \pm 6i$. (The sum $4 = -b$ gives $b = -4$, and the discriminant $16 - 160 = -144 < 0$ confirms the roots are non-real.)

A monic quadratic is $x^2 - (\text{sum})x + (\text{product})$.

$x^2 - 6x + 10 = 0$

Its discriminant is $36 - 40 = -4 < 0$, so the roots are indeed non-real: they are $3 \pm i$, with sum $6$ and product $9 + 1 = 10$.

Convert each factor to $i$-form before multiplying; the product rule for radicals fails when both radicands are negative.

$\sqrt{-9}\cdot\sqrt{-16} = (3i)(4i) = 12\,i^2 = -12$

Combining the radicands first would give $\sqrt{144} = 12$, which drops the factor $i^2 = -1$ and is wrong.

The discriminant is $\Delta = 36 - 180 = -144$, so $\sqrt{-144} = 12i$ and $2a = 18$.

$x = \frac{6 \pm 12i}{18} = \frac{1}{3} \pm \frac{2}{3}\,i$

Every term is divided by $18$, giving real part $\tfrac{1}{3}$ and imaginary part $\pm\tfrac{2}{3}$.

The sum of the pair is $\tfrac{1+i}{2} + \tfrac{1-i}{2} = 1$ and the product is

$\frac{(1 + i)(1 - i)}{4} = \frac{1 + 1}{4} = \frac{1}{2}.$

A monic version is $x^2 - x + \tfrac{1}{2}$; clearing the fraction by doubling gives $2x^2 - 2x + 1 = 0$. Check: its discriminant is $4 - 8 = -4$, so $x = \tfrac{2 \pm 2i}{4} = \tfrac{1 \pm i}{2}$.

The two roots are $z$ and $\overline{z}$, so the given quantities are exactly the sum and product of the roots. The monic quadratic is $x^2 - (\text{sum})x + (\text{product})$.

$x^2 - (-3)x + 5 = x^2 + 3x + 5 = 0$

Check: its discriminant is $9 - 20 = -11 < 0$, so the roots are indeed a non-real conjugate pair.

Two non-real roots require a strictly negative discriminant.

$\Delta = 36 - 4c < 0 \quad\Longleftrightarrow\quad 4c > 36 \quad\Longleftrightarrow\quad c > 9$

At $c = 9$ the discriminant is zero and the root $-3$ is real (repeated), so the boundary value is excluded.