Complex Roots of Quadratics: Additional Practice (Set 1)

With $a = 2$, $b = -6$, $c = 5$, the discriminant is $\Delta = 36 - 40 = -4$, so $\sqrt{-4} = 2i$ and $2a = 4$.

$x = \frac{6 \pm 2i}{4} = \frac{3}{2} \pm \frac{1}{2}\,i$

Dividing every term by $4$ gives real part $\tfrac{3}{2}$ and imaginary part $\pm\tfrac{1}{2}$.

Pull out the $i$ first, then extract the largest perfect square from $27$.

$\sqrt{-27} = i\sqrt{27} = i\sqrt{9 \cdot 3} = 3i\sqrt{3}$

Check: $\left(3i\sqrt{3}\right)^2 = 9 \cdot (-1) \cdot 3 = -27$.

Set the discriminant negative and solve for $m$.

$\Delta = 36 - 12m < 0 \quad\Longleftrightarrow\quad 12m > 36 \quad\Longleftrightarrow\quad m > 3$

At $m = 3$ the discriminant is zero (a repeated real root), so the boundary is excluded and the answer is $m > 3$.

Combine the fractions over the common denominator $r_1 r_2$, then use the sum and product of the roots, which hold for complex roots too.

$\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2} = \frac{4}{13}$

The sum is $-\tfrac{b}{a} = 4$ and the product is $\tfrac{c}{a} = 13$, so the answer is $\tfrac{4}{13}$, a real number even though each reciprocal is complex.

The roots are a conjugate pair $p \pm qi$ with $q \ne 0$, and their product is

$\frac{c}{a} = (p + qi)(p - qi) = p^2 + q^2 > 0,$

a sum of squares with $q \ne 0$. The discriminant agrees: $b^2 < 4ac$ forces $ac > \tfrac{b^2}{4} \ge 0$, so $a$ and $c$ share a sign and their ratio is positive.

First put the equation in standard form: $x^2 - 2x + 10 = 0$. The discriminant is $\Delta = 4 - 40 = -36$, so $\sqrt{-36} = 6i$.

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

The roots are the conjugate pair $1 + 3i$ and $1 - 3i$.

Conjugating the equation $az^2 + bz + c = 0$ term by term shows that whenever $z$ is a root, so is $\overline{z}$, because the real coefficients are their own conjugates.

$a\overline{z}^{\,2} + b\overline{z} + c = \overline{az^2 + bz + c} = \overline{0} = 0$

A non-real root therefore always brings its conjugate along, so it can never be paired with a single real root. The other three inventories are the $\Delta > 0$, $\Delta = 0$, and $\Delta < 0$ cases.

Set the discriminant negative and factor it.

$\Delta = k^2 - 4k = k(k - 4) < 0$

A product of two factors is negative exactly when they have opposite signs, which happens for $0 < k < 4$. At $k = 0$ and $k = 4$ the discriminant is zero, so the endpoints are excluded.

The discriminant is $\Delta = 16 - 28 = -12$, so $\sqrt{-12} = 2i\sqrt{3}$.

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

The parabola misses the real axis, but the equation still has its two roots: the conjugate pair $2 \pm i\sqrt{3}$, with real part on the axis of symmetry $x = 2$.

Convert the negative radicand to $i$-form before multiplying anything.

$i \cdot \sqrt{-4} \cdot \sqrt{9} = i \cdot 2i \cdot 3 = 6\,i^2 = -6$

The two factors of $i$ combine into $i^2 = -1$, so the result is the real number $-6$.

For a monic quadratic with a non-real conjugate pair, the constant term equals the product of the roots, which is the squared modulus.

$k = |z|^2 = \left(\sqrt{7}\right)^2 = 7$

Consistency check: with $k = 7$ the discriminant is $16 - 28 = -12 < 0$, so the roots $2 \pm i\sqrt{3}$ really are non-real, and $4 + 3 = 7$ confirms the modulus.

Treat the sum of squares as a difference of squares using $1 = -i^2$.

$(2x - i)(2x + i) = 4x^2 - i^2 = 4x^2 + 1$

The roots are $\pm\tfrac{i}{2}$, a conjugate pair, and the two linear factors are conjugates of each other.