Complex Roots of Quadratics: Additional Practice (Set 2)

The discriminant is $\Delta = 36 - 100 = -64$, so $\sqrt{-64} = 8i$ and $2a = 10$.

$x = \frac{6 \pm 8i}{10} = \frac{3}{5} \pm \frac{4}{5}\,i$

As a check, the product of the roots should be $\tfrac{c}{a} = 1$, and indeed $\left(\tfrac{3}{5}\right)^2 + \left(\tfrac{4}{5}\right)^2 = \tfrac{9 + 16}{25} = 1$: each root has modulus $1$.

The discriminant is $\Delta = 64 - 100 = -36$, so $\sqrt{-36} = 6i$.

$x = \frac{8 \pm 6i}{2} = 4 \pm 3i$

Both terms of the numerator are divided by $2$, giving the conjugate pair $4 \pm 3i$.

The discriminant is $\Delta = 16 - 24 = -8$, so $\sqrt{-8} = 2i\sqrt{2}$ and $2a = 6$.

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

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

If $2 - i$ is a root of a real-coefficient quadratic, so is $2 + i$. The sum of the pair is $4$ and the product is $4 + 1 = 5$, so the monic equation is $x^2 - 4x + 5 = 0$. Verify by substitution, using $(2 - i)^2 = 4 - 4i + i^2 = 3 - 4i$.

$(3 - 4i) - 4(2 - i) + 5 = 3 - 4i - 8 + 4i + 5 = 0$

The imaginary parts cancel, confirming the root.

Set the discriminant negative and solve for $k$.

$(k + 1)^2 - 16 < 0 \quad\Longleftrightarrow\quad (k + 1)^2 < 16 \quad\Longleftrightarrow\quad -4 < k + 1 < 4$

Subtracting $1$ throughout gives $-5 < k < 3$.

Use $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2$ with sum $-2$ and product $5$.

$r_1^2 + r_2^2 = (-2)^2 - 2(5) = 4 - 10 = -6$

A sum of squares can only be negative because the roots are non-real: they are $-1 \pm 2i$, whose squares $-3 \mp 4i$ add to $-6$.

First find the roots: $\Delta = 4 - 20 = -16$, so $x = \tfrac{2 \pm 4i}{2} = 1 \pm 2i$. Each root $r$ contributes a factor $(x - r)$.

$x^2 - 2x + 5 = \big(x - (1 + 2i)\big)\big(x - (1 - 2i)\big) = (x - 1 - 2i)(x - 1 + 2i)$

Expanding back: the sum of the roots is $2$ and the product is $1 + 4 = 5$, matching the coefficients.

Write $z = p + qi$. Then $z + \overline{z} = 2p = -4$ gives $p = -2$, and $z - \overline{z} = 2qi = 6i$ gives $q = 3$.

$x^2 - 2px + \left(p^2 + q^2\right) = x^2 + 4x + (4 + 9) = x^2 + 4x + 13 = 0$

Check: the discriminant is $16 - 52 = -36 < 0$ and the roots are $-2 \pm 3i$, matching both given conditions.

Use the symmetric-function identity $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2$ with sum $4$ and product $13$.

$r_1^2 + r_2^2 = 4^2 - 2(13) = 16 - 26 = -10$

A negative answer is impossible for real roots, but these roots are $2 \pm 3i$: their squares are $-5 \pm 12i$, and adding them gives $-10$ exactly.

For $\Delta < 0$ the roots are $p \pm qi$ with $q = \dfrac{\sqrt{|\Delta|}}{2a}$, and here $a = 1$.

$q = \frac{\sqrt{20}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$

So whatever the linear coefficient is, the roots sit $\sqrt{5}$ above and below the real axis.

Subtract the conjugates: the real parts cancel and the imaginary parts add.

$r_1 - r_2 = (3 + 2i) - (3 - 2i) = 4i, \qquad (4i)^2 = -16$

This equals the discriminant $\Delta = 36 - 52 = -16$, matching the identity $(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2 = 36 - 52$.

The two numbers are the roots of $x^2 - 10x + 40 = 0$. Its discriminant is $\Delta = 100 - 160 = -60$, so $\sqrt{-60} = 2i\sqrt{15}$.

$x = \frac{10 \pm 2i\sqrt{15}}{2} = 5 \pm i\sqrt{15}$

Check the product: $25 + 15 = 40$. This is Cardano's famous 1545 problem: no real pair works, but the conjugate pair does.