Complex Roots of Quadratics: Practice

A negative radicand comes out of the radical as a factor of $i$, and $\sqrt{-k} = i\sqrt{k}$ for positive $k$.

$\sqrt{-25} = i\sqrt{25} = 5i$

Check by squaring: $(5i)^2 = 25\,i^2 = -25$, as required.

Pull out the $i$ first, then simplify the real radical by extracting the largest perfect square.

$\sqrt{-18} = i\sqrt{18} = i\sqrt{9 \cdot 2} = 3i\sqrt{2}$

Squaring back gives $(3i\sqrt{2})^2 = 9 \cdot (-1) \cdot 2 = -18$, so the answer checks.

Read off $a = 1$, $b = 2$, $c = 5$ and substitute into $\Delta = b^2 - 4ac$.

$\Delta = 2^2 - 4(1)(5) = 4 - 20 = -16$

The discriminant is $-16$, which is negative, so this equation has a conjugate pair of non-real roots.

A negative discriminant means the square root in the quadratic formula is the square root of a negative number, which is a pure imaginary number.

$x = -\frac{b}{2a} \pm \frac{\sqrt{9}}{2a}\,i = p \pm qi$

The two roots share a real part and have opposite imaginary parts, so they are a pair of complex conjugates, neither of them real.

The discriminant is $\Delta = 4 - 20 = -16$, so $\sqrt{-16} = 4i$.

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

Both terms of the numerator are divided by $2$, giving the conjugate pair $-1 + 2i$ and $-1 - 2i$.

The conjugate flips only the sign of the imaginary part and leaves the real part alone.

$\overline{3 - 7i} = 3 + 7i$

The real part stays $3$; only the $-7i$ becomes $+7i$.

Compute the discriminant: $\Delta = (-2)^2 - 4(1)(2) = 4 - 8 = -4$, so $\sqrt{-4} = 2i$.

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

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

Isolate the square, then take both square roots of $-9$.

$x^2 = -9 \quad\Longrightarrow\quad x = \pm\sqrt{-9} = \pm 3i$

Both $3i$ and $-3i$ square to $-9$, so the solution set is the conjugate pair $\pm 3i$.

First convert the radical: $\sqrt{-16} = 4i$. Then divide both terms of the numerator by $2$.

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

Dividing only one of the two terms by $2$ produces the wrong answers $3 \pm 4i$ or $6 \pm 2i$.

A complex number written as $p + qi$ has real part $p$ and imaginary part $q$.

$-4 \pm 3i = \underbrace{-4}_{p} \pm \underbrace{3}_{q}\,i$

Both roots share the real part $-4$, which is the axis of symmetry $-\tfrac{b}{2a}$ of the parabola.

Write the root in standard form to see its parts: $6i = 0 + 6i$.

$\overline{0 + 6i} = 0 - 6i = -6i$

Real coefficients force the roots into a conjugate pair, so the other root is $-6i$.

Over the complex numbers every quadratic factors into two linear factors.

$ax^2 + bx + c = a(x - r_1)(x - r_2)$

If $\Delta > 0$ the two roots are distinct reals, if $\Delta = 0$ the one real root counts twice, and if $\Delta < 0$ the roots are a conjugate pair. In every case the count with multiplicity is exactly $2$; the discriminant only decides where the roots live, not how many there are.