Define the imaginary unit i and operate with complex numbers (add, subtract, multiply); use the discriminant to determine the nature of a quadratic's roots; and solve quadratics with complex roots using the quadratic formula or completing the square.

What this topic is asking

The Regents Algebra II exam (the Complex Number System, N-CN, and Reasoning with Equations, A-REI, clusters) wants you to define the imaginary unit $i$, operate with complex numbers (add, subtract, multiply), use the discriminant to classify a quadratic's roots, and solve quadratics that have complex roots. Complex numbers extend Algebra I quadratics to the case where the discriminant is negative.

The imaginary unit and complex numbers

The square root of a negative number is not real, so mathematics defines the imaginary unit.

So $\sqrt{-9} = \sqrt{9}\sqrt{-1} = 3i$, and $\sqrt{-20} = 2i\sqrt{5}$. A complex number $a + bi$ has a real part $a$ and an imaginary part $b$. Powers of $i$ cycle: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and repeat.

Operating with complex numbers

Adding and subtracting combines like parts; multiplying treats $i$ as a variable, then substitutes $i^2 = -1$.

For multiplication, expand and simplify:

The step that turns $-3i^2$ into $+3$ is where the real part grows, and skipping it is the most common error.

The discriminant and the nature of roots

The discriminant $b^2 - 4ac$ of a quadratic $ax^2 + bx + c = 0$ tells you how many real roots there are, and now whether the roots are complex.

A negative discriminant means the parabola never crosses the x-axis, so the roots are complex. They always come as a conjugate pair: if $2 + 3i$ is a root, so is $2 - 3i$.

Reading it for credit

A clarifying point worth stressing is that the quadratic formula works unchanged when the discriminant is negative; you simply rewrite the negative under the radical using $i$. The Regents often awards a credit for stating, from the discriminant alone, that a negative value produces two complex conjugate roots, before any solving. A second habit that protects credit is to always replace $i^2$ with $-1$ and present the answer in the standard $a + bi$ form, reduced; leaving $i^2$ in the expression or an unsimplified fraction loses the final credit even when the work is right.

A further idea connects complex roots back to graphs. Because a quadratic with a negative discriminant has no real zeros, its parabola never touches the x-axis: it sits entirely above (if it opens up) or below (if it opens down) the axis. This is the graphical meaning of "complex roots", and the Regents sometimes tests it by giving a parabola that misses the x-axis and asking what that says about the discriminant (it must be negative). Completing the square gives the same complex roots as the formula: writing $x^2 - 4x + 13 = (x - 2)^2 + 9 = 0$ leads to $(x - 2)^2 = -9$, so $x - 2 = \pm 3i$ and $x = 2 \pm 3i$, matching the worked example above.

Try this

Q1. Simplify $\sqrt{-49}$. [1 credit]

• Cue. $\sqrt{49}\sqrt{-1} = 7i$.

Q2. What does a discriminant of $-12$ tell you about the roots? [1 credit]

• Cue. Negative discriminant means two complex conjugate roots.