Key Terms, Definitions, and Rules
| Term |
Definition |
Rules / Outcomes |
|
Rational Exponent
\(b^{m/n}\)
|
A fractional exponent. The denominator is the index of the root; the numerator is the power applied to the base.
|
\(b^{m/n} = \left(\sqrt[n]{(b)}\right)^m = \sqrt[n]{(b^m)}\)
Order: root and power can be applied in either order — same result.
|
|
Discriminant
\(b^2 - 4ac\)
|
The expression under the radical in the quadratic formula. Determines the number and type of solutions to \(ax^2+bx+c=0\).
|
If \(b^2-4ac > 0\): 2 real solutions
If \(b^2-4ac = 0\): 1 real solution
If \(b^2-4ac < 0\): 0 real solutions (2 complex)
|
|
Imaginary Unit
\(i\)
|
A number defined so that \(i^2 = -1\). It is the square root of \(-1\). It is not a real number.
|
\(\sqrt{(-a)} = \sqrt{(a)} \cdot i\) for \(a > 0\)
\(i^2 = -1\) — replace every \(i^2\) in a product with \(-1\)
|
|
Complex Number
\(a + bi\)
|
Standard form with real part \(a\) and imaginary part \(bi\), where \(a\) and \(b\) are real numbers.
|
Add/Subtract: combine real parts, combine imaginary parts
Multiply: FOIL, then replace \(i^2\) with \(-1\)
|
|
Square Root Equation
\(\sqrt{(f(x))} = k\)
|
An equation with a variable inside a square root. Solve by isolating the radical, then squaring both sides.
|
If \(k \geq 0\): possible solution — always check for extraneous
If \(k < 0\): no real solution (\(\sqrt{\phantom{x}}\) is never negative)
|
|
Cube Root Equation
\(\sqrt[3]{(f(x))} = k\)
|
An equation with a variable inside a cube root. Solve by isolating the radical, then cubing both sides.
|
Always exactly 1 real solution for any value of \(k\)
No extraneous solutions — cube root equations are always valid
|
|
Completing the Square
\((x+p)^2 = q\)
|
Rewriting a quadratic so one side is a perfect square. Then take square roots of both sides to solve.
|
If \(q > 0\): 2 real solutions \(x = -p \pm \sqrt{(q)}\)
If \(q = 0\): 1 real solution \(x = -p\)
If \(q < 0\): 2 complex solutions \(x = -p \pm i\sqrt{(-q)}\)
|
|
Quadratic Formula
\(\dfrac{-b \pm \sqrt{(b^2-4ac)}}{2a}\)
|
A formula that gives all solutions to \(ax^2+bx+c=0\) directly from its coefficients.
|
Works for every quadratic equation
Sign of \(b^2-4ac\) determines solution type (see Discriminant row)
When \(b^2-4ac < 0\): \(\sqrt{(b^2-4ac)} = i\sqrt{(|b^2-4ac|)}\)
|
Problem 1 — Rational Exponents
Compute \(4^{3/2}\) using two methods.
Worked Solution
1
Read the exponent: denominator = index of the root; numerator = power to apply.
In \(4^{3/2}\): denominator 2 = square root; numerator 3 = cube the result.
4^(3/2) — Root First
\(4^{3/2} = \left(\sqrt{(4)}\right)^3 = 2^3 = 8\)
4^(3/2) — Power First
\(4^{3/2} = \sqrt{(4^3)} = \sqrt{(64)} = 8\)
Answer
\(4^{3/2} = 8\). Both methods produce the same result.
Key Insight
When we see a fraction as the exponent — like \(\frac{3}{2}\) — the numerator on top applies a power (3) to the base (4) to make \(4^3\). The denominator on the bottom applies a root: since 2 is on the bottom, it is a square root. So \(4^{3/2} = \sqrt{(4^3)}\).
Problem 2 — Discriminant
How many real solutions does \(x^2 + 2x + 5 = 0\) have?
Worked Solution
1
Identify a, b, c.
\(x^2 + 2x + 5 = 0 \implies a = 1,\; b = 2,\; c = 5\)
2
Compute the discriminant.
\(b^2 - 4ac = (2)^2 - 4(1)(5)\)
\(= 4 - 20\)
\(= -16\)
The discriminant is the expression under the radical in the quadratic formula. It tells you the number and type of solutions.
3
Interpret the sign.
Since \(-16 < 0\), the square root \(\sqrt{(-16)}\) is not a real number.
A negative discriminant means no real solutions — both solutions are complex (imaginary).
Answer
0 real solutions. The discriminant is \(-16 < 0\), so both solutions are complex.
Key Insight — Discriminant Summary
\(b^2 - 4ac > 0\) → two real solutions |
\(b^2 - 4ac = 0\) → one real solution |
\(b^2 - 4ac < 0\) → zero real solutions (two complex)
Problem 3 — Complex Solutions from a Perfect Square
Find all solutions, real or complex, to \((x - 1)^2 = -4\).
Worked Solution
1
Take the square root of both sides.
\((x-1)^2 = -4\)
\(x - 1 = \pm\sqrt{(-4)}\)
The two solutions come from the ± because squaring a number and squaring its negative both give the same result.
2
Simplify the square root of a negative number.
\(\sqrt{(-4)} = \sqrt{(4)} \cdot i = 2i\)
Factor out \(i\) and simplify the positive part. \(\sqrt{(4)} = 2\).
3
Substitute and solve for x.
\(x - 1 = \pm 2i\)
\(x - 1 + 1 = \pm 2i + 1\)
\(x = 1 \pm 2i\)
4
Write both solutions separately.
\(x = 1 + 2i \qquad \text{and} \qquad x = 1 - 2i\)
5
Verify by checking one solution.
Check \(x = 1 + 2i\):
\((x-1)^2 = (1+2i-1)^2 = (2i)^2 = 4i^2 = 4(-1) = -4\) ✓
Answer
\(x = 1 + 2i\) and \(x = 1 - 2i\).
Problem 4 — Complex Number Arithmetic
Let \(p = 3 + 2i\) and \(q = 1 - 5i\). Write each expression in the form \(a + bi\).
Part a — p + q
Worked Solution
1
Write the addition.
\(p + q = (3 + 2i) + (1 - 5i)\)
2
Group real parts and imaginary parts.
\(= (3 + 1) + (2 + (-5))i\)
Real parts: 3 and 1. Imaginary parts: 2i and −5i.
Answer\(p + q = 4 - 3i\)
Part b — p − q
Worked Solution
1
Write the subtraction — distribute the negative sign.
\(p - q = (3 + 2i) - (1 - 5i) = (3 + 2i) + (-1 + 5i)\)
Subtracting a complex number is the same as adding its opposite. Flip the sign of every term in q.
2
Group and combine.
\(= (3 + (-1)) + (2 + 5)i\)
\(= 2 + 7i\)
Answer\(p - q = 2 + 7i\)
Part c — pq
Worked Solution
1
Write the multiplication and distribute (FOIL).
\(pq = (3 + 2i)(1 - 5i)\)
\(= (3)(1) + (3)(-5i) + (2i)(1) + (2i)(-5i)\)
2
Compute each term.
\((3)(1) = 3\)
\((3)(-5i) = -15i\)
\((2i)(1) = 2i\)
\((2i)(-5i) = -10i^2\)
3
Replace i² with −1.
\(-10i^2 = -10(-1) = +10\)
This is the key step. Every \(i^2\) in the product becomes \(-1\), which turns an imaginary term into a real term.
4
Combine real parts and imaginary parts.
Real: \(3 + 10 = 13\)
Imaginary: \(-15i + 2i = -13i\)
\(pq = 13 - 13i\)
Answer\(pq = 13 - 13i\)
Key Insight
The term \(i^2 = -1\) is what makes complex multiplication different. Without it, you would just be distributing variables. Replacing \(i^2\) with \(-1\) converts one of the imaginary terms into a real number, which changes the real part of the answer.
Problem 5 — Solving Radical Equations
Part a
Solve \(\sqrt{(2x+3)} - 5 = -2\).
Worked Solution
1
Isolate the radical — move the constant to the right side.
\(\sqrt{(2x+3)} - 5 = -2\)
\(\sqrt{(2x+3)} - 5 + 5 = -2 + 5\)
\(\sqrt{(2x+3)} = 3\)
2
Square both sides to eliminate the radical.
\(\left(\sqrt{(2x+3)}\right)^2 = 3^2\)
\(2x + 3 = 9\)
Squaring undoes the square root on the left side.
3
Solve the resulting linear equation.
\(2x + 3 - 3 = 9 - 3\)
\(2x = 6\)
\(x = 3\)
4
Check the solution in the original equation.
\(\sqrt{(2(3)+3)} - 5 = \sqrt{(9)} - 5 = 3 - 5 = -2\) ✓
Always check when solving radical equations — squaring both sides can introduce extraneous (false) solutions.
Answer\(x = 3\)
Part b
Explain why \(\sqrt{(2x+3)} + 5 = -2\) has no real solution.
Reasoning
1
Isolate the radical.
\(\sqrt{(2x+3)} + 5 = -2\)
\(\sqrt{(2x+3)} + 5 - 5 = -2 - 5\)
\(\sqrt{(2x+3)} = -7\)
2
Ask: can a square root equal a negative number?
For any real number \(x\), the square root \(\sqrt{(2x+3)} \geq 0\).
There is no real value of \(x\) that makes \(\sqrt{(2x+3)} = -7\), because \(-7 < 0\).
The square root of a real number is always zero or positive. It can never equal a negative number.
Try Saying
This equation has no real solution because \(\sqrt{(2x+3)}\) can never equal \(-7\), since the square root of any real number is always greater than or equal to zero.
Key Insight — Why These Two Equations Differ
Both equations have the same radicand \(\sqrt{(2x+3)}\) and the same right-hand side \(-2\), but the sign on 5 is different.
In Part a, subtracting 5 isolates the radical as \(\sqrt{(2x+3)} = 3\). Since 3 is positive, a solution exists.
In Part b, adding 5 isolates the radical as \(\sqrt{(2x+3)} = -7\). Since \(-7\) is negative and square roots are never negative, no solution exists.
Problem 6 — Cube Roots
Part a
The graph of \(g(x) = \sqrt[3]{(x)}\) is a strictly increasing curve — it always goes up from left to right, and it passes through every height exactly once.
Use this to explain why there is only one real solution to every equation of the form \(\sqrt[3]{(x)} = a\).
Reasoning
1
Interpret the equation as an intersection.
The equation \(\sqrt[3]{(x)} = a\) asks: at what \(x\)-value does the graph of \(g(x) = \sqrt[3]{(x)}\) reach height \(a\)?
2
Use the strictly increasing property.
Because the graph always increases, a horizontal line at any height \(y = a\) crosses the graph at exactly one point.
A function that is strictly increasing never repeats the same output. Each horizontal line hits it once and only once.
Try Saying
The equation \(\sqrt[3]{(x)} = a\) has exactly one real solution because the graph of \(g(x) = \sqrt[3]{(x)}\) is strictly increasing — it crosses every horizontal line exactly once.
Part b
Use the meaning of cube roots to find an exact solution to \(\sqrt[3]{(x+1)} = 2\).
Worked Solution
1
State what the cube root means.
If \(\sqrt[3]{(x+1)} = 2\), then \(x + 1\) is the number whose cube root is \(2\).
2
Apply the inverse — cube both sides.
\(\left(\sqrt[3]{(x+1)}\right)^3 = 2^3\)
\(x + 1 = 8\)
Cubing is the inverse of taking a cube root. It removes the radical.
3
Solve for x.
\(x + 1 - 1 = 8 - 1\)
\(x = 7\)
Answer\(x = 7\)
Part c
Use the meaning of cube roots to find an exact solution to \(\sqrt[3]{(x)} - 1 = 2\).
Notice the Difference
In Part b, the "+1" was inside the cube root: \(\sqrt[3]{(x+1)}\).
In Part c, the "−1" is outside the cube root: \(\sqrt[3]{(x)} - 1\).
The steps are slightly different because of where the constant is located.
Worked Solution
1
Isolate the cube root.
\(\sqrt[3]{(x)} - 1 = 2\)
\(\sqrt[3]{(x)} - 1 + 1 = 2 + 1\)
\(\sqrt[3]{(x)} = 3\)
2
Cube both sides.
\(\left(\sqrt[3]{(x)}\right)^3 = 3^3\)
\(x = 27\)
Answer\(x = 27\)
Key Insight — Cube Roots vs. Square Roots
Cube root equations always have exactly one real solution — even when the right side is negative — because every real number has exactly one real cube root. (This is different from square root equations, where a negative right side means no real solution.)
Problem 7 — Solving x² − 4x + 5 = 0 Two Ways
Solve \(x^2 - 4x + 5 = 0\) using two different methods.
Part a — Method 1: Completing the Square
Worked Solution
1
Move the constant to the right side.
\(x^2 - 4x + 5 = 0\)
\(x^2 - 4x + 5 - 5 = 0 - 5\)
\(x^2 - 4x = -5\)
2
Complete the square — add \(\left(\dfrac{-4}{2}\right)^2 = 4\) to both sides.
\(x^2 - 4x + 4 = -5 + 4\)
\((x - 2)^2 = -1\)
Half of −4 is −2. Squaring gives 4. Adding 4 to both sides keeps the equation balanced.
3
Take the square root of both sides.
\(x - 2 = \pm\sqrt{(-1)} = \pm i\)
\(\sqrt{(-1)} = i\) by definition.
4
Add 2 to both sides and write both solutions.
\(x - 2 + 2 = 2 \pm i\)
\(x = 2 + i \qquad \text{and} \qquad x = 2 - i\)
Answer\(x = 2 + i\) and \(x = 2 - i\)
Part b — Method 2: Quadratic Formula
Worked Solution
1
Identify a, b, and c.
\(x^2 - 4x + 5 = 0 \Rightarrow a = 1,\ b = -4,\ c = 5\)
2
Substitute into the quadratic formula.
\(x = \dfrac{-(-4) \pm \sqrt{((-4)^2 - 4(1)(5))}}{2(1)}\)
3
Compute the discriminant.
\((-4)^2 - 4(1)(5) = 16 - 20 = -4\)
So: \(x = \dfrac{4 \pm \sqrt{(-4)}}{2}\)
4
Simplify \(\sqrt{(-4)}\) using i.
\(\sqrt{(-4)} = \sqrt{(4)} \cdot i = 2i\)
So: \(x = \dfrac{4 \pm 2i}{2}\)
5
Divide every term in the numerator by 2.
\(x = \dfrac{4}{2} \pm \dfrac{2i}{2} = 2 \pm i\)
Answer\(x = 2 + i\) and \(x = 2 - i\)
Key Insight — Both Methods, Same Answer
Completing the square and the quadratic formula are equivalent — one is derived from the other. Both reach the same solutions: \(x = 2 + i\) and \(x = 2 - i\). These are complex conjugates — same real part, opposite imaginary parts. Complex solutions always come in conjugate pairs.
Try Saying
The solutions are complex because the discriminant \(b^2 - 4ac = -4\) is negative, which means the square root in the formula produces an imaginary number.