Chapter 5: Radical Functions & Rational Exponents
Radicals and rational exponents extend our ability to work with powers beyond whole numbers. Whether you are solving a physics equation involving the period of a pendulum ($T = 2\pi\sqrt{\frac{L}{g}}$), computing the side length of a square from its area, or modeling population growth with fractional exponents, these tools are indispensable. In this chapter we develop a rigorous understanding of rational exponents, learn to simplify and operate on radical expressions, solve radical equations (watching carefully for extraneous solutions), graph the family of radical functions, and conclude with an introduction to complex numbers—the number system that finally allows every polynomial equation to have a solution.
5.1 Rational Exponents
Up to this point, you have worked with integer exponents: $x^2$, $x^{-3}$, $x^0 = 1$. Now we extend the definition to allow exponents that are fractions. The goal is to define $a^{m/n}$ in a way that preserves all the familiar exponent rules—the product rule, power rule, and quotient rule—while also connecting cleanly to radicals.
Definition: Rational Exponent
For any real number $a \geq 0$ and positive integers $m$ and $n$ (with $n \geq 2$), the rational exponent is defined as:
$$a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$$
In particular, $a^{1/n} = \sqrt[n]{a}$ is the principal $n$th root of $a$.
When $a < 0$ and $n$ is odd, we define $a^{1/n} = -\sqrt[n]{|a|}$. When $a < 0$ and $n$ is even, $a^{1/n}$ is not a real number.
Why does this definition make sense? Consider $a^{1/2}$. If the usual exponent rules hold, then $(a^{1/2})^2 = a^{(1/2) \cdot 2} = a^1 = a$. So $a^{1/2}$ must be the number whose square is $a$—which is exactly $\sqrt{a}$. The same reasoning extends: $(a^{1/n})^n = a$, confirming that $a^{1/n} = \sqrt[n]{a}$.
Properties of Rational Exponents
For $a, b > 0$ and rational exponents $r$ and $s$:
$$a^r \cdot a^s = a^{r+s} \qquad \frac{a^r}{a^s} = a^{r-s} \qquad (a^r)^s = a^{rs}$$
$$(ab)^r = a^r \cdot b^r \qquad \left(\frac{a}{b}\right)^r = \frac{a^r}{b^r} \qquad a^{-r} = \frac{1}{a^r}$$
Example 5.1.1: Converting Between Radicals and Rational Exponents
Rewrite each expression in the other form:
(a) $\sqrt[3]{x^5}$ (b) $y^{-2/3}$ (c) $\left(\sqrt[4]{7}\right)^3$
Solution:
(a) $\sqrt[3]{x^5} = x^{5/3}$
(b) $y^{-2/3} = \dfrac{1}{y^{2/3}} = \dfrac{1}{\sqrt[3]{y^2}}$
(c) $\left(\sqrt[4]{7}\right)^3 = 7^{3/4}$
Example 5.1.2: Simplifying Expressions with Rational Exponents
Simplify completely. Assume all variables represent positive real numbers.
(a) $\dfrac{x^{3/4} \cdot x^{1/2}}{x^{1/4}}$ (b) $(8a^6)^{2/3}$ (c) $\left(\dfrac{27}{64}\right)^{-2/3}$
Solution:
(a) Add exponents in the numerator, then subtract the denominator exponent:
$$\frac{x^{3/4} \cdot x^{1/2}}{x^{1/4}} = \frac{x^{3/4 + 2/4}}{x^{1/4}} = \frac{x^{5/4}}{x^{1/4}} = x^{5/4 - 1/4} = x^{4/4} = x$$
(b) Apply the power to each factor:
$$(8a^6)^{2/3} = 8^{2/3} \cdot (a^6)^{2/3} = (\sqrt[3]{8})^2 \cdot a^{6 \cdot 2/3} = 2^2 \cdot a^4 = 4a^4$$
(c) First handle the negative exponent, then simplify:
$$\left(\frac{27}{64}\right)^{-2/3} = \left(\frac{64}{27}\right)^{2/3} = \frac{64^{2/3}}{27^{2/3}} = \frac{(\sqrt[3]{64})^2}{(\sqrt[3]{27})^2} = \frac{4^2}{3^2} = \frac{16}{9}$$
Tip: When simplifying $a^{m/n}$, it is usually easier to take the root first, then raise to the power: $(\sqrt[n]{a})^m$ rather than $\sqrt[n]{a^m}$. Taking the root first keeps the numbers smaller and more manageable.
5.2 Simplifying Radicals
Simplifying radical expressions relies on two fundamental properties that follow directly from the exponent rules established in Section 5.1. These properties let us break apart and recombine radicals, just as the distributive property lets us break apart products and sums.
Product Rule for Radicals
For $a \geq 0$ and $b \geq 0$ (and $a, b$ real when $n$ is odd):
$$\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$$
Quotient Rule for Radicals
For $a \geq 0$ and $b > 0$:
$$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$$
A radical expression is in simplest form when:
- No perfect $n$th-power factors remain under the radical sign.
- No fractions appear under the radical sign.
- No radicals appear in the denominator of a fraction.
Example 5.2.1: Simplifying Square Roots
Simplify: (a) $\sqrt{72}$ (b) $\sqrt{50x^3y^4}$ (c) $\sqrt[3]{54a^5}$
Solution:
(a) Factor out the largest perfect square: $72 = 36 \cdot 2$
$$\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}$$
(b) Separate into perfect-square and remaining factors:
$$\sqrt{50x^3y^4} = \sqrt{25 \cdot 2 \cdot x^2 \cdot x \cdot y^4} = 5xy^2\sqrt{2x}$$
(c) For cube roots, look for perfect cube factors: $54 = 27 \cdot 2$ and $a^5 = a^3 \cdot a^2$
$$\sqrt[3]{54a^5} = \sqrt[3]{27 \cdot 2 \cdot a^3 \cdot a^2} = 3a\sqrt[3]{2a^2}$$
Rationalizing the Denominator
When a radical appears in the denominator, we rationalize the denominator by multiplying numerator and denominator by an appropriate expression that eliminates the radical from the denominator. This process does not change the value of the expression—it only rewrites it in an equivalent, standard form.
Example 5.2.2: Rationalizing with a Single Radical
Rationalize the denominator: (a) $\dfrac{5}{\sqrt{3}}$ (b) $\dfrac{2}{\sqrt[3]{9}}$
Solution:
(a) Multiply by $\dfrac{\sqrt{3}}{\sqrt{3}}$:
$$\frac{5}{\sqrt{3}} = \frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$$
(b) We need the radicand to become a perfect cube. Since $9 = 3^2$, multiply by $\dfrac{\sqrt[3]{3}}{\sqrt[3]{3}}$:
$$\frac{2}{\sqrt[3]{9}} = \frac{2}{\sqrt[3]{3^2}} \cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}} = \frac{2\sqrt[3]{3}}{\sqrt[3]{3^3}} = \frac{2\sqrt[3]{3}}{3}$$
Rationalizing with Conjugates
When the denominator contains a sum or difference involving a square root, such as $a + \sqrt{b}$, we multiply by its conjugate $a - \sqrt{b}$. This exploits the difference-of-squares pattern: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$, which contains no radical.
Definition: Conjugate
The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$, and vice versa. More generally, the conjugate of $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$. The product of conjugate pairs always produces a rational expression.
Example 5.2.3: Rationalizing with Conjugates
Rationalize the denominator: $\dfrac{3}{2 + \sqrt{5}}$
Solution: Multiply numerator and denominator by the conjugate $2 - \sqrt{5}$:
$$\frac{3}{2 + \sqrt{5}} \cdot \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{3(2 - \sqrt{5})}{(2)^2 - (\sqrt{5})^2} = \frac{3(2 - \sqrt{5})}{4 - 5} = \frac{3(2 - \sqrt{5})}{-1}$$
$$= -3(2 - \sqrt{5}) = -6 + 3\sqrt{5}$$
5.3 Operations with Radicals
Adding and subtracting radicals is analogous to combining like terms in polynomial algebra. We can combine $3x + 5x = 8x$ because the variable parts match; similarly, we can combine $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$ because the radical parts match. Radicals with the same index and the same radicand are called like radicals.
Adding and Subtracting Like Radicals
$$a\sqrt[n]{c} + b\sqrt[n]{c} = (a + b)\sqrt[n]{c}$$
Only radicals with the same index and same radicand can be combined directly. Always simplify each radical first, since terms that initially look different may become like radicals after simplification.
Example 5.3.1: Adding and Subtracting Radicals
Simplify: (a) $3\sqrt{12} + 5\sqrt{27} - 2\sqrt{48}$ (b) $\sqrt[3]{16} + 5\sqrt[3]{2}$
Solution:
(a) Simplify each radical first:
$$3\sqrt{12} = 3\sqrt{4 \cdot 3} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}$$
$$5\sqrt{27} = 5\sqrt{9 \cdot 3} = 5 \cdot 3\sqrt{3} = 15\sqrt{3}$$
$$2\sqrt{48} = 2\sqrt{16 \cdot 3} = 2 \cdot 4\sqrt{3} = 8\sqrt{3}$$
Now combine like radicals:
$$6\sqrt{3} + 15\sqrt{3} - 8\sqrt{3} = (6 + 15 - 8)\sqrt{3} = 13\sqrt{3}$$
(b) Simplify: $\sqrt[3]{16} = \sqrt[3]{8 \cdot 2} = 2\sqrt[3]{2}$. Then:
$$2\sqrt[3]{2} + 5\sqrt[3]{2} = 7\sqrt[3]{2}$$
Multiplying Radical Expressions
Multiplying radicals uses the product rule in the forward direction, combined with the distributive property (or FOIL for binomials). After multiplying, always simplify the result.
Example 5.3.2: Multiplying Radical Expressions
Expand and simplify: (a) $\sqrt{3}(\sqrt{6} + 2\sqrt{3})$ (b) $(3 + \sqrt{5})(2 - \sqrt{5})$
Solution:
(a) Distribute $\sqrt{3}$:
$$\sqrt{3} \cdot \sqrt{6} + \sqrt{3} \cdot 2\sqrt{3} = \sqrt{18} + 2\sqrt{9} = \sqrt{9 \cdot 2} + 2 \cdot 3 = 3\sqrt{2} + 6$$
(b) Apply FOIL:
$$(3 + \sqrt{5})(2 - \sqrt{5}) = 6 - 3\sqrt{5} + 2\sqrt{5} - (\sqrt{5})^2$$
$$= 6 - 3\sqrt{5} + 2\sqrt{5} - 5 = 1 - \sqrt{5}$$
Example 5.3.3: Special Products with Radicals
Expand: $(4 - \sqrt{7})^2$
Solution: Use $(a - b)^2 = a^2 - 2ab + b^2$:
$$(4 - \sqrt{7})^2 = 4^2 - 2(4)(\sqrt{7}) + (\sqrt{7})^2 = 16 - 8\sqrt{7} + 7 = 23 - 8\sqrt{7}$$
Tip: The product of conjugate radical expressions always yields a rational number: $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$. This is the key idea behind rationalizing denominators and is also useful for simplifying products.
5.4 Radical Equations
A radical equation is an equation in which the variable appears under a radical sign. The primary strategy for solving such equations is to isolate the radical on one side and then raise both sides to the appropriate power to eliminate it. However, this squaring (or cubing) step can introduce extraneous solutions—values that satisfy the transformed equation but not the original. Checking all solutions in the original equation is therefore mandatory.
Solving Radical Equations
Step 1: Isolate the radical on one side of the equation.
Step 2: Raise both sides to the $n$th power (where $n$ is the index of the radical).
Step 3: Solve the resulting equation.
Step 4: Check every solution in the original equation to eliminate extraneous solutions.
Caution: Squaring both sides of an equation is not a reversible operation. If $a = b$, then $a^2 = b^2$, but the converse is false (for example, $-3 \neq 3$ even though $(-3)^2 = 3^2$). This is why extraneous solutions can appear, and why checking is essential—not optional.
Example 5.4.1: A Basic Radical Equation
Solve: $\sqrt{2x + 3} = 5$
Solution: The radical is already isolated. Square both sides:
$$(\sqrt{2x + 3})^2 = 5^2$$
$$2x + 3 = 25$$
$$2x = 22 \implies x = 11$$
Check: $\sqrt{2(11) + 3} = \sqrt{25} = 5$. Confirmed.
$$\boxed{x = 11}$$
Example 5.4.2: A Radical Equation with an Extraneous Solution
Solve: $\sqrt{x + 6} = x$
Solution: Square both sides:
$$x + 6 = x^2$$
$$x^2 - x - 6 = 0$$
$$(x - 3)(x + 2) = 0 \implies x = 3 \text{ or } x = -2$$
Check $x = 3$: $\sqrt{3 + 6} = \sqrt{9} = 3 = x$. Valid.
Check $x = -2$: $\sqrt{-2 + 6} = \sqrt{4} = 2 \neq -2$. Extraneous.
$$\boxed{x = 3}$$
Equations with Two Radicals
When an equation contains two radical terms, isolate one radical first, square both sides, and then repeat the process if a radical remains.
Example 5.4.3: An Equation with Two Radicals
Solve: $\sqrt{3x + 1} - \sqrt{x + 4} = 1$
Solution: Isolate one radical:
$$\sqrt{3x + 1} = 1 + \sqrt{x + 4}$$
Square both sides:
$$3x + 1 = 1 + 2\sqrt{x + 4} + (x + 4)$$
$$3x + 1 = x + 5 + 2\sqrt{x + 4}$$
$$2x - 4 = 2\sqrt{x + 4}$$
$$x - 2 = \sqrt{x + 4}$$
Square again:
$$x^2 - 4x + 4 = x + 4$$
$$x^2 - 5x = 0$$
$$x(x - 5) = 0 \implies x = 0 \text{ or } x = 5$$
Check $x = 0$: $\sqrt{1} - \sqrt{4} = 1 - 2 = -1 \neq 1$. Extraneous.
Check $x = 5$: $\sqrt{16} - \sqrt{9} = 4 - 3 = 1$. Valid.
$$\boxed{x = 5}$$
Example 5.4.4: A Cube Root Equation
Solve: $\sqrt[3]{2x - 1} = 3$
Solution: Cube both sides (no extraneous solution risk for odd roots):
$$2x - 1 = 27$$
$$2x = 28 \implies x = 14$$
Check: $\sqrt[3]{2(14) - 1} = \sqrt[3]{27} = 3$. Confirmed.
$$\boxed{x = 14}$$
Tip: When you square both sides and a radical remains, do not panic. Simply isolate the remaining radical and square again. With practice this two-step squaring procedure becomes routine. Also remember: cube root equations (odd index) do not produce extraneous solutions, since cubing is a one-to-one operation.
5.5 Graphing Radical Functions
A radical function is a function that contains a variable under a radical. The most fundamental radical function is $f(x) = \sqrt{x}$, whose graph is the top half of a sideways parabola. In this section we study how transformations shift, stretch, and reflect the graph of $y = \sqrt{x}$ to produce the general family $y = a\sqrt{x - h} + k$, and we extend our discussion to higher-index roots.
The Parent Function $y = \sqrt{x}$
The graph of $y = \sqrt{x}$ has the following key features:
Domain: $[0, \infty)$ — we can only take the square root of non-negative numbers.
Range: $[0, \infty)$ — the principal square root is always non-negative.
Key points: $(0, 0)$, $(1, 1)$, $(4, 2)$, $(9, 3)$.
The graph starts at the origin, increases throughout its domain, and is concave down (it rises more and more slowly).
Transformations of $y = a\sqrt{x - h} + k$
Starting from $y = \sqrt{x}$:
$h$ shifts the graph horizontally: right if $h > 0$, left if $h < 0$.
$k$ shifts the graph vertically: up if $k > 0$, down if $k < 0$.
$a$ controls the vertical stretch/compression and reflection:
- $|a| > 1$: vertical stretch by factor $|a|$
- $0 < |a| < 1$: vertical compression by factor $|a|$
- $a < 0$: reflection across the $x$-axis
The starting point (vertex) of the graph is $(h, k)$.
Domain: $[h, \infty)$ when $a > 0$; $[h, \infty)$ when $a < 0$. Range: $[k, \infty)$ when $a > 0$; $(-\infty, k]$ when $a < 0$.
Example 5.5.1: Graphing a Transformed Square Root Function
Describe the transformations and state the domain and range of $f(x) = -2\sqrt{x + 3} - 1$.
Solution: Rewrite in standard form: $f(x) = -2\sqrt{x - (-3)} + (-1)$, so $a = -2$, $h = -3$, $k = -1$.
- Shift left 3 units (since $h = -3$)
- Shift down 1 unit (since $k = -1$)
- Vertical stretch by factor 2 and reflection across the $x$-axis (since $a = -2$)
Starting point: $(-3, -1)$
Domain: $[-3, \infty)$ Range: $(-\infty, -1]$
Plot additional points: when $x = -2$, $f(-2) = -2\sqrt{1} - 1 = -3$. When $x = 1$, $f(1) = -2\sqrt{4} - 1 = -5$.
Use the interactive graph below to explore how the parameters $a$, $h$, and $k$ transform the square root function.
Higher-Index Root Functions
The cube root function $y = \sqrt[3]{x}$ behaves differently from the square root: its domain and range are both $(-\infty, \infty)$ because cube roots of negative numbers are defined. The graph passes through the origin and is symmetric about it (it is an odd function). More generally, $y = x^{1/n}$ for even $n$ resembles the square root (restricted domain), while for odd $n$ it resembles the cube root (unrestricted domain).
5.6 Complex Numbers
Throughout your algebra courses, certain equations have had "no real solution." The equation $x^2 = -1$ is the simplest example: no real number squared gives $-1$. Mathematicians resolved this by inventing a new kind of number. The resulting complex number system extends the real numbers and guarantees that every polynomial equation has a solution (the Fundamental Theorem of Algebra).
Definition: The Imaginary Unit
The imaginary unit $i$ is defined by:
$$i = \sqrt{-1} \qquad \text{equivalently,} \qquad i^2 = -1$$
For any positive real number $b$, we define $\sqrt{-b} = i\sqrt{b}$.
Definition: Complex Number
A complex number is a number of the form
$$z = a + bi$$
where $a$ and $b$ are real numbers. The number $a$ is called the real part (written $\operatorname{Re}(z) = a$) and $b$ is called the imaginary part (written $\operatorname{Im}(z) = b$).
If $b = 0$, the number is real. If $a = 0$ and $b \neq 0$, the number is pure imaginary.
The powers of $i$ follow a repeating cycle of length 4:
$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \quad i^6 = -1, \quad \ldots$$
To evaluate $i^n$ for any positive integer $n$, divide $n$ by 4 and use the remainder: $i^n = i^{n \bmod 4}$ (with $i^0 = 1$).
Operations with Complex Numbers
Arithmetic of Complex Numbers
Let $z_1 = a + bi$ and $z_2 = c + di$.
Addition: $(a + bi) + (c + di) = (a + c) + (b + d)i$
Subtraction: $(a + bi) - (c + di) = (a - c) + (b - d)i$
Multiplication: $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$
(Use FOIL and replace $i^2$ with $-1$.)
Example 5.6.1: Adding, Subtracting, and Multiplying Complex Numbers
Let $z_1 = 3 + 2i$ and $z_2 = 1 - 4i$. Compute:
(a) $z_1 + z_2 = (3 + 1) + (2 + (-4))i = 4 - 2i$
(b) $z_1 - z_2 = (3 - 1) + (2 - (-4))i = 2 + 6i$
(c) $z_1 \cdot z_2 = (3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i - 8(-1) = 11 - 10i$
Complex Conjugates and Division
Definition: Complex Conjugate
The complex conjugate of $z = a + bi$ is $\bar{z} = a - bi$. The product of a complex number and its conjugate is always a non-negative real number:
$$z \cdot \bar{z} = (a + bi)(a - bi) = a^2 + b^2$$
To divide complex numbers, we multiply numerator and denominator by the conjugate of the denominator. This is exactly the same strategy we used for rationalizing denominators with radical conjugates in Section 5.2.
Example 5.6.2: Dividing Complex Numbers
Write $\dfrac{5 + 3i}{2 - i}$ in $a + bi$ form.
Solution: Multiply numerator and denominator by the conjugate $2 + i$:
$$\frac{5 + 3i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{(5 + 3i)(2 + i)}{(2 - i)(2 + i)}$$
Numerator: $(5 + 3i)(2 + i) = 10 + 5i + 6i + 3i^2 = 10 + 11i + 3(-1) = 7 + 11i$
Denominator: $(2)^2 + (1)^2 = 4 + 1 = 5$
$$\frac{7 + 11i}{5} = \frac{7}{5} + \frac{11}{5}i$$
Example 5.6.3: Simplifying Square Roots of Negative Numbers
Simplify: (a) $\sqrt{-48}$ (b) $\sqrt{-3} \cdot \sqrt{-12}$
Solution:
(a) $\sqrt{-48} = i\sqrt{48} = i\sqrt{16 \cdot 3} = 4i\sqrt{3}$
(b) Important: Convert to $i$-form first, then multiply. Do not use $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ when $a$ and $b$ are both negative.
$$\sqrt{-3} \cdot \sqrt{-12} = (i\sqrt{3})(i\sqrt{12}) = i^2 \cdot \sqrt{36} = (-1)(6) = -6$$
Caution: The product rule $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ is only valid when $a \geq 0$ and $b \geq 0$. Writing $\sqrt{-3} \cdot \sqrt{-12} = \sqrt{36} = 6$ is incorrect. Always express square roots of negative numbers in terms of $i$ before performing any operations.
Tip: Complex numbers connect many topics in this chapter. The conjugate technique for dividing complex numbers is the same idea as rationalizing radical denominators with conjugates. Recognizing these structural parallels deepens your understanding and makes both procedures feel natural.
5.7 Practice Problems
Work through each problem, then expand the solution to check your work. For full benefit, attempt each problem on paper before looking at the answer.
Problem 1
Simplify: $\left(16x^8 y^{12}\right)^{3/4}$
Show Solution
Apply the exponent to each factor:
$$16^{3/4} \cdot (x^8)^{3/4} \cdot (y^{12})^{3/4}$$
$$= (\sqrt[4]{16})^3 \cdot x^{8 \cdot 3/4} \cdot y^{12 \cdot 3/4}$$
$$= 2^3 \cdot x^6 \cdot y^9$$
$$\boxed{8x^6 y^9}$$
Problem 2
Simplify: $\dfrac{x^{-1/3} \cdot x^{5/6}}{x^{1/2}}$
Show Solution
Combine exponents using a common denominator of 6:
$$\frac{x^{-1/3} \cdot x^{5/6}}{x^{1/2}} = \frac{x^{-2/6 + 5/6}}{x^{3/6}} = \frac{x^{3/6}}{x^{3/6}} = x^{3/6 - 3/6} = x^0 = 1$$
$$\boxed{1}$$
Problem 3
Simplify: $\sqrt{200a^5 b^2}$
Show Solution
Factor out perfect squares: $200 = 100 \cdot 2$, $a^5 = a^4 \cdot a$, $b^2$ is already a perfect square.
$$\sqrt{200a^5 b^2} = \sqrt{100 \cdot 2 \cdot a^4 \cdot a \cdot b^2} = 10a^2 b\sqrt{2a}$$
$$\boxed{10a^2 b\sqrt{2a}}$$
Problem 4
Rationalize the denominator: $\dfrac{6}{3 - \sqrt{7}}$
Show Solution
Multiply by the conjugate $\dfrac{3 + \sqrt{7}}{3 + \sqrt{7}}$:
$$\frac{6}{3 - \sqrt{7}} \cdot \frac{3 + \sqrt{7}}{3 + \sqrt{7}} = \frac{6(3 + \sqrt{7})}{9 - 7} = \frac{6(3 + \sqrt{7})}{2}$$
$$= 3(3 + \sqrt{7})$$
$$\boxed{9 + 3\sqrt{7}}$$
Problem 5
Simplify: $2\sqrt{45} - 3\sqrt{20} + \sqrt{80}$
Show Solution
Simplify each radical:
$2\sqrt{45} = 2\sqrt{9 \cdot 5} = 6\sqrt{5}$
$3\sqrt{20} = 3\sqrt{4 \cdot 5} = 6\sqrt{5}$
$\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}$
Combine: $6\sqrt{5} - 6\sqrt{5} + 4\sqrt{5} = 4\sqrt{5}$
$$\boxed{4\sqrt{5}}$$
Problem 6
Expand and simplify: $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})$
Show Solution
This is a difference of squares: $(a + b)(a - b) = a^2 - b^2$.
$$(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$$
$$\boxed{1}$$
Problem 7
Solve: $\sqrt{5x - 1} = x - 1$
Show Solution
Square both sides:
$$5x - 1 = (x - 1)^2 = x^2 - 2x + 1$$
$$0 = x^2 - 7x + 2$$
Wait—let us redo this carefully:
$$5x - 1 = x^2 - 2x + 1$$
$$0 = x^2 - 2x + 1 - 5x + 1 = x^2 - 7x + 2$$
Using the quadratic formula: $x = \dfrac{7 \pm \sqrt{49 - 8}}{2} = \dfrac{7 \pm \sqrt{41}}{2}$
$x_1 = \dfrac{7 + \sqrt{41}}{2} \approx 6.70$ and $x_2 = \dfrac{7 - \sqrt{41}}{2} \approx 0.30$
Check $x_1 \approx 6.70$: $\sqrt{5(6.70) - 1} = \sqrt{32.5} \approx 5.70$ and $6.70 - 1 = 5.70$. Valid.
Check $x_2 \approx 0.30$: $\sqrt{5(0.30) - 1} = \sqrt{0.5} \approx 0.707$ and $0.30 - 1 = -0.70$. Extraneous (LHS is positive, RHS is negative).
$$\boxed{x = \frac{7 + \sqrt{41}}{2}}$$
Problem 8
Solve: $\sqrt{x + 7} + \sqrt{x} = 7$
Show Solution
Isolate one radical: $\sqrt{x + 7} = 7 - \sqrt{x}$
Square both sides:
$$x + 7 = 49 - 14\sqrt{x} + x$$
$$7 = 49 - 14\sqrt{x}$$
$$14\sqrt{x} = 42$$
$$\sqrt{x} = 3 \implies x = 9$$
Check: $\sqrt{9 + 7} + \sqrt{9} = \sqrt{16} + 3 = 4 + 3 = 7$. Valid.
$$\boxed{x = 9}$$
Problem 9
State the domain, range, and starting point of $g(x) = 3\sqrt{x - 2} + 4$.
Show Solution
This is in the form $y = a\sqrt{x - h} + k$ with $a = 3$, $h = 2$, $k = 4$.
Starting point: $(h, k) = (2, 4)$
Domain: $x - 2 \geq 0 \implies x \geq 2$, so the domain is $[2, \infty)$.
Range: Since $a = 3 > 0$, the function opens upward from the starting point: $[4, \infty)$.
$$\boxed{\text{Starting point: } (2, 4); \quad \text{Domain: } [2, \infty); \quad \text{Range: } [4, \infty)}$$
Problem 10
Simplify: $(4 - 3i)(2 + 5i)$
Show Solution
Use FOIL:
$$(4)(2) + (4)(5i) + (-3i)(2) + (-3i)(5i)$$
$$= 8 + 20i - 6i - 15i^2$$
$$= 8 + 14i - 15(-1)$$
$$= 8 + 14i + 15$$
$$\boxed{23 + 14i}$$
Problem 11
Write in $a + bi$ form: $\dfrac{2 + i}{3 - 2i}$
Show Solution
Multiply numerator and denominator by the conjugate $3 + 2i$:
$$\frac{(2 + i)(3 + 2i)}{(3 - 2i)(3 + 2i)} = \frac{6 + 4i + 3i + 2i^2}{9 + 4} = \frac{6 + 7i + 2(-1)}{13}$$
$$= \frac{4 + 7i}{13} = \frac{4}{13} + \frac{7}{13}i$$
$$\boxed{\frac{4}{13} + \frac{7}{13}i}$$
Problem 12
Evaluate $i^{43}$.
Show Solution
Divide the exponent by 4: $43 = 4 \cdot 10 + 3$, so the remainder is 3.
$$i^{43} = i^3 = -i$$
$$\boxed{-i}$$