Radical Functions and Equations - Ninth Grade Math
Introduction to Radical Functions
Radical Function: A function that contains a variable inside a radical (square root, cube root, etc.)
Square Root Function: A function with variable under a square root: $f(x) = \sqrt{x}$
Cube Root Function: A function with variable under a cube root: $f(x) = \sqrt[3]{x}$
Parent Functions: The simplest forms serving as building blocks
Square Root Function: A function with variable under a square root: $f(x) = \sqrt{x}$
Cube Root Function: A function with variable under a cube root: $f(x) = \sqrt[3]{x}$
Parent Functions: The simplest forms serving as building blocks
1. Evaluate Square Root Functions
Evaluating: Finding the output (y-value) for a given input (x-value)
Basic Square Root Function:
$$f(x) = \sqrt{x}$$
General Form:
$$f(x) = a\sqrt{x - h} + k$$
where:
• $a$ = vertical stretch/compression and reflection
• $h$ = horizontal shift
• $k$ = vertical shift
$$f(x) = \sqrt{x}$$
General Form:
$$f(x) = a\sqrt{x - h} + k$$
where:
• $a$ = vertical stretch/compression and reflection
• $h$ = horizontal shift
• $k$ = vertical shift
Steps to Evaluate:
Step 1: Substitute the given x-value into the function
Step 2: Simplify inside the radical first
Step 3: Take the square root
Step 4: Complete any operations outside the radical
Step 5: Simplify if needed
Step 1: Substitute the given x-value into the function
Step 2: Simplify inside the radical first
Step 3: Take the square root
Step 4: Complete any operations outside the radical
Step 5: Simplify if needed
Example 1: Evaluate $f(x) = \sqrt{x}$ when $x = 25$
$f(25) = \sqrt{25} = 5$
Answer: $5$
$f(25) = \sqrt{25} = 5$
Answer: $5$
Example 2: Evaluate $f(x) = \sqrt{x - 4}$ when $x = 13$
$f(13) = \sqrt{13 - 4} = \sqrt{9} = 3$
Answer: $3$
$f(13) = \sqrt{13 - 4} = \sqrt{9} = 3$
Answer: $3$
Example 3: Evaluate $g(x) = 2\sqrt{x} + 3$ when $x = 16$
$g(16) = 2\sqrt{16} + 3 = 2(4) + 3 = 8 + 3 = 11$
Answer: $11$
$g(16) = 2\sqrt{16} + 3 = 2(4) + 3 = 8 + 3 = 11$
Answer: $11$
Example 4: Evaluate $h(x) = \sqrt{2x + 1}$ when $x = 4$
$h(4) = \sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3$
Answer: $3$
$h(4) = \sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3$
Answer: $3$
Example 5: Evaluate $f(x) = -\sqrt{x - 3} + 7$ when $x = 12$
$f(12) = -\sqrt{12 - 3} + 7 = -\sqrt{9} + 7 = -3 + 7 = 4$
Answer: $4$
$f(12) = -\sqrt{12 - 3} + 7 = -\sqrt{9} + 7 = -3 + 7 = 4$
Answer: $4$
Important: The value under the square root must be non-negative (≥ 0) for real numbers!
2-3. Domain and Range of Square Root Functions
Domain: All possible input values (x-values) that make the function defined
Range: All possible output values (y-values) the function can produce
Range: All possible output values (y-values) the function can produce
For Parent Function $f(x) = \sqrt{x}$:
Domain: $[0, \infty)$ or $\{x | x \geq 0\}$
• Cannot take square root of negative numbers (in real numbers)
Range: $[0, \infty)$ or $\{y | y \geq 0\}$
• Square root always gives non-negative result
For General Form $f(x) = a\sqrt{x - h} + k$:
Domain: Find where expression under radical ≥ 0
$$x - h \geq 0 \implies x \geq h$$
Domain: $[h, \infty)$
Range: Depends on vertical shift and reflection
• If $a > 0$: Range is $[k, \infty)$ (opens up)
• If $a < 0$: Range is $(-\infty, k]$ (opens down)
Domain: $[0, \infty)$ or $\{x | x \geq 0\}$
• Cannot take square root of negative numbers (in real numbers)
Range: $[0, \infty)$ or $\{y | y \geq 0\}$
• Square root always gives non-negative result
For General Form $f(x) = a\sqrt{x - h} + k$:
Domain: Find where expression under radical ≥ 0
$$x - h \geq 0 \implies x \geq h$$
Domain: $[h, \infty)$
Range: Depends on vertical shift and reflection
• If $a > 0$: Range is $[k, \infty)$ (opens up)
• If $a < 0$: Range is $(-\infty, k]$ (opens down)
Finding Domain from Equations
Steps to Find Domain:
Step 1: Set expression inside radical ≥ 0
Step 2: Solve the inequality
Step 3: Write domain in interval notation or set notation
Step 1: Set expression inside radical ≥ 0
Step 2: Solve the inequality
Step 3: Write domain in interval notation or set notation
Example 1: Find domain and range of $f(x) = \sqrt{x - 5}$
Domain:
$x - 5 \geq 0$
$x \geq 5$
Domain: $[5, \infty)$
Range:
No vertical shift ($k = 0$), opens up ($a = 1 > 0$)
Range: $[0, \infty)$
Domain:
$x - 5 \geq 0$
$x \geq 5$
Domain: $[5, \infty)$
Range:
No vertical shift ($k = 0$), opens up ($a = 1 > 0$)
Range: $[0, \infty)$
Example 2: Find domain and range of $g(x) = \sqrt{x + 3} - 2$
Domain:
$x + 3 \geq 0$
$x \geq -3$
Domain: $[-3, \infty)$
Range:
Vertical shift down 2 ($k = -2$), opens up
Range: $[-2, \infty)$
Domain:
$x + 3 \geq 0$
$x \geq -3$
Domain: $[-3, \infty)$
Range:
Vertical shift down 2 ($k = -2$), opens up
Range: $[-2, \infty)$
Example 3: Find domain and range of $h(x) = -\sqrt{x - 1} + 4$
Domain:
$x - 1 \geq 0$
$x \geq 1$
Domain: $[1, \infty)$
Range:
Reflected over x-axis ($a = -1 < 0$), shifted up 4
Opens down, maximum at $y = 4$
Range: $(-\infty, 4]$
Domain:
$x - 1 \geq 0$
$x \geq 1$
Domain: $[1, \infty)$
Range:
Reflected over x-axis ($a = -1 < 0$), shifted up 4
Opens down, maximum at $y = 4$
Range: $(-\infty, 4]$
Example 4: Find domain of $f(x) = \sqrt{2x + 6}$
$2x + 6 \geq 0$
$2x \geq -6$
$x \geq -3$
Domain: $[-3, \infty)$
$2x + 6 \geq 0$
$2x \geq -6$
$x \geq -3$
Domain: $[-3, \infty)$
Example 5: Find domain of $g(x) = \sqrt{5 - x}$
$5 - x \geq 0$
$-x \geq -5$
$x \leq 5$ (inequality flips when dividing by negative)
Domain: $(-\infty, 5]$
$5 - x \geq 0$
$-x \geq -5$
$x \leq 5$ (inequality flips when dividing by negative)
Domain: $(-\infty, 5]$
4. Graph Square Root Functions
Shape: Half of a sideways parabola
Starting Point: Begins at the leftmost point of the domain
Direction: Increases to the right (if $a > 0$)
Starting Point: Begins at the leftmost point of the domain
Direction: Increases to the right (if $a > 0$)
Parent Function Graph $f(x) = \sqrt{x}$:
• Starting point: $(0, 0)$
• Domain: $[0, \infty)$
• Range: $[0, \infty)$
• Key points: $(0, 0)$, $(1, 1)$, $(4, 2)$, $(9, 3)$
• Shape: Curves upward and to the right
• Increases slowly as x increases
• Starting point: $(0, 0)$
• Domain: $[0, \infty)$
• Range: $[0, \infty)$
• Key points: $(0, 0)$, $(1, 1)$, $(4, 2)$, $(9, 3)$
• Shape: Curves upward and to the right
• Increases slowly as x increases
Steps to Graph Square Root Functions:
Step 1: Find the domain (starting x-value)
Step 2: Find the starting point (leftmost point)
Step 3: Create a table of values (use perfect squares)
Step 4: Plot the points
Step 5: Connect with smooth curve
Step 6: Draw arrow to show continuation
Step 1: Find the domain (starting x-value)
Step 2: Find the starting point (leftmost point)
Step 3: Create a table of values (use perfect squares)
Step 4: Plot the points
Step 5: Connect with smooth curve
Step 6: Draw arrow to show continuation
Example 1: Graph $f(x) = \sqrt{x - 2} + 1$
Domain: $x \geq 2$, so start at $x = 2$
Starting point: $(2, 1)$
Table of values:
Key points: $(2, 1)$, $(3, 2)$, $(6, 3)$, $(11, 4)$
Transformation: Shifted right 2, up 1 from parent
Domain: $x \geq 2$, so start at $x = 2$
Starting point: $(2, 1)$
Table of values:
| x | 2 | 3 | 6 | 11 |
|---|---|---|---|---|
| y | 1 | 2 | 3 | 4 |
Key points: $(2, 1)$, $(3, 2)$, $(6, 3)$, $(11, 4)$
Transformation: Shifted right 2, up 1 from parent
Example 2: Graph $g(x) = -\sqrt{x} + 3$
Starting point: $(0, 3)$
Reflected over x-axis, shifted up 3
Key points:
$(0, 3)$, $(1, 2)$, $(4, 1)$, $(9, 0)$
Opens downward to the right
Starting point: $(0, 3)$
Reflected over x-axis, shifted up 3
Key points:
$(0, 3)$, $(1, 2)$, $(4, 1)$, $(9, 0)$
Opens downward to the right
5. Graph Cube Root Functions
Cube Root Function: $f(x) = \sqrt[3]{x}$
Key Difference: Can accept negative inputs (domain is all real numbers)
Key Difference: Can accept negative inputs (domain is all real numbers)
Parent Cube Root Function:
$$f(x) = \sqrt[3]{x}$$
Properties:
• Domain: $(-\infty, \infty)$ (all real numbers)
• Range: $(-\infty, \infty)$ (all real numbers)
• Passes through origin: $(0, 0)$
• Shape: S-curve through origin
• Increasing everywhere
General Form:
$$f(x) = a\sqrt[3]{x - h} + k$$
$$f(x) = \sqrt[3]{x}$$
Properties:
• Domain: $(-\infty, \infty)$ (all real numbers)
• Range: $(-\infty, \infty)$ (all real numbers)
• Passes through origin: $(0, 0)$
• Shape: S-curve through origin
• Increasing everywhere
General Form:
$$f(x) = a\sqrt[3]{x - h} + k$$
Key Points for Parent $f(x) = \sqrt[3]{x}$:
Shape: Steeper near origin, flatter away from origin
| x | -8 | -1 | 0 | 1 | 8 |
|---|---|---|---|---|---|
| y | -2 | -1 | 0 | 1 | 2 |
Shape: Steeper near origin, flatter away from origin
Example 1: Graph $f(x) = \sqrt[3]{x + 2}$
Transformation: Shift LEFT 2 units
Domain: All real numbers
Center point: $(-2, 0)$ (where radicand = 0)
Key points:
$(-10, -2)$, $(-3, -1)$, $(-2, 0)$, $(-1, 1)$, $(6, 2)$
Transformation: Shift LEFT 2 units
Domain: All real numbers
Center point: $(-2, 0)$ (where radicand = 0)
Key points:
$(-10, -2)$, $(-3, -1)$, $(-2, 0)$, $(-1, 1)$, $(6, 2)$
Example 2: Graph $g(x) = \sqrt[3]{x} - 1$
Transformation: Shift DOWN 1 unit
Center point: $(0, -1)$
Key points:
$(-8, -3)$, $(-1, -2)$, $(0, -1)$, $(1, 0)$, $(8, 1)$
Transformation: Shift DOWN 1 unit
Center point: $(0, -1)$
Key points:
$(-8, -3)$, $(-1, -2)$, $(0, -1)$, $(1, 0)$, $(8, 1)$
Difference from Square Root:
• Cube root extends in BOTH directions (left and right)
• No domain restrictions
• Passes through negative x-values
• Cube root extends in BOTH directions (left and right)
• No domain restrictions
• Passes through negative x-values
6-7. Solve Radical Equations
Radical Equation: An equation with variable inside a radical
Goal: Isolate variable and solve
Key Method: Eliminate radical by raising both sides to appropriate power
Goal: Isolate variable and solve
Key Method: Eliminate radical by raising both sides to appropriate power
Power Property of Equality:
If $a = b$, then $a^n = b^n$
For square roots:
If $\sqrt{x} = a$, then $(\sqrt{x})^2 = a^2$, so $x = a^2$
IMPORTANT: Squaring both sides can introduce extraneous solutions!
Always check your answers in the original equation!
If $a = b$, then $a^n = b^n$
For square roots:
If $\sqrt{x} = a$, then $(\sqrt{x})^2 = a^2$, so $x = a^2$
IMPORTANT: Squaring both sides can introduce extraneous solutions!
Always check your answers in the original equation!
Steps to Solve Radical Equations:
Step 1: Isolate the radical on one side
Step 2: Raise both sides to the power of the index (square both sides for square roots)
Step 3: Solve the resulting equation
Step 4: Check all solutions in the original equation
Step 5: Discard any extraneous solutions
Step 1: Isolate the radical on one side
Step 2: Raise both sides to the power of the index (square both sides for square roots)
Step 3: Solve the resulting equation
Step 4: Check all solutions in the original equation
Step 5: Discard any extraneous solutions
Type 1: Single Radical (Simple)
Example 1: Solve $\sqrt{x} = 7$
Step 1: Radical already isolated
Step 2: Square both sides
$(\sqrt{x})^2 = 7^2$
$x = 49$
Step 3: Check
$\sqrt{49} = 7$ ✓
Answer: $x = 49$
Step 1: Radical already isolated
Step 2: Square both sides
$(\sqrt{x})^2 = 7^2$
$x = 49$
Step 3: Check
$\sqrt{49} = 7$ ✓
Answer: $x = 49$
Example 2: Solve $\sqrt{x - 3} = 5$
Square both sides:
$(\sqrt{x - 3})^2 = 5^2$
$x - 3 = 25$
$x = 28$
Check: $\sqrt{28 - 3} = \sqrt{25} = 5$ ✓
Answer: $x = 28$
Square both sides:
$(\sqrt{x - 3})^2 = 5^2$
$x - 3 = 25$
$x = 28$
Check: $\sqrt{28 - 3} = \sqrt{25} = 5$ ✓
Answer: $x = 28$
Example 3: Solve $\sqrt{2x + 1} = 3$
$2x + 1 = 9$
$2x = 8$
$x = 4$
Check: $\sqrt{2(4) + 1} = \sqrt{9} = 3$ ✓
Answer: $x = 4$
$2x + 1 = 9$
$2x = 8$
$x = 4$
Check: $\sqrt{2(4) + 1} = \sqrt{9} = 3$ ✓
Answer: $x = 4$
Type 2: Radical with Other Terms
Example 4: Solve $\sqrt{x} + 3 = 7$
Step 1: Isolate radical
$\sqrt{x} = 4$
Step 2: Square both sides
$x = 16$
Check: $\sqrt{16} + 3 = 4 + 3 = 7$ ✓
Answer: $x = 16$
Step 1: Isolate radical
$\sqrt{x} = 4$
Step 2: Square both sides
$x = 16$
Check: $\sqrt{16} + 3 = 4 + 3 = 7$ ✓
Answer: $x = 16$
Example 5: Solve $\sqrt{x - 2} - 5 = 0$
Isolate: $\sqrt{x - 2} = 5$
Square: $x - 2 = 25$
Solve: $x = 27$
Check: $\sqrt{27 - 2} - 5 = \sqrt{25} - 5 = 5 - 5 = 0$ ✓
Answer: $x = 27$
Isolate: $\sqrt{x - 2} = 5$
Square: $x - 2 = 25$
Solve: $x = 27$
Check: $\sqrt{27 - 2} - 5 = \sqrt{25} - 5 = 5 - 5 = 0$ ✓
Answer: $x = 27$
Type 3: Equations with Variables on Both Sides
Example 6: Solve $\sqrt{x} = x - 2$
Square both sides:
$(\sqrt{x})^2 = (x - 2)^2$
$x = x^2 - 4x + 4$
Rearrange:
$0 = x^2 - 5x + 4$
$0 = (x - 1)(x - 4)$
$x = 1$ or $x = 4$
Check $x = 1$: $\sqrt{1} = 1$ and $1 - 2 = -1$ → $1 \neq -1$ ✗
Check $x = 4$: $\sqrt{4} = 2$ and $4 - 2 = 2$ → $2 = 2$ ✓
Answer: $x = 4$ (x = 1 is extraneous)
Square both sides:
$(\sqrt{x})^2 = (x - 2)^2$
$x = x^2 - 4x + 4$
Rearrange:
$0 = x^2 - 5x + 4$
$0 = (x - 1)(x - 4)$
$x = 1$ or $x = 4$
Check $x = 1$: $\sqrt{1} = 1$ and $1 - 2 = -1$ → $1 \neq -1$ ✗
Check $x = 4$: $\sqrt{4} = 2$ and $4 - 2 = 2$ → $2 = 2$ ✓
Answer: $x = 4$ (x = 1 is extraneous)
Example 7: Solve $\sqrt{3x + 1} = x - 3$
Square:
$3x + 1 = (x - 3)^2$
$3x + 1 = x^2 - 6x + 9$
$0 = x^2 - 9x + 8$
$0 = (x - 1)(x - 8)$
$x = 1$ or $x = 8$
Check $x = 1$: $\sqrt{4} = 2$ and $1 - 3 = -2$ → $2 \neq -2$ ✗
Check $x = 8$: $\sqrt{25} = 5$ and $8 - 3 = 5$ → $5 = 5$ ✓
Answer: $x = 8$
Square:
$3x + 1 = (x - 3)^2$
$3x + 1 = x^2 - 6x + 9$
$0 = x^2 - 9x + 8$
$0 = (x - 1)(x - 8)$
$x = 1$ or $x = 8$
Check $x = 1$: $\sqrt{4} = 2$ and $1 - 3 = -2$ → $2 \neq -2$ ✗
Check $x = 8$: $\sqrt{25} = 5$ and $8 - 3 = 5$ → $5 = 5$ ✓
Answer: $x = 8$
Type 4: Two Radicals
Example 8: Solve $\sqrt{x} = \sqrt{x + 12}$
Square both sides:
$x = x + 12$
$0 = 12$
This is false! No solution.
Answer: No solution
Square both sides:
$x = x + 12$
$0 = 12$
This is false! No solution.
Answer: No solution
Example 9: Solve $\sqrt{x + 7} = \sqrt{2x - 3}$
Square:
$x + 7 = 2x - 3$
$10 = x$
Check: $\sqrt{17} = \sqrt{17}$ ✓
Answer: $x = 10$
Square:
$x + 7 = 2x - 3$
$10 = x$
Check: $\sqrt{17} = \sqrt{17}$ ✓
Answer: $x = 10$
Type 5: More Complex (Radical Equations II)
Example 10: Solve $\sqrt{x + 5} + 3 = x$
Isolate radical:
$\sqrt{x + 5} = x - 3$
Square both sides:
$x + 5 = (x - 3)^2$
$x + 5 = x^2 - 6x + 9$
$0 = x^2 - 7x + 4$
Use quadratic formula:
$x = \frac{7 \pm \sqrt{49 - 16}}{2} = \frac{7 \pm \sqrt{33}}{2}$
$x = \frac{7 + \sqrt{33}}{2} \approx 6.37$ or $x = \frac{7 - \sqrt{33}}{2} \approx 0.63$
Must check both in original equation!
Isolate radical:
$\sqrt{x + 5} = x - 3$
Square both sides:
$x + 5 = (x - 3)^2$
$x + 5 = x^2 - 6x + 9$
$0 = x^2 - 7x + 4$
Use quadratic formula:
$x = \frac{7 \pm \sqrt{49 - 16}}{2} = \frac{7 \pm \sqrt{33}}{2}$
$x = \frac{7 + \sqrt{33}}{2} \approx 6.37$ or $x = \frac{7 - \sqrt{33}}{2} \approx 0.63$
Must check both in original equation!
Square Root vs Cube Root Functions
| Feature | Square Root $f(x) = \sqrt{x}$ | Cube Root $f(x) = \sqrt[3]{x}$ |
|---|---|---|
| Domain | $[0, \infty)$ (non-negative only) | $(-\infty, \infty)$ (all real numbers) |
| Range | $[0, \infty)$ | $(-\infty, \infty)$ |
| Starting Point | $(0, 0)$ | Passes through $(0, 0)$ |
| Shape | Half sideways parabola | S-curve |
| Extends | Only to the right | Both left and right |
| Key Points | $(0,0)$, $(1,1)$, $(4,2)$, $(9,3)$ | $(-8,-2)$, $(-1,-1)$, $(0,0)$, $(1,1)$, $(8,2)$ |
Domain and Range Quick Reference
| Function | Domain | Range |
|---|---|---|
| $f(x) = \sqrt{x}$ | $[0, \infty)$ | $[0, \infty)$ |
| $f(x) = \sqrt{x - h}$ | $[h, \infty)$ | $[0, \infty)$ |
| $f(x) = \sqrt{x} + k$ | $[0, \infty)$ | $[k, \infty)$ |
| $f(x) = \sqrt{x - h} + k$ | $[h, \infty)$ | $[k, \infty)$ |
| $f(x) = -\sqrt{x - h} + k$ | $[h, \infty)$ | $(-\infty, k]$ |
| $f(x) = \sqrt[3]{x}$ | $(-\infty, \infty)$ | $(-\infty, \infty)$ |
Solving Radical Equations Checklist
| Step | Action | Example |
|---|---|---|
| 1 | Isolate the radical | $\sqrt{x} + 3 = 7$ → $\sqrt{x} = 4$ |
| 2 | Square both sides | $(\sqrt{x})^2 = 4^2$ → $x = 16$ |
| 3 | Solve resulting equation | If quadratic, factor or use formula |
| 4 | CHECK in original! | $\sqrt{16} + 3 = 4 + 3 = 7$ ✓ |
| 5 | Discard extraneous | If check fails, solution is extraneous |
Success Tips for Radical Functions and Equations:
✓ For square root functions: domain starts where radicand ≥ 0
✓ Cube root functions have no domain restrictions
✓ Range depends on vertical shift and reflection
✓ Square root graphs curve upward, cube roots make S-shape
✓ Always isolate radical BEFORE squaring
✓ Squaring can create extraneous solutions - MUST CHECK!
✓ Check ALL solutions in the ORIGINAL equation
✓ For domain: set expression under radical ≥ 0 and solve
✓ Use perfect squares/cubes for easier graphing points
✓ Remember: $\sqrt{x^2} = |x|$ but $\sqrt[3]{x^3} = x$
✓ For square root functions: domain starts where radicand ≥ 0
✓ Cube root functions have no domain restrictions
✓ Range depends on vertical shift and reflection
✓ Square root graphs curve upward, cube roots make S-shape
✓ Always isolate radical BEFORE squaring
✓ Squaring can create extraneous solutions - MUST CHECK!
✓ Check ALL solutions in the ORIGINAL equation
✓ For domain: set expression under radical ≥ 0 and solve
✓ Use perfect squares/cubes for easier graphing points
✓ Remember: $\sqrt{x^2} = |x|$ but $\sqrt[3]{x^3} = x$