Literal Equations, Inverse Operations, and Solving Radical Equations

Factoring and Literal Equations

• Common Denominators and Unfactorable Expressions:

  • A common denominator must be factored first before finding the least common denominator.

  • If an algebraic expression within an equation cannot be factored (for example, $2x + 7$), treat the entire expression as a single prime unit by enclosing it in parentheses $(2x + 7)$.

  • To form a common denominator when an expression is unfactorable, multiply that entire grouped quantity by the other denominator terms.

• Solving Literal Equations (Rearranging Formulas):

  • To isolate a specified variable in a literal formula such as $s = p + prt$ solved for $t$:

  • Step 1: Isolate the term containing the target variable by subtracting non-target terms from both sides:       $s - p = prt$

  • Step 2: Divide both sides by all factors multiplied by the target variable (in this case, $pr$):       $t = \frac{s - p}{pr}$

Inverse Operations and Radical Equations

• Fundamental Principle of Inverse Operations:

  • Every algebraic operation possesses an inverse operation that undoes it:

  • Multiplication is undone by division (e.g., $3x = 9 \implies x = 3$).

  • Subtraction is undone by addition (e.g., $x - 7 = 1 \implies x = 8$).

  • Taking a square root is undone by squaring (raising the expression to the second power).

• Concept of Squaring a Radical:

  • Multiplying any algebraic or numeric term by itself is equivalent to squaring that term:

  • $5 \times 5 = 5^2$

  • $7 \times 7 = 7^2$

  • $x \times x = x^2$

  • $\sqrt{2} \times \sqrt{2} = (\sqrt{2})^2 = 2$

  • Any square root expression multiplied by itself cancels out the radical sign, leaving only the radicand (the expression underneath the radical):

  • $\sqrt{47911380024} \times \sqrt{47911380024} = 47911380024$

  • $(\sqrt{\text{elephant}})^2 = \text{elephant}$

  • $(\sqrt{x - 3})^2 = x - 3$

Solving Basic Radical Equations

• General Procedure for Solving Equations with One Radical:

  • Step 1: Isolate the radical expression on one side of the equation using standard addition, subtraction, multiplication, or division operations.

  • Step 2: Square both sides of the equation to cancel the square root.

  • Step 3: Solve the resulting linear or quadratic equation for the variable.

  • Step 4: Check candidate solutions in the original equation to eliminate extraneous solutions.

• Worked Example 1: Basic Radical Equation:

  • Equation to solve:     $3\sqrt{x + 2} - 4 = 5$

  • Step 1: Add $4$ to both sides to isolate the radical term:     $3\sqrt{x + 2} = 9$

  • Step 2: Divide both sides by $3$ to leave the radical alone on the left side:     $\sqrt{x + 2} = 3$

  • Step 3: Square both sides of the equation:     $(\sqrt{x + 2})^2 = 3^2$     $x + 2 = 9$

  • Step 4: Subtract $2$ from both sides to find the candidate solution:     $x = 7$

Extraneous Solutions and Mandatory Checking

• The Nature of Extraneous Solutions:

  • Squaring both sides of an equation is a mathematically necessary operation to eliminate radicals, but it creates the potential for extraneous solutions—candidate values that emerge from correct algebraic steps but do not satisfy the original equation.

  • Checking candidate solutions is a mandatory step for every radical equation.

  • Candidate solutions must be substituted back into the original equation before transformation. Checking in intermediate steps will fail to identify extraneous solutions created by squaring.

• Verification of Worked Example 1:

  • Substitute $x = 7$ into the original equation $3\sqrt{x + 2} - 4 = 5$:     $3\sqrt{7 + 2} - 4 = 5$     $3\sqrt{9} - 4 = 5$     $3(3) - 4 = 5$     $9 - 4 = 5$     $5 = 5$

  • Since $5 = 5$ is a true identity, $x = 7$ is confirmed as a valid solution.

Advanced Radical Equations with Variable Expressions

• Rule for Squaring Binomials:

  • Squaring a binomial (an expression with two terms) always yields a trinomial (an expression with three terms):     $(a - b)^2 = a^2 - 2ab + b^2$

  • Never square terms individually inside a binomial. Always expand using FOIL ($\text{First}, \text{Outer}, \text{Inner}, \text{Last}$):     $(x - 2)^2 = (x - 2)(x - 2) = x^2 - 2x - 2x + 4 = x^2 - 4x + 4$

• Worked Example 2: Radical Equation Equated to a Variable Expression:

  • Equation to solve:     $\sqrt{2x - 1} + 2 = x$

  • Step 1: Isolate the square root by subtracting $2$ from both sides:     $\sqrt{2x - 1} = x - 2$

  • Step 2: Square both sides of the entire equation:     $(\sqrt{2x - 1})^2 = (x - 2)^2$     $2x - 1 = x^2 - 4x + 4$

  • Step 3: Rearrange the quadratic equation to equal zero ($ax^2 + bx + c = 0$):

  • Subtract $2x$ from both sides:       $-1 = x^2 - 6x + 4$

  • Add $1$ to both sides:       $0 = x^2 - 6x + 5$

  • Step 4: Solve the quadratic equation by factoring:

  • Identify two numbers that multiply to $5$ and add to $-6$: $-1$ and $-5$     $(x - 5)(x - 1) = 0$

  • Apply the zero-product property:       $x - 5 = 0 \implies x = 5$       $x - 1 = 0 \implies x = 1$

• Verification of Candidate Solutions for Example 2:

  • Test candidate solution $x = 5$ in original equation $\sqrt{2x - 1} + 2 = x$:     $\sqrt{2(5) - 1} + 2 = 5$     $\sqrt{10 - 1} + 2 = 5$     $\sqrt{9} + 2 = 5$     $3 + 2 = 5$     $5 = 5$

  • Result: $x = 5$ is a valid solution.

  • Test candidate solution $x = 1$ in original equation $\sqrt{2x - 1} + 2 = x$:     $\sqrt{2(1) - 1} + 2 = 1$     $\sqrt{2 - 1} + 2 = 1$     $\sqrt{1} + 2 = 1$     $1 + 2 = 1$     $3 = 1$

  • Result: False ($3 \neq 1$). Therefore, $x = 1$ is an extraneous solution and must be discarded.

  • Final solution set: $x = 5$

Classification of Solution Sets for Radical Equations

• Possible Outcomes When Solving Radical Equations:

  • Single Valid Solution: One candidate solution satisfies the original equation, while any second candidate solution fails (is extraneous).

  • Two Valid Solutions: Both candidate solutions satisfy the original equation upon testing.

  • No Solution: All produced candidate solutions fail the verification test when substituted into the original equation.