Key Precalculus Equations and Problem Types: Quadratics, Radicals, Inequalities, and Absolute Value

Equations and Inequalities

Quadratic Equations with Complex Numbers

Quadratic equations are equations of the form ax2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. When the discriminant (b2 - 4ac) is negative, the solutions are complex numbers.

• Quadratic Formula: The solutions to ax2 + bx + c = 0 are given by: $x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a }$

• Complex Solutions: If b^2 - 4ac < 0, then the solutions are: $x = \frac{ -b \pm i\sqrt{ |b^2 - 4ac| } }{ 2a }$ where i is the imaginary unit, defined as i = \sqrt{ -1 }.

• Example: Solve x2 + 4x + 5 = 0.

  • Discriminant: 16 - 20 = -4

  • Solutions: $x = \frac{ -4 \pm i\sqrt{4} }{ 2 } = -2 \pm i$

Radical Equations

Radical equations are equations in which the variable appears under a radical sign (most commonly a square root). Solving these equations often involves isolating the radical and then raising both sides to a power to eliminate the radical.

• General Form: $\sqrt{ f(x) } = g(x)$

• Solving Steps:

  1. Isolate the radical on one side of the equation.

  2. Raise both sides to the appropriate power to eliminate the radical.

  3. Solve the resulting equation.

  4. Check all solutions in the original equation to avoid extraneous solutions.

• Example: Solve $\sqrt{ x + 3 } = x - 1$

  • Square both sides: $x + 3 = (x - 1)^2$

  • Expand: $x + 3 = x^2 - 2x + 1$

  • Rearrange: $0 = x^2 - 3x - 2$

  • Factor: $(x - 2)(x - 1) = 0$

  • Potential solutions: $x = 2, x = 1$

  • Check in original: Only $x = 2$ is valid.

Compound Linear Inequalities

Compound inequalities involve two or more inequalities joined by the words "and" or "or." They are solved by finding the solution set that satisfies all (for "and") or at least one (for "or") of the inequalities.

• "And" Compound Inequality: $a < x < b$ (intersection of solutions)

• "Or" Compound Inequality: $x < a$ or $x > b$ (union of solutions)

• Solving Steps:

  1. Solve each inequality separately.

  2. Combine the solution sets appropriately.

• Example: Solve $-2 < 3x + 1 < 7$

  • Subtract 1: $-3 < 3x < 6$

  • Divide by 3: $-1 < x < 2$

Absolute Value Equations

Absolute value equations involve expressions of the form |A| = B. The absolute value of a number is its distance from zero on the number line, so these equations often have two solutions.

• General Principle: $|A| = B$ implies $A = B$ or $A = -B$ (for $B \geq 0$)

• Solving Steps:

  1. Isolate the absolute value expression.

  2. Set up two equations: one with the positive value, one with the negative.

  3. Solve both equations.

• Example: Solve $|2x - 5| = 7$

  • Case 1: $2x - 5 = 7 \implies x = 6$

  • Case 2: $2x - 5 = -7 \implies x = -1$

  • Solutions: $x = 6, x = -1$

Word Problems Involving Equations and Inequalities

Word problems require translating a real-world scenario into an equation or inequality, then solving for the unknown. This process involves identifying variables, setting up relationships, and interpreting the solution in context.

• Steps for Solving:

  1. Read the problem carefully and define variables.

  2. Write an equation or inequality that models the situation.

  3. Solve the equation or inequality.

  4. Check the solution in the context of the problem.

• Example: A number increased by 7 is equal to twice the number minus 3. Find the number.

  • Let the number be x.

  • Equation: $x + 7 = 2x - 3$

  • Solve: $x = 10$