Quadratic, Polynomial, Radical, and Absolute Value Equations: Precalculus Study Guide

Quadratic Equations

Solving Quadratic Equations by the Square Root Property

Quadratic equations can often be solved by isolating the squared term and applying the square root property.

• Square Root Property: If $x^2 = a$, then $x = \sqrt{a}$ or $x = -\sqrt{a}$.

• Example: Solve $x^2 = 100$.

  • Take the square root of both sides: $x = \sqrt{100}$ or $x = -\sqrt{100}$

  • Thus, $x = 10$ or $x = -10$

Checking Solutions: Substitute each solution into the original equation to verify.

Solving Quadratic Equations by the Quadratic Formula

The quadratic formula provides solutions to any quadratic equation of the form $ax^2 + bx + c = 0$.

• Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Example: Solve $7x^2 + 6x + 1 = 0$.

  • Identify $a = 7$, $b = 6$, $c = 1$

  • Plug into the formula: $x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 7 \cdot 1}}{2 \cdot 7}$

  • Simplify: $x = \frac{-6 \pm \sqrt{36 - 28}}{14} = \frac{-6 \pm \sqrt{8}}{14}$

Factoring and the Zero-Product Principle

Factoring is a method for solving polynomial equations by expressing the equation as a product of factors and applying the zero-product principle.

• Zero-Product Principle: If $ab = 0$, then $a = 0$ or $b = 0$.

• Example: $25x^2 - 36 = 0$ can be factored as $(5x - 6)(5x + 6) = 0$.

• Set each factor to zero: $5x - 6 = 0$ or $5x + 6 = 0$

• Solutions: $x = \frac{6}{5}$ or $x = -\frac{6}{5}$

Solving Rational Equations

Equations with Rational Expressions

Rational equations involve fractions with polynomials in the numerator and denominator. To solve, obtain a common denominator and simplify.

• Example: $\frac{2x}{x^2 - 8x} + \frac{16}{x^2 - 8x} = \frac{158}{x^2 - 64}$

• Find a common denominator, combine terms, and set the equation to zero.

• Factor and apply the zero-product principle to solve for $x$.

Applications of Quadratic Equations

Geometric Applications

Quadratic equations can model geometric problems, such as finding the side length of a square given its area after a change.

• Example: If each side of a square is lengthened by 6 inches and the new area is 49 square inches, find the original side length.

• Let $x$ be the original side length. $(x + 6)^2 = 49$

• Solve for $x$: $x + 6 = 7$ or $x + 6 = -7$; $x = 1$ (since side length cannot be negative)

Physical Applications

Quadratic equations can be used to solve real-world problems, such as determining the depth of a gutter with a given cross-sectional area.

• Example: The cross-sectional area of a gutter is $40$square inches. The width is $30 - 2x$, and the depth is $x$.

• Set up the equation: $x(30 - 2x) = 40$

• Rewrite in standard form: $-2x^2 + 30x - 40 = 0$

• Use the quadratic formula to solve for $x$.

Polynomial Equations and Factoring

Factoring by Grouping

Some polynomial equations can be solved by grouping terms and factoring.

• Example: $16y^2 - 5y - 80y^2 = 0$

• Group and factor: $16y^2 - 80y^2 - 5y = 0$

• Factor out $y$ and solve using the zero-product principle.

Radical Equations

Solving Equations with Square Roots

Radical equations involve variables inside a root. Isolate the radical and square both sides to eliminate it.

• Example: $\sqrt{9x} = 9x + 5$

• Isolate and square both sides: $9x = (9x + 5)^2$

• Expand and solve the resulting quadratic equation.

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

Equations with Rational Exponents

Solving Equations with Fractional Powers

Equations with rational exponents can be solved by raising both sides to the reciprocal of the exponent.

• Example: $(x - 7)^{3/2} = 64$

• Raise both sides to the $2/3$ power: $x - 7 = 64^{2/3}$

• Solve for $x$.

Equations Quadratic in Form

Substitution Method

Some equations are quadratic in form, where a variable and its square appear. Substitute $u = x^2$ to reduce to a quadratic equation.

• Example: $25x^4 = 106x^2 - 81$

• Let $u = x^2$, rewrite as $25u^2 - 106u + 81 = 0$

• Solve for $u$, then back-substitute to solve for $x$.

Absolute Value Equations

Solving Absolute Value Equations

Isolate the absolute value expression and solve for the variable. If the absolute value equals a positive number, set up two equations.

• Example: $|9x - 4| = 12$

• Set $9x - 4 = 12$ or $9x - 4 = -12$

• Solve each equation for $x$.

Summary Table: Methods for Solving Equations

Equation Type	Method	Key Formula
Quadratic	Factoring, Square Root, Quadratic Formula	$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Rational	Common Denominator, Zero-Product Principle	Set denominator equal, solve numerator
Radical	Isolate Radical, Square Both Sides	$(\sqrt{a})^2 = a$
Rational Exponent	Raise Both Sides to Reciprocal Power	$(x^a)^b = x^{ab}$
Absolute Value	Set Up Two Equations	$|x| = a \implies x = a$ or $x = -a$

Additional info: These notes expand on the original problems by providing definitions, formulas, and general methods for solving each type of equation, suitable for Precalculus students preparing for exams.