	Algebra Tutorials!

Thursday 21st of November



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	Use of Parentheses or Brackets (The Distributive Law)
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	Use of Parentheses or Brackets (The Distributive Law)
	Simplifying Complex Fractions 1
	Adding Fractions
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	Solutions to Linear Equations in Two Variables
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	Rise and Run
	Graphing Exponential Functions
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	The Cartesian Coordinate System
	Writing the Terms of a Polynomial in Descending Order
	Fractions
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	Solving Inequalities
	Solving Rational Inequalities with a Sign Graph
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	Simplifying Rational Expressions
	Exponents
	Intercepts of a Line
	Completing the Square
	Order of Operations
	Factoring Trinomials
	Solving Linear Equations
	Solving Multi-Step Inequalities
	Solving Quadratic Equations Graphically and Algebraically
	Collecting Like Terms
	Solving Equations with Radicals and Exponents
	Percent of Change
	Powers of ten (Scientific Notation)
	Comparing Integers on a Number Line
	Solving Systems of Equations Using Substitution
	Factoring Out the Greatest Common Factor
	Families of Functions
	Monomial Factors
	Multiplying and Dividing Complex Numbers
	Properties of Exponents
	Multiplying Square Roots
	Radicals
	Adding or Subtracting Rational Expressions with Different Denominators
	Expressions with Variables as Exponents
	The Quadratic Formula
	Writing a Quadratic with Given Solutions
	Simplifying Square Roots
	Adding and Subtracting Square Roots
	Adding and Subtracting Rational Expressions
	Combining Like Radical Terms
	Solving Systems of Equations Using Substitution
	Dividing Polynomials
	Graphing Functions
	Product of a Sum and a Difference
	Solving First Degree Inequalities
	Solving Equations with Radicals and Exponents
	Roots and Powers
	Multiplying Numbers

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Solving Equations with Radicals and Exponents

Raising Each Side to a Power

Example 1

Raising each side to a power to eliminate radicals

Solve each equation.

a)		Original equation
		Cube each side.
3x + 5	= x - 1
2x	= -6
x	= -3

Check x = -3 in the original equation:

Note that is a real number. The solution set is {-3}. In this example we checked for arithmetic mistakes. There was no possibility of extraneous solutions here because we raised each side to an odd power.

b)	= x				Original equation
3x + 18	= x²				Square both sides.
-x² + 3x + 18	= x²				Simplify.
x² - 3x - 18	= 0				Subtract x² from each side to get zero on one side.
(x - 6)(x + 3)	= 0				Multiply each side by -1 for easier factoring.
x - 6	= 0				Factor.
x	= 0	or	x + 3	= 0	Zero factor property
	= 6	or	x	= -3

Because we squared both sides, we must check for extraneous solutions. If x = -3 in the original equation , we get

which is not correct. If x = 6 in the original equation, we get which is correct. The solution set is {6}.