	Algebra Tutorials!

	Saturday 21st of December



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	What the Standard Form of a Quadratic can tell you about the graph
	Simplifying Radical Expressions Containing One Term
	Adding and Subtracting Fractions
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	Adding and Subtracting Fractions
	Multiplying and Dividing With Square Roots
	Graphing Linear Inequalities
	Absolute Value Function
	Real Numbers and the Real Line
	Monomial Factors
	Raising an Exponential Expression to a Power
	Rational Exponents
	Multiplying Two Fractions Whose Numerators Are Both 1
	Multiplying Rational Expressions
	Building Up the Denominator
	Adding and Subtracting Decimals
	Solving Quadratic Equations
	Scientific Notation
	Like Radical Terms
	Graphing Parabolas
	Subtracting Reverses
	Solving Linear Equations
	Dividing Rational Expressions
	Complex Numbers
	Solving Linear Inequalities
	Working with Fractions
	Graphing Linear Equations
	Simplifying Expressions That Contain Negative Exponents
	Rationalizing the Denominator
	Decimals
	Estimating Sums and Differences of Mixed Numbers
	Algebraic Fractions
	Simplifying Rational Expressions
	Linear Equations
	Dividing Complex Numbers
	Simplifying Square Roots That Contain Variables
	Simplifying Radicals Involving Variables
	Compound Inequalities
	Factoring Special Quadratic Polynomials
	Simplifying Complex Fractions
	Rules for Exponents
	Finding Logarithms
	Multiplying Polynomials
	Using Coordinates to Find Slope
	Variables and Expressions
	Dividing Radicals
	Using Proportions and Cross
	Solving Equations with Radicals and Exponents
	Natural Logs
	The Addition Method
	Equations

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Solving Quadratic Equations

Solving by extracting square roots

EXAMPLE:

Solve by extracting square roots

a: Solve 5x = 500

First divide both sides by 5! Then notice that the equation only involves two terms! Also notice that the value left on the right hand side of the equation is a perfect square number!

5x = 500 Given equation

x = 100 Divide by 5

Thus x = Â± 10 meaning x can be +10 or x can be -10!

b: Solve x = 7

Notice that the equation only involves two terms! Also notice that the value left on the right hand side of the equation is NOT a perfect square number! So what !!! You can still solve this by extracting square roots.

x = 7

So x = Â± 7

c: Solve (x- 2) = 64

METHOD 1: Notice that the left hand side of the equation is still a squared term. It just happens to be x -2! It can be solved by taking the square root of both sides as follows:

(x- 2)= 64

So x - 2 = Â±8

Thus x - 2 = 8 or x - 2 = -8

x = 10 or x = -6

METHOD 2: Clear the parentheses, collect like terms on left, set = 0, factor!

(x- 2) = 64

x -4x + 4 = 64

x -4x + 4 - 64 = 0

x -4x - 60 = 0

(x - 10)(x + 6) = 0

Either x - 10 = 0 or x + 6 = 0

x = 10 or x = -6