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Scientific Notation
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Linear Equations and Inequalities in One Variable
Graphing Equations in Three Variables
What the Standard Form of a Quadratic can tell you about the graph
Simplifying Radical Expressions Containing One Term
Adding and Subtracting Fractions
Multiplying Radical Expressions
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
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
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Solving Quadratic Equations

Solving by extracting square roots


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 = ± 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 = ± 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 = ±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

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