collegealgebra 04 equations and inequalities...absolute value equations an absolute value equationin...
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Equations and InequalitiesCollege Algebra
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Radical Equations
Radical Equations: are equations that contain variables in the radicand
How to Solve a Radical Equation:1. Isolate the radical expression on one side of the equal sign. Put all remaining
terms on the other side2. If the radical is a square root, then square both sides of the equation. If it is
a cube root, then raise both sides of the equation to the third power. In other words, for an ππ‘β root radical, raise both sides to the ππ‘β power. Doing so eliminates the radical symbol
3. Solve the remaining equation4. If a radical term still remains, repeat steps 1β25. Confirm solutions by substituting them into the original equation
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Extraneous Solutions to Radical Equations
We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not solutions to the equation.Example:
15 β 2π₯οΏ½ = π₯( 15 β 2π₯οΏ½ )-= π₯-
π₯- + 2π₯ β 15 = 0π₯ + 5 π₯ β 3 = 0
The two proposed solutions are π₯ = β5 and π₯ = 3. Substituting back into the original equation, we get 25οΏ½ = β5 and 9οΏ½ = 3. Therefore, π₯ = β5 is an extraneous solution and π₯ = 3 is the only solution.
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Rational Exponents
A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:
π34 = (π
54)3= (π3)
54= π36 = π6 3
Example:8-8 = (8
58)-= ( 89 )-= 2- = 4
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Polynomial Equations
A polynomial of degree n is an expression of the type
π4π₯4 + π4;5π₯4;5 +=== +π-π₯- + π5π₯ + π>
where π is a positive integer and π4,β¦,π> are real numbers and π4 β 0
Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent π
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Absolute Value Equations
An absolute value equation in the form ππ₯ + π = π has the following properties:
If π < 0 the equation has no solution.If π = 0 the equation has one solution.If π > 0 the equation has two solutions.
To solve an absolute value equation, isolate the absolute value expression on one side of the equal sign. If π > 0, write and solve two equations: ππ₯ +π = π and ππ₯ + π = βπ
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Solving Equations in Quadratic FormIf the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form
How to solve an equation in quadratic form:1. Identify the exponent on the leading term and determine whether it is
double the exponent on the middle term2. If it is, substitute a variable, such as u, for the variable portion of the
middle term3. Rewrite the equation so that it takes on the standard form of a quadratic4. Solve using one of the usual methods for solving a quadratic5. Replace the substitution variable with the original term6. Solve the remaining equation
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Modelling a Linear Equation to Fit aReal-World Problem
1. Identify known quantities2. Assign a variable to represent the unknown quantity3. If there is more than one unknown quantity, find a way to write the
second unknown in terms of the first4. Write an equation interpreting the words as mathematical operations5. Solve the equation. Be sure the solution can be explained in words,
including the units of measure
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Common Verbal Expressions and their Equivalent Mathematical Expressions
Verbal Translation to Math OperationsOne number exceeds another by π π₯, π₯ + πTwice a number 2π₯One number is π more than another number π₯, π₯ + πOne number is π less than twice another number π₯, 2π₯ β πThe product of a number andπ, decreased by π ππ₯ β πThe quotient of a number and the number plus π is three times the number
π₯π₯ + 3 = 3π₯
The product of three times a number and the number decreased by π is π
3π₯ π₯ β π = π
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Models and Applications
β’ A linear equation can be used to solve for an unknown in a number problemβ’ Applications can be written as mathematical problems by identifying
known quantities and assigning a variable to unknown quantitiesβ’ There are many known formulas that can be used to solve applications.
Distance problems, for example, are solved using the π = ππ‘ formulaβ’ Many geometry problems are solved using the perimeter formula π = 2πΏ +2π, the area formula π΄ = πΏπ, or the volume formula π = πΏππ»
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The Zero-Product Property and Quadratic Equations
The zero-product property statesIf π = π = 0,then π = 0 or π = 0
where π and π are real numbers or algebraic expressions
A quadratic equation is an equation containing a second-degree polynomial; for example
ππ₯- + ππ₯ + π = 0where π, π, and π are real numbers, and if π β 0, it is in standard form
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Solving Quadratics with a Leading Coefficient of 1
In the quadratic equation π₯- + π₯ β 6 = 0, the leading coefficient, or the coefficient of π₯-, is 1.
For a Quadratic Equation with the Leading Coefficient of 1, Factor it1. Find 2 numbers whose product equals π and whose sum equals π2. Use those numbers to write two factors of the form (π₯ + π) or (π₯ β π),
where π is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and β2, the factors are (π₯ + 1)(π₯ β 2)
3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable
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Using the Square Root Property
With the π₯- term isolated, the square root property states that:if π₯- = π, then π₯ = Β± ποΏ½
where π is a nonzero real number
For a Quadratic Equation with an ππterm but no π term, use the Square Root Property to solve it1. Isolate the π₯- term on one side of the equal sign2. Take the square root of both sides of the equation, putting a Β± sign
before the expression on the side opposite the squared term3. Simplify the numbers on the side with the Β± sign
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Completing the Square
We can solve some quadratic equations by adding or subtracting terms to both sides of the equation until we have a perfect square trinomial on one side. We can then apply the square root property.
Example: Solve π₯- + 6π₯ + 1 = 0
π₯- + 6π₯ = β1π₯- + 6π₯ + 9 = β1 + 9
π₯ + 3 - = 8π₯ + 3 = Β± 8οΏ½
π₯ = β3 + 8οΏ½ , π₯ = β3 β 8οΏ½
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Using the Pythagorean Theorem
π- + π- = π-
where π and π refer to the legs of a right triangle adjacent to the 90Β° angle, and π refers to the hypotenuse
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Using the Quadratic FormulaWritten in standard form, ππ₯- + ππ₯ + π = 0, any quadratic equation can be solved using the quadratic formula:
π₯ =βπ Β± π- β 4πποΏ½
2πwhere π, π, and π are real numbers and π β 0
To Solve using the Quadratic Formula:1. Make sure the equation is in standard form: ππ₯- + ππ₯ + π = 02. Make note of the values of the coefficients and constant term, π, π, and π3. Carefully substitute the values noted in step 2 into the equation. To avoid
needless errors, use parentheses around each number input into the formula.4. Calculate and solve
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The Discriminant
For ππ₯- + ππ₯ + π = 0, where π, π, and π are real numbers, the discriminant is the expression under the radical in the quadratic formula: π- β 4ππ. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect:
β’ If π- β 4ππ = 0, there is one rational solutionβ’ If π- β 4ππ > 0 and a perfect square, there are two rational solutionsβ’ If π- β 4ππ > 0 and not a perfect square, there are two irrational solutionsβ’ If π- β 4ππ < 0, there are two complex solutions
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Set-Builder Notation and Interval Notation
The solution to an inequality such as π₯ β₯ 4 can be written in several ways.
In set-builder notation, braces are used to indicate a set of real numbers, such as π₯ π₯ β₯ 4 .In interval notation, parentheses or brackets indicate whether the interval includes the endpoint(s), such as [4,β).
π₯ π < π₯ < π in set-builder notation is π, π in interval notation
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The Properties of Inequalities
Addition Property If π < π, then π + π < π + π
Multiplication Property If π < π and π > 0, then ππ < ππIf π < π and π < 0, then ππ > ππ
These properties also apply to π β€ π, π > π, and π β₯ π
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Desmos Interactive
Topic: writing and graphing inequalities
https://www.desmos.com/calculator/4529rytfef
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Absolute Value of Inequalities
For an algebraic expression π, and π > 0, an absolute value inequality is an inequality of the form
|π| < π is equivalent to βπ < π < π|π| > π is equivalent to π < βππππ > π
These statements also apply to |π| β€ π and |π| β₯ π
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Quick Review
β’ How do you solve a radical equation?β’ What is an extraneous solution?β’ How many solutions to a polynomial equation are there?β’ What is the standard form of a quadratic equation?β’ How do you solve a quadratic equation by completing the square?β’ What is the quadratic formula?β’ What is the discriminant?β’ What two notations can be used to describe the solution to an inequality?β’ What is the multiplication property of inequalities?