College Algebra Linear and Quadratic Functions (Chapter2) L:13 1 University of Palestine IT-College
Transcript
Slide 1
C ollege A lgebra Linear and Quadratic Functions (Chapter2)
L:13 1 University of Palestine IT-College
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Definition of a Quadratic Equation A quadratic equation in x is
an equation that can be written in the standard form ax 2 bx c 0
where a, b, and c are real numbers with a not equal to 0. A
quadratic equation in x is also called a second-degree polynomial
equation in x.
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The Zero-Product Principle If the product of two algebraic
expressions is zero, then at least one of the factors is equal to
zero. If AB 0, then A 0 or B 0.
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Solving a Quadratic Equation by Factoring 1.If necessary,
rewrite the equation in the form ax 2 bx c 0, moving all terms to
one side, thereby obtaining zero on the other side. 2.Factor. 3.
Apply the zero product principle, setting each factor equal to
zero. 4. Solve the equations in step 3. 5.Check the solutions in
the original equation.
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Text Example Solve 2x 2 7x 4 by factoring and then using the
zero product principle. Step 1 Move all terms to one side and
obtain zero on the other side. Subtract 4 from both sides and write
the equation in standard form. 2x 2 7x 4 4 4 2x 2 7x 4 Step 2
Factor. 2x 2 7x 4 (2x 1)(x 4) 0
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Solution cont. Solve 2x 2 7x 4 by factoring and then using the
zero product principle. Steps 3 and 4 Set each factor equal to zero
and solve each resulting equation. 2 x 1 or x 4 2 x 1x 4 x = 1/2
Steps 5 check your solution
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Example 1. Solve for x: (2x + -3)(2x + 1) = 5 4x 2 - 4x - 3 = 5
4x 2 - 4x - 8 = 0 4(x 2 - x - 2)=0 4(x - 2)(x + 1) = 0 x - 2 = 0,
and x + 1 = 0 So x = 2, or -1
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The Square Root Method If u is an algebraic expression and d is
a positive real number, then u 2 = d has exactly two solutions. If
u 2 = d, then u = d or u = - d Equivalently, If u 2 = d then u =
d
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Completing the Square If x 2 + bx is a binomial, then by adding
(b/2) 2, which is the square of half the coefficient of x, a
perfect square trinomial will result. That is, x 2 + bx + (b/2) 2 =
(x + b/2) 2
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10 Example : Solve by completing the square Step 1:Make sure
that the coefficient on the term is equal to 1.coefficient on the
The coefficient of the term is already 1. Step 2: Isolate the and x
terms.Isolate the The and x terms are already isolated. Step 3:
Complete the square.Complete the square
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11 Step 4: Factor the perfect square trinomialFactor the
perfect square trinomial Step 5: Solve the equation in step 4 by
using the square root method.square root method There are two
solutions to this quadratic equation: x = 9 and x = 1.
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Text Example What term should be added to the binomial x 2 + 8x
so that it becomes a perfect square trinomial? Then write and
factor the trinomial. The term that should be added is the square
of half the coefficient of x. The coefficient of x is 8. Thus,
(8/2) 2 = 4 2. A perfect square trinomial is the result. x 2 + 8x +
4 2 = x 2 + 8x + 16 = (x + 4) 2
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Quadratic Equation
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Quadratic Formula
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Example 2. Solve for x using the Quadratic Formula:
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No x-intercepts No real solution; two complex imaginary
solutions b 2 4ac < 0 One x-intercept One real solution (a
repeated solution) b 2 4ac = 0 Two x-intercepts Two unequal real
solutionsb 2 4ac > 0 Graph of y = ax 2 + bx + c Kinds of
solutions to ax 2 + bx + c = 0 Discriminant b 2 4ac The
Discriminant and the Kinds of Solutions to ax 2 + bx +c = 0
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19 Example 11: Solve by using the quadratic formula.
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Quadratic Equations Questions ??
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21 Section 2.6 Other Types of Equations
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22 Polynomial Equation A Polynomial equation is an equation in
the form: Example:
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23 Solving a Polynomial Equation by Factoring 1.Move all terms
to one side and obtain zero on the other side. 2.Factor. 3. Apply
the zero product principle, setting each factor equal to zero. 4.
Solve the equations in step 3. 5.Check the solutions in the
original equation.
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24 Text Example (p. 132) Solve by factoring: 3x 4 = 27x 2. Step
1 Move all terms to one side and obtain zero on the other side.
Subtract 27x 2 from both sides 3x 4 x 2 27x 2 27x 2 3x 4 27x 2 Step
2 Factor. 3x 4 27x 2 3x 2 (x 2 - 9) 0
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25 Solution cont. Solve by factoring: 3x 4 = 27x 2. Steps 3 and
4 Set each factor equal to zero and solve each resulting equation.
3x 2 (x 2 - 9) 0 3x 2 = 0orx 2 - 9 = 0 x 2 = 0x 2 = 9 x = 0x = 9 x
= 0x = 3 Steps 5 check your solution
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26 Example 3: Solve by factoring Step 1: Simplify each side if
needed.Simplify This polynomial equation is already simplified.
Step 2: Write in standard form,, if needed. Step 3:
Factor.Factor
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27 Step 4: Use the Zero-Product PrincipleZero-Product Principle
Step 5: Solve for the equation (s) set up in step 4.
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28 Radical Equations and Equations Involving Rational
Exponents
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29 Example 1 Solve: Answer: Collect like terms
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30 Example 1 Solve: Answer: Square both sides of the
equation.
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31 Example 1 Solve: Answer: Add 3 to both sides of the equation
Check:
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32 Example 2 Solve: Answer: Isolate the radical
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33 Example 2 (continued) Solve: Answer: Square both sides of
the equation.
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34 Example 2 (continued) Solve: Answer: Collect all of the
terms on one side of the equation.
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35 Example 2 (continued) Solve: Answer: Factor
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36 Example 2 (continued) Solve: Answer: Use the principle of
zero product to solve
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37 Example 2 (continued) Solve: Check: x = 5x = 6 Therefore, x
= 5 is the only solution. True Not True
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38 Example : Solve the radical equation Step 1: Isolate one of
the radicals. Isolate one of the radicals. Step 2: Get rid of your
radical sign.Get rid of your radical sign.
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39 Step 3: If you still have a radical left, repeat steps 1 and
2.
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40 Step 4: Solve the remaining equation. Step 5: Check for
extraneous solutions.extraneous solutions. Lets check to see if y =
3 is an extraneous solution:
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41 Since we got a true statement, y = 3 is a solution. Lets
check to see if y = -1 is an extraneous solution: Since we got a
true statement, y = -1 is a solution. There are two solutions to
this radical equation: y = 3 and y = -1.
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42 Solving Radical Equations of the Form x m/n = k Assume that
m and n are positive integers, m/n is in lowest terms, and k is a
real number. 1.Isolate the expression with the rational exponent.
2.Raise both sides of the equation to the n/m power.
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43 Solving Radical Equations of the Form x m/n = k cont. If m
is even:If m is odd: x m/n = k x m/n = k (x m/n ) n/m = k(x m/n )
n/m = k n/m x = k n/m x = k n/m It is incorrect to insert the when
the numerator of the exponent is odd. An odd index has only one
root. 3. Check all proposed solutions in the original equation to
find out if they are actual solutions or extraneous solutions.
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44 Text Example (page 137) Solve: x 2/3 - 3/4 = -1/2. Isolate x
2/3 by adding 3/4 to both sides of the equation: x 2/3 = 1/4. Raise
both sides to the 3/2 power: (x 2/3 ) 3/2 = (1/4) 3/2. x =
1/8.
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45 Example : Solve the rational exponent equation. Step 1:
Isolate the base with the rational exponent.Isolate the base with
the rational exponent. The base with the rational exponent is
already isolated. Step 2: Get rid of the rational exponent. Get rid
of the rational exponent.
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46 Step 3: Solve the remaining equation. Step 4: Check for
extraneous solutions.extraneous solutions.
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47 Example: Solve the rational exponent equation. Step 1:
Isolate the base with the rational exponent.Isolate the base with
the rational exponent. Step 2: Get rid of the rational exponent.
Get rid of the rational exponent.
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48 Step 3: Solve the remaining equation. Step 4: Check for
extraneous solutions.extraneous solutions. There is no solution to
this rational exponent equation.
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49 Some equations that are not quadratic can be written as
quadratic equations using an appropriate substitution. Here are
some examples. 5t 2 + 11t + 2 = 0t = x 1/3 5x 2/3 + 11x 1/3 + 2 = 0
or 5(x 1/3 ) 2 + 11x 1/3 + 2 = 0 t 2 8t 9 = 0t = x 2 x 4 8x 2 9 = 0
or (x 2 ) 2 8x 2 9 = 0 New EquationSubstitutionGiven Equation
Equations That Are Quadratic in Form