C ollege A lgebra Linear and Quadratic Functions (Chapter2) L:13 1 University of Palestine IT-College.

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C ollege A lgebra Linear and Quadratic Functions (Chapter2) L:13 1 University of Palestine IT-College

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.

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.

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.

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

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

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

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

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

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

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.

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

Example 2. Solve for x using the Quadratic Formula:

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

19 Example 11: Solve by using the quadratic formula.

Quadratic Equations Questions ??

21 Section 2.6 Other Types of Equations

22 Polynomial Equation A Polynomial equation is an equation in the form: Example:

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.

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

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

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

27 Step 4: Use the Zero-Product PrincipleZero-Product Principle Step 5: Solve for the equation (s) set up in step 4.

28 Radical Equations and Equations Involving Rational Exponents

29 Example 1 Solve: Answer: Collect like terms

30 Example 1 Solve: Answer: Square both sides of the equation.

31 Example 1 Solve: Answer: Add 3 to both sides of the equation Check:

32 Example 2 Solve: Answer: Isolate the radical

33 Example 2 (continued) Solve: Answer: Square both sides of the equation.

34 Example 2 (continued) Solve: Answer: Collect all of the terms on one side of the equation.

35 Example 2 (continued) Solve: Answer: Factor

36 Example 2 (continued) Solve: Answer: Use the principle of zero product to solve

37 Example 2 (continued) Solve: Check: x = 5x = 6 Therefore, x = 5 is the only solution. True Not True

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.

39 Step 3: If you still have a radical left, repeat steps 1 and 2.

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:

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.

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.

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.

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.

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.

46 Step 3: Solve the remaining equation. Step 4: Check for extraneous solutions.extraneous solutions.

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.

48 Step 3: Solve the remaining equation. Step 4: Check for extraneous solutions.extraneous solutions. There is no solution to this rational exponent equation.

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

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