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3.  Learn the definition of quadratic equation.  Multiply two binomials using the FOIL method.  Factor trinomials.  Solve quadratic equation by factoring.  Solve quadratic equation using the quadratic formula  Use quadratic equations in applications.

4.  Binomial: has 2 terms  E.g., x + 3, 3x – 5  FOIL Method

5.  Multiply: (x + 3)(x + 4)  Solution: F: First terms= x x = x² O: Outside terms= x 4 = 4x I: Inside terms= 3 x = 3x L: Last terms= 3 4 = 12 (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12

6.  Trinomial: -- has 3 terms  Note: Factored Form F O I L Trinomial Form (x + 3)( x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12  Factoring an algebraic expression: --finding expression that is a product

8. Factor: x² + 6x + 8. Solution: 1. Enter x as the first term of each factor: x² + 6x + 8 = (x )(x ) 2. List all pairs of Factors of constant, 8. 3. Try various combinations of these factors. The correct factorization is the one in which the sum of the Outside and Inside products is equal to 6x. Factors of 88, 14, 2-8, -1-4, -3

9. Possible factorization o x 2 + 6x + 8 Sum of Outside and Inside Products should equal 6x (x + 8)(x + 1)X + 8x = 9x (x + 4)(x + 2)4x + 2x = 6x (x – 8)(x – 1)-8x – x = -9x (x – 4)(x – 2)-2x – 4x = -6x Thus, x² + 6x + 8 = (x + 4 )(x + 2 ).

10.  Factor: 3x²  20x + 28  Solution: 1. Factor: 3x²  20x + 28 3x²  20x + 28 = (3x )(x ) 2. List all pairs of factors of 28. Because the middle term,  20x, is negative, both factors must be negative. The negative factorizations are: z  1(  28),  2(  14),  4(  7) 3. Try various combinations of these factors:

11. Possible Factorizatins of 3x2 – 20x + 28 Sum of Outside and Inside Products: Should Equal -20x (3x – 1)(x – 28)  84x – x =  85x (3x – 28)(x – 1)  3x – 28x =  31x (3x – 2)(x – 14)  42x  2x =  44x (3x – 14)(x – 2  6x  14x =  20x (3x – )(x – 7)  21x – 4x =  25x (3x – 7)(x – 4)  12x  7x =  19x Thus, 3x²  20x + 28 = (3x  14 )(x  2). Step 4: Verify using the FOIL method: (3x  14)(x  2) = 3x²  6x  14x + 28 = 3x²  20x + 28.

12. 1. Rewrite (in necessary) the equation in the form ax 2 + bx + c = 0. 2. Factor. 3. Let each factor equal to 0. 4. Solve the equations in step 3. 5. Check the solutions in the original equation.

13.  Solve: x²  2x = 35  Solution:  x²  2x  35= 0  (x – 7)(x + 5) = 0  Set each factor equal to zero. x – 7 = 0 or x + 5 = 0  Solve each equation. x = 7 x =  5 Check: 7²  2· 7 = 35(  5) ²  2(  5) = 35 49  14 = 3525 +10 = 3535 = 35

14.  Given: ax 2 + bx + c = 0  Solve for x.

16.  Given: ax 2 + bx + c = 0, with a ≠ 0  Solution is given by the Quadratic Formula:

17. Solve: 2x² + 9x – 5 = 0 Solution: a = 2, b = 9, c =  5

18.  Solve 2x² = 4x + 1  Solution: 2x²  4x – 1 = 0 a = 2, b =  4 and c =  1

19.  Solve the following quadratic equations using the quadratic formula. 1. 3x 2 + 2x – 3 = 0 2. 4x 2 – 3x = 7 3. 5x 2 – 1x = – 4

20.  Use the quadratic formula to solve the following equations: 1. X 2 + 3x – 4 = 0 2. x(x – 4) = 8 3. 2x 2 – 4x – 3 = 0 4. 9x 2 + 12x + 4 = 0

21. A person's normal systolic blood pressure measured in millimeters of mercury depends on his or her age – for men, according to the formula: P = 0.006A²  0.02A + 120 Find the age, to the nearest year, of a man whose systolic blood pressure is 125 mm Hg.

22. Solution: From the formula, 125 = 0.006A² – 0.02A + 120. Rewriting, we get 0.006A² – 0.02A – 5 = 0z where a = 0.006, b =  0.02 and c =  5 Therefore, 31 is the approximate age for a man with a normal systolic blood pressure of 125 mm Hg.