# § 8.2

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## § 8.2

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2. The Quadratic Formula See page 575 of your textbook to see how the quadratic formula is derived using ‘completing the square’. Blitzer, Intermediate Algebra, 4e – Slide #24

3. The Quadratic Formula EXAMPLE Solve using the quadratic formula: SOLUTION The given equation is in standard form. Begin by identifying the values for a, b, and c. a = 1 b = 8 c = 15 Substituting these values into the quadratic formula and simplifying gives the equation’s solutions. Blitzer, Intermediate Algebra, 4e – Slide #25

4. The Quadratic Formula CONTINUED Use the quadratic formula. Substitute the values for a, b, and c: a = 1, b = 8, c = 15. Simplify. Subtract. The square root of 4 is 2. Blitzer, Intermediate Algebra, 4e – Slide #26

5. The Quadratic Formula CONTINUED Now we will evaluate this expression in two different ways to obtain the two solutions. On the left, we will add 2 to -8. On the right, we will subtract 2 from -8. The solutions are -3 and -5. The solution set is {-3,-5}. Blitzer, Intermediate Algebra, 4e – Slide #27

6. The Quadratic Formula EXAMPLE Solve using the quadratic formula: SOLUTION The quadratic equation must be in standard form to identify the values for a, b, and c. To move all terms to one side and obtain zero on the right, we subtract -4x + 5 from both sides. Then we can identify the values for a, b, and c. This is the given equation. Subtract -4x + 5 from both sides. a = 2 b = 4 c = -5 Blitzer, Intermediate Algebra, 4e – Slide #28

7. The Quadratic Formula CONTINUED Substituting these values into the quadratic formula and simplifying gives the equation’s solutions. Use the quadratic formula. Substitute the values for a, b, and c: a = 2, b = 4, c = -5. Simplify. Add. Blitzer, Intermediate Algebra, 4e – Slide #29

8. The Quadratic Formula CONTINUED Factor out 2 from the numerator. Divide the numerator and denominator by 2. The solutions are , and the solution set is Blitzer, Intermediate Algebra, 4e – Slide #30

9. Quadratic Formula – The Discriminant Blitzer, Intermediate Algebra, 4e – Slide #31

10. Quadratic Formula – The Discriminant CONTINUED Blitzer, Intermediate Algebra, 4e – Slide #32

11. Quadratic Formula – The Discriminant CONTINUED Blitzer, Intermediate Algebra, 4e – Slide #33

12. Quadratic Formula – The Discriminant EXAMPLE For each equation, compute the discriminant. Then determine the number and type of solutions: SOLUTION Begin by identifying the values for a, b, and c in each equation. Then compute , the discriminant. a = 2 b = -4 c = 3 Substitute and compute the discriminant: Blitzer, Intermediate Algebra, 4e – Slide #34

13. Quadratic Formula – The Discriminant CONTINUED The discriminant, -8, shows that there are two imaginary solutions. These solutions are complex conjugates of each other. We must first put the quadratic equation in standard form. Subtract 20x – 25 from both sides. a = 4 b = -20 c = 25 Substitute and compute the discriminant: Blitzer, Intermediate Algebra, 4e – Slide #35

14. Quadratic Formula – The Discriminant CONTINUED The discriminant, 0, shows that there is only one real solution. This real solution is a rational number. Blitzer, Intermediate Algebra, 4e – Slide #36

15. Solving Quadratic Equations Blitzer, Intermediate Algebra, 4e – Slide #37

16. Solving Quadratic Equations CONTINUED Blitzer, Intermediate Algebra, 4e – Slide #38

17. Solving Quadratic Equations CONTINUED a = 1 b = -2 c = -6 Blitzer, Intermediate Algebra, 4e – Slide #39

18. The Zero-Product Principle Blitzer, Intermediate Algebra, 4e – Slide #40

19. The Zero-Product Principle EXAMPLE Write a quadratic equation with the given solution set: SOLUTION Because the solution set is , then Obtain zero on one side of each equation. Clear fractions, multiplying by 6 and 3 respectively. Blitzer, Intermediate Algebra, 4e – Slide #41

20. The Zero-Product Principle CONTINUED Use the zero-product principle in reverse. Use the FOIL method to multiply. Combine like terms. Thus, one equation is . Many other quadratic equations have for their solution sets. Blitzer, Intermediate Algebra, 4e – Slide #42

21. Quadratic Formula in Application EXAMPLE The hypotenuse of a right triangle is 6 feet long. One leg is 2 feet shorter than the other. Find the lengths of the legs. Round to the nearest tenth of a foot. SOLUTION Since the hypotenuse is 6 feet long, and one leg of the triangle, x, is 2 feet longer than the other leg, x - 2, the triangle can be represented as follows. 6 x - 2 x Blitzer, Intermediate Algebra, 4e – Slide #43

22. Quadratic Formula in Application CONTINUED Now we can use the Pythagorean Theorem to create an equation that contains the information provided. This is the Pythagorean Theorem. Evaluate the exponents. Simplify. Subtract 36 from both sides. Factor 2 out of all terms on left side. Divide both sides by 2. Determine a, b, and c to use the quadratic formula. a = 1 b = -2 c = -16 Blitzer, Intermediate Algebra, 4e – Slide #44

23. Quadratic Formula in Application CONTINUED Substitute the values for a, b, and c into the quadratic formula. Simplify. Simplify. Rewrite the radicand. Rewrite as two radicals. Blitzer, Intermediate Algebra, 4e – Slide #45

24. Quadratic Formula in Application CONTINUED Simplify. Factor 2 out of the numerator. Divide the numerator and denominator by 2. Now and to the nearest tenth of a foot. The answer -3.1 feet is of course impossible. Therefore, the length of the side labeled x must be 5.1 feet. Therefore, the side labeled x – 2 must be 5.1 – 2 = 3.1 feet. Blitzer, Intermediate Algebra, 4e – Slide #46