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1. 1. A QUADRATIC is a polynomialwhose highestexponent is 2.
2. 2. A quadratic equation is a second-orderpolynomial equation in a single variable x ax2+bx+c=0with a ≠ 0. Because it is a second-orderpolynomial equation, the fundamentaltheorem of algebra guarantees that ithas two solutions. These solutions may be both real, or both complex.
3. 3. The roots can be found by completing the square, Solving for then givesThis equation is known as the quadratic formula.
4. 4. The plus-minus sign states that you have two numbers and
5. 5. What do we mean by a root of a quadratic? A solution to the quadratic equation. For example, the roots of this quadratic x² + 2x − 8 are the solutions to x² + 2x − 8 = 0.
6. 6. To find the roots, we can factor that quadratic as (x + 4)(x − 2).Now, if x = −4, then the first factorwill be 0. While if x = 2, the secondfactor will be 0. But if any factor is 0, then the entire product will be 0. Therefore, if x = −4 or 2, then x² + 2x − 8 = 0.
7. 7. −4 and 2are the solutions to the quadraticequation. They are the roots of that quadratic.
8. 8. A root of a quadraticis also called a zero. Because, as we will see, at each root, the value of the graph is 0.
9. 9. How many roots has a quadratic?Always two. Because a quadratic (with leading coefficient 1, at least) can always be factored as (x − a)(x − b), and a, b are the two roots.Note that if a factor is (x + q), then the root is −q. For, (x + q) can take the form (x − a): (x + q) = [x − (−q)]. −q is the root,
10. 10. What do we mean by a double root? The two roots are equal. The factors are(x − a)(x − a), so that the two roots are a, a. For example, this quadratic x² − 10x + 25 can be factored as (x − 5)(x − 5). If x = 5, then each factor will be 0, andtherefore the quadratic will be 0. 5 is called a double root.
11. 11. When will a quadratic have a double root? When the quadratic is a perfect square trinomial.
12. 12. Find the roots of eachquadratic by factoring.
13. 13. a) x² − 3x + 2 (x − 1)(x − 2) x = 1 or 2.b) x² + 7x + 12 (x + 3)(x + 4) x = −3 or −4.
14. 14. c) x² + 3x − 10 (x + 5)(x − 2)x = −5 or 2.d) x² − x – 30 (x + 5)(x − 6)x = −5 or 6.
15. 15. e) 2x² + 7x + 3 (2x + 1)(x + 3)x = − 1 or −3. 2 f) 3x² + x − 2 (3x − 2)(x + 1)x= 2 or −1. 3
16. 16. g) x² + 12x + 36 (x + 6)² x = −6, −6.A double root. h) x² − 2x + 1 (x − 1)² x = 1, 1.A double root.
17. 17. Notice that we usethe conjunction "or," because x takes on only one value at a time.
18. 18. c = 0. Solve thisquadratic equation: ax² + bx = 0
19. 19. Since there is no constant term -- c = 0 -- x is a common factor: x(ax + b) = 0. This implies: x= 0 or x = b a. Those are the two roots.
20. 20. a) x² − 5x x(x − 5)x = 0 or 5. b) x² + x x(x + 1)x = 0 or −1.
21. 21. c) 3x² + 4x x(3x + 4)x = 0 or −4 3 d) 2x² − x x(2x − 1)x = 0 or ½
22. 22. b = 0. Solve thisquadratic equation: ax² − c = 0.
23. 23. In the case where there is no middle term, we can write: ax²=c. This implies: x²=c a x=
24. 24. However, if the form is the difference of two squares -- x² − 16 -- then we can factor it as: (x + 4)(x −4). The roots are ±4. In fact, if the quadratic is x² − c, then we could factor it as: (x + )(x − ), so that the roots are ± .
25. 25. a) x² − 3 x² = 3 x=± .b) x² − 25(x + 5)(x − 5) x = ±5. c) x² − 10(x + )(x − ) x=± .
26. 26. Solve each equation for x.
27. 27. a) x² = 5x − 6 x² − 5x + 6 = 0(x − 2)(x − 3) = 0 x = 2 or 3.b) x² + 12 = 8xx² − 8x + 12 = 0(x − 2)(x − 6) = 0 x = 2 or 6.
28. 28. c) 3x² + x = 10 3x² + x − 10 = 0 (3x − 5)(x + 2) = 0 x = 5/3 or − 2. d) 2x² = x 2x² − x = 0 x(2x − 1) = 0 x = 0 or 1/2.
29. 29. Solve this equation
30. 30. We can put this equation in thestandard form by changing all the signs on both sides. 0 will not change. We have the standard form: 5 3− x − 3x² = 0 2
31. 31. Next, we can get rid of the fraction by multiplying both sides by 2. Again, 0 will not change. 6x² + 5x − 6 = 0 (3x − 2)(2x + 3) = 0. 2 3 The roots are 3 and − 2
32. 32. THE END.