5 2nd degree equations and the quadratic formula

2nd degree equations and the quadratic formula

1. Solving 2nd Degree Equations ax 2 + bx + c = 0

2. Solving 2nd Degree Equations ax 2 + bx + c = 0 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing.

3. Solving 2nd Degree Equations ax 2 + bx + c = 0 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials.

4. Solving 2nd Degree Equations ax 2 + bx + c = 0 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

5. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

6. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

7. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d. To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

8. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d. II. Then x = ±  d are the roots. To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

9. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d. II. Then x = ±  d are the roots. Example A. Solve 3x 2 – 7 = 0 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

10. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d. II. Then x = ±  d are the roots. Example A. Solve 3x 2 – 7 = 0 3x 2 – 7 = 0 solve for x 2 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

11. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d. II. Then x = ±  d are the roots. Example A. Solve 3x 2 – 7 = 0 3x 2 – 7 = 0 solve for x 2 3x 2 = 7 x 2 = 7 3 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

12. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d. II. Then x = ±  d are the roots. Example A. Solve 3x 2 – 7 = 0 3x 2 – 7 = 0 solve for x 2 3x 2 = 7 x 2 = take square root x = ±  7/3 7 3 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

13. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d II. Then x = ±  d are the roots. Example A. Solve 3x 2 – 7 = 0 3x 2 – 7 = 0 solve for x 2 3x 2 = 7 x 2 = take square root x = ±  7/3  ±1.53 7 3 To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).

14. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d II. Then x = ±  d are the roots. Example A. Solve 3x 2 – 7 = 0 3x 2 – 7 = 0 solve for x 2 3x 2 = 7 x 2 = take square root x = ±  7/3  ±1.53 7 3 exact answers To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF).

15. Solving 2nd Degree Equations ax 2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x 2 and get the form x 2 = d II. Then x = ±  d are the roots. Example A. Solve 3x 2 – 7 = 0 3x 2 – 7 = 0 solve for x 2 3x 2 = 7 x 2 = take square root x = ±  7/3  ±1.53 7 3 exact answers approx. answers To solve ax 2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd : Use the quadratic formula (QF). In this course, all approx. answers are given with three significant digits, i.e. starting from the first nonzero digit, round off the answer to a three digit number.

16. 2. Solve by the factoring m ethod Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0

17. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 a. 2x 2 – 8x =

18. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 a. 2x 2 – 8x = 2x(x – 4) =

19. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 a. 2x 2 – 8x = 2x(x – 4) = h ence x = 0 or x – 4 = 0 so x = 4

20. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 3x 2 – 3x + x – 8 = a. 2x 2 – 8x = 2x(x – 4) = h ence x = 0 or x – 4 = 0 so x = 4 set one side to be

21. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 3x 2 – 3x + x – 8 = 3x 2 – 2x – 8 = a. 2x 2 – 8x = 2x(x – 4) = h ence x = 0 or x – 4 = 0 so x = 4 set one side to be

22. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 3x 2 – 3x + x – 8 = 3x 2 – 2x – 8 = 0 (3x + 4)(x – 2) = a. 2x 2 – 8x = 2x(x – 4) = h ence x = 0 or x – 4 = 0 so x = 4 set one side to be factor

23. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 3x 2 – 3x + x – 8 = 3x 2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = –4/3 , x = 2 a. 2x 2 – 8x = 2x(x – 4) = h ence x = 0 or x – 4 = 0 so x = 4 set one side to be factor and extract the solutions

24. Example B. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax 2 + bx + c = 0 b . x(3x – 3 ) = – x + 8 3x 2 – 3x + x – 8 = 3x 2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = –4/3 , x = 2 a. 2x 2 – 8x = 2x(x – 4) = h ence x = 0 or x – 4 = 0 so x = 4 set one side to be factor and extract the solutions If the equation is not easily factorable, use the formula below.

25. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x =

26. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are.

27. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots,

28. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

29. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

30. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

31. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b 2 – 4ac = (–2) 2 – 4(3)(–2) = 28. 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

32. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b 2 – 4ac = (–2) 2 – 4(3)(–2) = 28. Not factorable! 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

33. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b 2 – 4ac = (–2) 2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) ±  28 2(3) 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

34. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b 2 – 4ac = (–2) 2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) ±  28 2(3) = 2 ±  28 6 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

35. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b 2 – 4ac = (–2) 2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) ±  28 2(3) = 2 ±  28 6 = 2 ± 2  7 6 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

36. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b 2 – 4ac = (–2) 2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) ±  28 2(3) = 2 ±  28 6 = 2 ± 2  7 6 = 1 ±  7 3 3. Use the quadratic formula ( QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

37. –b ±  b 2 – 4ac 2a Solving 2nd Degree Equations ax 2 + bx + c = 0 Example C. Solve 3x 2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b 2 – 4ac = (–2) 2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) ±  28 2(3) = 2 ±  28 6 = 2 ± 2  7 6 = 1 ±  7 3  { 1.22 –0.549 3. Use the quadratic formula (QF) The roots for the equation ax 2 + bx + c = 0 are x = b 2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b 2 – 4ac is a perfect square, we have fractional roots, if b 2 – 4ac < 0 there is no real roots.

38. Solving 2nd Degree Equations ax 2 + bx + c = 0 Example D. Solve 3x 2 – 2x + 2 =

39. Solving 2nd Degree Equations ax 2 + bx + c = 0 Example D. Solve 3x 2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b 2 – 4ac = (–2) 2 – 4(3 )(2 ) = –20

40. Solving 2nd Degree Equations ax 2 + bx + c = 0 Example D. Solve 3x 2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b 2 – 4ac = (–2) 2 – 4(3 )(2 ) = –20 and  –20 does not exist. Hence it does not have any real number solution.

41. Solving 2nd Degree Equations ax 2 + bx + c = 0 Example D. Solve 3x 2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b 2 – 4ac = (–2) 2 – 4(3 )(2 ) = –20 and  –20 does not exist. Hence it does not have any real number solution. Following is a theorem that demonstrates the central role of the 1 st and 2 nd degree equations in algebra .

42. Solving 2nd Degree Equations ax 2 + bx + c = 0 Example D. Solve 3x 2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b 2 – 4ac = (–2) 2 – 4(3 )(2 ) = –20 and  –20 does not exist. Hence it does not have any real number solution. i.e. P(x ) = A(x – #)( x – #).. (x 2 + #x + #)( x 2 + #x + #)... Each gives a real root. Following is a theorem that demonstrates the central role of the 1 st and 2 nd degree equations in algebra. Factorization Theorem of Real Polynomials Let P(x) = Ax n +… be a real polynomial, then P(x ) = AL 1 (x)L 2 (x).. Q 1 (x)Q 2 (x).. where each L i (x)’s is of the linear form (x – r i ) with r i a real number, and each Q i (x ) is an irreducible real quadratic, Each gives a pair of complex roots.

43. Rules of Radicals Exercise 3. A . Solve for x. Give both the exact and approximate answers. If the answer does not exist, state so. 1. x 2 = 1 2. x 2 – 5 = 4 3. x 2 + 5 = 4 4. 2x 2 = 31 5. 4x 2 – 5 = 4 6. 5 = 3x 2 + 1 7. 4x 2 = 1 8. x 2 – 3 2 = 4 2 9. x 2 + 6 2 = 10 2 10. 2x 2 + 7 = 11 11. 2x 2 – 5 = 6 12. 4 = 3x 2 + 5 B. Solve the following equations by factoring. 5. x 2 – 3x = 10 9. x 3 – 2x 2 = 0 6. x 2 = 4 7. 2x(x – 3) + 4 = 2x – 4 10. 2x 2 (x – 3) = –4x 8. x(x – 3) + x + 6 = 2x 2 + 3x 1. x 2 – 3x – 4 = 0 2. x 2 – 2x – 15 = 0 3. x 2 + 7x + 12 = 0 4. –x 2 – 2x + 8 = 0 11. 4x 2 = x 4 12. 7x 2 = –4x 3 – 3x 13. 5 = (x + 2)(2x + 1) 14. (x + 1) 2 = x 2 + (x – 1) 2 15. (x + 3) 2 – (x + 2) 2 = (x + 1) 2

44. C. Solve the following equations by the quadratic formula. If the answers are not real numbers, just state so. 1. x 2 – x + 1 = 0 2. x 2 – x – 1 = 0 3. x 2 – 3x – 2 = 0 4. x 2 – 2x + 3 = 0 5. 2x 2 – 3x – 1 = 0 6. 3x 2 = 2x + 3 Equations

45. (Answers to odd problems ) Exercise C . 1. No real solution 3. 5.   1. 3. 5. 7. 9. 11. 13. 15.   Equations 1. x = ±1 3. Doesn’t exist. 7. x = ± 1/2 9. x = ±8 11. x = ±√11/2 Exercise A. 5. x = ±3/2 Exercise B.