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Chapter 4

Chapter 4 . Quadratic Equations. 4.1 – Graphical solutions of quadratic equations. Chapter 4: Quadratic Equations. Quadratic equations. A quadratic equation is a second-degree equation with standard form ax 2 + bx + c = 0, where a ≠ 0.

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Chapter 4

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  1. Chapter 4 Quadratic Equations

  2. 4.1 – Graphical solutions of quadratic equations Chapter 4: Quadratic Equations

  3. Quadratic equations A quadratic equation is a second-degree equation with standard form ax2 + bx + c = 0, where a ≠ 0. Root(s) of an equation are the solution(s) to an equation. The zero(s) of a function are the value(s) of x for which f(x) = 0.

  4. example What are the roots of the equation –x2 + 8x – 16 = 0. We can either create our own graph by creating a table of values, or we can use our calculators to find the solution. Method 1: We can see that the vertex is (4,0). So, the root of the equation is x = 4.

  5. example What are the roots of the equation –x2 + 8x – 16 = 0. Method 2: By “LEFT” and “RIGHT” we mean that you must push the left key until the target is to the left/right side of the zero. In your Y= enter the equation. Then go to the graph. Buttons to press: ENTER 2 LEFT TRACE 2ND ENTER RIGHT ENTER Is our solution the same as our other method? The solution is down on the bottom left side of the graph, where it says x = ___.

  6. The manager of Jasmine’s Fine Fashions is investigating the effect that raising or lowering dress prices has on the daily revenue from dress sales. The function R(x) = 100 + 15x – x2 gives the store’s revenue R, in dollars, from dress sales, where x is the price change, in dollars. What price changes will result in no revenue? • Example of Creating a Table: • Substitute in x = -6: • R(x) = 100 + 15x – x2 • R(-6) = 100 + 15(-6) – (-6)2 • R(-6) = 100 – 90 – 36 • R(-6) = -26 From the graph, and the table of values, we can see that price change of x = $20, and x = $–5, will result in no revenue. Try finding the roots on your calculator. Are they the same?

  7. example Solve 2x2 + x = –2 by graphing. • Firstly, rearrange the equation so that it is in the form ax2+ bx + c = 0. • 2x2 + x + 2 = 0 • Now, graph like usual. Try it on your calculator! What do you find? What does this mean? If an equation has no zeros (no x-intercepts), then it has no solution!

  8. Pg. 215-217, # 3, 6, 7, 9, 11, 13, 14, 15, 16, 18 Independent Practice

  9. 4.2 – Factoring Quadratic equations Chapter 4: Quadratic Equations

  10. Handout There are four questions on the handout that involve factoring. Answer allof them to the best of your abilities. Review of Factoring: Example: x2 + x – 6 (x + __)(x - __) How do I know that one of the factored terms will be a subtraction? (x + 3)(x – 2) This is our factored form! What relationship did you notice between the factored form of a quadratic equations and its roots?

  11. Review Recall factoring “Perfect Squares”: 4x2 + 12x + 9 = (2x + 3)(2x + 3) = (2x + 3)2 9x2 – 24x + 16 = (3x – 4)(3x – 4) = (3x – 4)2 Recall factoring “A Difference of Squares”: 4x2 – 16y2 = (2x – 4y)(2x + 4y)

  12. Factoring polynomials having a quadratic pattern If you have a polynomial that looks like this: 3(x + 2)2 – 13(x + 12) + 12 How can you factor it? Notice that it has the pattern of a typical quadratic equation. You could multiply it all out, and work backwards, but it is easier to simply substitute in another variable. Let r = x + 2 3(x + 2)2 – 13(x + 12) + 12 = 3r2 – 13r + 12 = (3r – ___)(r – ___) = (3r – 4)(r – 3) Now, substitute the original r = x + 2 back in: (3r – 4)(r – 3) = [3(x + 2) – 4][(x+2) – 3] = (3x + 6 – 4)(x – 1) = (3x + 2)(x – 1)

  13. example Factor. a) 2x2 – 2x – 12 b) 9x2 – 0.64y2 If there’s a common factor in all three terms, remove it first! 2x2 – 2x – 12 = 2(x2 – x – 6) = 2(x - __)(x + __) = 2(x – 3)(x + 2) b) If there are two terms, check if it’s a difference of squares. When there is a difference of squares, you just need to take the square root of each term. 9x2 – 0.64y2 = (3x – 0.8y)(3x + 0.8y) How do I know that it’s - / +?

  14. Try it Factor:

  15. example Factor: 9(2t + 1)2 – 4(s – 2)2 Notice that both terms are perfect squares. So, this is a difference of squares. We just need the square roots of each term. 9(2t + 1)2 – 4(s – 2)2 = [3(2t + 1) – 2(s – 2)][3(2t + 1) + 2(s-2)] = (6t + 3 – 2s + 4)(6t + 3 + 2s – 4) = (6t – 2s + 7)(6t + 2s – 1)

  16. example Determine the roots of each quadratic equation. Verify your solution. a) x2 + 6x + 9 = 0 b) x2 + 4x – 21 = 0 c) 2x2 – 9x – 5 = 0 • x2 + 6x + 9 = 0 • (x + 3)(x + 3) = 0 • With quadratic equations, the roots can be found by factoring. Once we get to this stage you can split it into two separate equations. Why can we do this? • (x + 3) = 0 • (x + 3) = 0 (In this case, there’s really only one equation) • x = –3 • Verify: (–3)2 + 6(–3) + 9 = 9 – 18 + 9 = 0 • So, the only solution is x = –3.

  17. examples b) x2 + 4x – 21 = 0 c) 2x2 – 9x – 5 = 0 • x2 + 4x – 21 = (x - __)(x + __) • x2 + 4x – 21 = (x – 3)(x + 7) • (x – 3) = 0 • x =3 • (x + 7) = 0 • x = –7 • 2x2 – 9x – 5 = (2x + __)(x - __) • 2x2 – 9x – 5 = (2x + 1)(x – 5) • (2x + 1) = 0 • x = –1/2 • (x – 5) = 0 • x = 5 Verify on your calculator. The solutions are x = -1/2 and x = 5. • Verify: • (3)2 + 4(3) – 21 = 9 + 12 – 21 = 0 • (-7)2 + 4(-7) – 21 = 49 – 28 – 21 = 0 • They both work! So the solutions are x = 3 and x = –7.

  18. P. 229-233 # 3-6(a,c), 10, 11, 13, 15, 17, 19, 21, 25, 27, 28, 30 Independent Practice

  19. 4.3 – Solving quadratic equations by completing the square Chapter 4: Quadratic Equations

  20. Recall: Completing the square Completing the square involves adding a value to and subtracting a value from a quadratic polynomial so that it contains a perfect square trinomial. You can then rewrite this trinomial as the square of a binomial (in other words, rewrite it in vertex form). Step 1: Put brackets around the x2- and the x-term. Example: y = x2 – 8x + 5 y = (x2 – 8x) + 5 y = (x2 – 8x + 16 – 16) + 5 y = (x2 – 8x + 16) – 16 +5 y = (x – 4)2 – 16 + 5 y = (x – 4)2 – 11 Step 2: Add/Subtract (b/2)2 inside of the brackets. Step 3: Pull the term you don’t need outside of the brackets. Step 4: Factor what’s in the brackets. Step 5: Simplify!

  21. Example A wide-screen television has a diagonal measure of 42 in. The width of the screen is 16 in. more than the height. Determine the dimensions of the screen, to the nearest tenth of an inch. • Complete the square! • (16/2)2 = 64 • h2 + 16h + 64 – 64 = 754 • h2 + 16h + 64 = 818 • (h + 8)2 = 818 Diagram: • Use the Pythagorean Theorem: • h2 + (h + 16)2 = 422 • h2 + (h2 + 32h + 256) = 1764 • 2h2 + 32h = 1508 • h2 + 16h = 754 Can the height be -36.6 inches? Why not? So the TV must be 20.6 inches high, and 36.6 inches wide.

  22. example Determine the roots of –2x2 – 3x + 7 = 0, to the nearest hundredth. Then, use technology to verify your answers. Try factoring the expression! If we can’t factor, it may be helpful to complete the square. To verify, put the equation into the Y= in your calculator, and check to see if the zeros match up.

  23. example A defender kicks a soccer ball away from her goal. The path of the kicked soccer ball can be approximated by the quadratic function h(x) = –0.06x2 + 3.168x – 35.34, where x is the horizontal distance travelled, in metres, from the goal line and h is the height, in metres. You can determine the distance the soccer ball is from the goal line by solving the corresponding equation, –0.06x2 + 3.168x – 35.34 = 0. How far is the soccer ball from the goal line when it is kicked? Express your answer to the nearest tenth of a metre. How far does the soccer ball travel before it hits the ground? • –0.06x2 + 3.168x – 35.34 = 0 • x2 – 52.8x + 589 = 0 • x2 – 52.8x = –589 • x2 – 52.8x + 26.42 = –589 + 26.42 • x2 – 52.8x + 696.96 = 107.96 • (x – 26.4)2 = 107.96 • x – 26.4 = The ball is kicked from the smallest root, so it was kicked 16.0 m from the goal line.

  24. example b) How far does the soccer ball travel before it hits the ground? From part a) we know that the ball is kicked from 16.0 m from the goal line, and lands 36.8 m from the goal line. So, it travels: 36.8 – 16.0 = 20.8 m The ball travels 20.8 m before it hits the ground.

  25. P. 240-243 # 5, 6, 9, 11, 12, 13, 14, 16, 17, 19 Independent Practice

  26. 4.4 – The Quadratic formula Chapter 4:Quadratic Equations

  27. Quadratic formula The method used for Completing the Square can help us to develop the quadratic formula—an extremely important formula that can help us to solve any quadratic equation.

  28. handout In the handout you will develop the Quadratic Formula by following the steps typically used in completing the square. You will then investigate the ways that the parameters a, b, and c have on the number of real solutions for an equation. The Quadratic Formula:

  29. The quadratic formula The discriminant is the part underneath the square root sign: so that means that it is b2 – 4ac. It can be used to tell how many roots there are for an equation. What happens if the discriminant is zero? What happens if the discriminant is negative? What happens if the discriminant is positive? • If b2 – 4ac < 0 then there are no real roots. • If b2 – 4ac > 0 then there are two real roots. • If b2 – 4ac = 0 then there is one real root.

  30. example Use the discriminant to determine the nature of the roots for 3x2 – 5x = –9. Check by graphing. • Firstly, we need to manipulate the equation into standard form: • 3x2 – 5x = –9 • 3x2 – 5x + 9 = 0 • a = 3, b = –5, c = 9 • The discriminant, b2 – 4ac = (–5)2 – 4(3)(9) = 25 – 108 = –83 • Since the discriminant is negative, there are no real roots. How many roots does the following equation have? 0.25x2 – 3x + 9 = 0

  31. example Solve using the quadratic formula, express your answer to the nearest hundredth: 5x2 – 7x – 1 = 0 a = 5, b = –7, c = –1 To the nearest hundredth, the two roots are x = 1.53 and x = –0.130.

  32. Try it! 3x2 + 5x – 2 = 0

  33. example Leah wants to frame an oil original painted on canvas measuring 50 cm by 60 cm. Before framing, she places the painting on a rectangular mat so that a uniform strip of the mat shows on all sides of the painting. The area of the mat is twice the area of the painting. How wide is the strip of exposed mat showing on all sides of the painting, to the nearest tenth of a centimetre. • Area of painting = 50 x 60 = 3000 • Area of the mat = 2(3000) = 6000 • Length of the mat = 2x + 60 • Width of the mat = 2x + 50 • Area = Length x Width • 6000 = (2x + 60)(2x + 50) • 6000 = 4x2 + 220x + 3000 • 0 = 4x2 + 220x – 3000 • 0 = 4(x2 + 55x – 750) • 0 = x2 + 55x – 750 • a = 1, b = 55, c = –750 The width of the exposed strip of mat is 11.3 centimetres. Can x be negative?

  34. P. 254-257 # 2, 5, 7, 9, 11, 13, 15, 17, 18, 19, 21, 22. Independent Practice

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