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1 Topic 7.2.1 Completing the Square. 2 Topic 7.2.1 Completing the Square California Standard: 14.0 Students solve a quadratic equation by factoring or.

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Presentation on theme: "1 Topic 7.2.1 Completing the Square. 2 Topic 7.2.1 Completing the Square California Standard: 14.0 Students solve a quadratic equation by factoring or."— Presentation transcript:

1 1 Topic 7.2.1 Completing the Square

2 2 Topic 7.2.1 Completing the Square California Standard: 14.0 Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll form perfect square trinomials by adding numbers to binomial expressions. Key words: completing the square perfect square trinomial binomial

3 3 Topic 7.2.1 Completing the Square “Completing the square” is another method for solving quadratic equations — but before you solve any equations, you need to know how completing the square actually works.

4 4 Topic 7.2.1 Writing Perfect Square Trinomials as Perfect Squares Completing the Square An expression such as ( x + 1) 2 is called a perfect square — because it’s (something) 2. In a similar way, an expression such as x 2 + 2 x + 1 is called a perfect square trinomial (“trinomial” because it has 3 terms). Any trinomial of the form x 2 + 2 dx + d 2 is a perfect square trinomial, since it can be written as the square of a binomial: x 2 + 2 dx + d 2 = ( x + d ) 2 This is because it can be written as a perfect square: x 2 + 2 x + 1 = ( x + 1) 2

5 5 Topic 7.2.1 Converting Binomials to Perfect Squares Completing the Square The binomial expression x 2 + 4 x is not a perfect square — it can’t be written as the square of a binomial. However, it can be turned into a perfect square trinomial if you add a constant (a number) to the expression.

6 6 Topic 7.2.1 Example 1 Solution follows… Completing the Square Convert x 2 + 4 x to a perfect square trinomial. Solution To do this you have to add a number to the original expression. First look at the form of perfect square trinomials, and compare the coefficient of x with the constant term (the number not followed by x or x 2 ): x 2 + 2 dx + d 2 = ( x + d ) 2 The coefficient of x is 2 d, while the constant term is d 2. To convert x 2 + 4 x to a perfect square trinomial, add the square of half of 4 — that is, add 2 2 = 4, to give x 2 + 4 x + 4 = ( x + 2) 2 The constant term is the square of half of the coefficient of x.

7 7 Topic 7.2.1 Guided Practice Solution follows… Completing the Square By adding a constant, convert each of these binomials into a perfect square trinomial. 1. x 2 + 14 x 2. x 2 – 12 x 3. x 2 + 2 x 4. x 2 – 8 x 5. y 2 + 20 y 6. p 2 – 16 p x 2 + 14 x + 49 x 2 – 12 x + 36 x 2 + 2 x + 1 x 2 – 8 x + 16 y 2 + 20 y + 100 p 2 – 16 p + 64

8 8 Topic 7.2.1 Completing the Square for x 2 + bx Completing the Square Completing the square for x 2 + bx To convert x 2 + bx into a perfect square trinomial, add. b 2 2 The resulting trinomial is x +. b 2 2

9 9 Topic 7.2.1 Example 2 Solution follows… Completing the Square Form a perfect square trinomial from x 2 + 8 x. Solution This gives you x 2 + 8 x + 16. Here, b = 8, so to complete the square you add. 2 8 2

10 10 Topic 7.2.1 Example 3 Solution follows… Completing the Square What must be added to y 2 – 12 y to make it a perfect square trinomial? Solution This time, b = –12 So 36 must be added (giving y 2 – 12 y + 36). To complete the square you add = (–6) 2 = 36. 2 –12 2

11 11 Topic 7.2.1 Example 4 Solution follows… Completing the Square Suppose x 2 – 10 x + c is a perfect square trinomial, and is equal to ( x + k ) 2. What are the values of c and k ? Solution Here the coefficient of x is –10. So to form a perfect square trinomial, the constant term has to be the square of half of –10, so c = (–5) 2 = 25. Therefore x 2 – 10 x + c = x 2 – 10 x + 25 = ( x + k ) 2. Now multiply out the parentheses of ( x + k ) 2. ( x + k ) 2 = x 2 + 2 kx + k 2 Solution continues…

12 12 Topic 7.2.1 Example 4 Completing the Square Suppose x 2 – 10 x + c is a perfect square trinomial, and is equal to ( x + k ) 2. What are the values of c and k ? Solution (continued) ( x + k ) 2 has to equal x 2 – 10 x + 25, which gives x 2 – 10 x + 25 = x 2 + 2 kx + k 2. Equate the coefficients of x : the coefficient of x on the left-hand side is –10, while on the right-hand side it is 2 k. So –10 = 2 k, or k = –5. Comparing the constant terms in a similar way, you find that 25 = k 2, which is also satisfied by k = –5.

13 13 Topic 7.2.1 Guided Practice Solution follows… Completing the Square Find the value of k that will make each expression below a perfect square trinomial. 7. x 2 – 7 x + k 8. q 2 + 5 q + k 9. x 2 + 6 mx + k 10. d 2 – 2 md + k k = = 49 4 2 –7 2 k = = 25 4 2 5 2 k = = 9 m 2 2 6m6m 2 k = = m 2 2 2m2m 2

14 14 Topic 7.2.1 Guided Practice Solution follows… Completing the Square Form a perfect square trinomial from the following expressions by adding a suitable term. 11. x 2 + 10 x 12. x 2 – 16 x 13. y 2 + 2 y 14. x 2 + bx 15. y 2 – 18 y 16. a 2 – 2 a 17. y 2 + 12 y 18. y 2 + 36 y Add a constant equal to (10 ÷ 2) 2  x 2 + 10 x + 25 Add a constant equal to (12 ÷ 2) 2  y 2 + 12 y + 36 Add a constant equal to (–2 ÷ 2) 2  a 2 – 2 a + 1 Add a constant equal to (–18 ÷ 2) 2  y 2 – 18 y + 81 Add a constant equal to (2 ÷ 2) 2  y 2 + 2 y + 1 Add a constant equal to (–16 ÷ 2) 2  x 2 – 16 x + 64 Add a constant equal to ( b ÷ 2) 2  x 2 + bx + b2b2 4 Add a constant equal to (36 ÷ 2) 2  y 2 + 36 y + 324

15 15 Topic 7.2.1 Guided Practice Solution follows… Completing the Square Form a perfect square trinomial from the following expressions by adding a suitable term. 19. 6 x + 9 20. 1 – 8 x 21. 25 – 20 y Add a squared term, so that ax 2 + 6 x + 9 = ( bx + 3) 2 Find a and b, ax 2 + 6 x + 9 = b 2 x 2 + 6 bx + 9 Comparing x terms shows b = 1  a = b 2 = 1 So the trinomial is x 2 + 6 x + 9 Add a squared term, so that ax 2 – 8 x + 1 = ( bx – 1) 2 Find a and b, ax 2 – 8 x + 1 = b 2 x 2 – 2 bx + 1 Comparing x terms shows b = 4  a = b 2 = 16 So the trinomial is 16 x 2 – 8 x + 1 Add a squared term, so that ay 2 – 20 y + 25 = ( by – 5) 2 Find a and b, ay 2 – 20 y + 25 = b 2 y 2 – 10 by + 25 Comparing y terms shows b = 2  a = b 2 = 4 So the trinomial is 4 y 2 – 20 y + 25

16 16 Topic 7.2.1 Guided Practice Solution follows… Completing the Square Form a perfect square trinomial from the following expressions by adding a suitable term. 22. 4 y 2 + 4 yb 23. 4 a 2 + 12 ab 24. 9 a 2 + 16 b 2 Add a b 2 term, so that 4 y 2 + 4 yb + a 2 b 2 = (2 y + ab ) 2 Find a, 4 y 2 + 4 yb + a 2 b 2 = 4 y 2 + 4 aby + a 2 b 2 Comparing y terms shows a = 1  a 2 = 1 So the trinomial is 4 y 2 + 4 yb + b 2 Add a b 2 term, so that 4 a 2 + 12 ab + c 2 b 2 = (2 a + cb ) 2 Find c, 4 a 2 + 12 ab + c 2 b 2 = 4 a 2 + 4 cab + c 2 b 2 Comparing a terms shows c = 3  b 2 = 9 So the trinomial is 4 a 2 + 12 ab + 9 b 2 Add an ab term, so that 9 a 2 + cab + 16 b 2 = (3 a + 4 b ) 2 Find c, 9 a 2 + cab + 16 b 2 = 9 a 2 + 24 ab + 16 b 2 Comparing ab terms shows c = 24 So the trinomial is 9 a 2 + 24 ab + 16 b 2

17 17 Topic 7.2.1 Guided Practice Solution follows… Completing the Square The quadratics below are perfect square trinomials. Find the value of c and k to make each statement true. 25. x 2 – 6 x + c = ( x + k ) 2 26. x 2 + 16 x + c = ( x + k ) 2 27. 4 x 2 + 12 x + c = (2 x + k ) 2 28. 9 x 2 + 30 x + c = (3 x + k ) 2 29. 4 a 2 – 4 ab + cb 2 = (2 a + kb ) 2 30. 9 a 2 – 12 ab + cb 2 = (3 a + kb ) 2 c = 9, k = –3 c = 4, k = –2 c = 1, k = –1 c = 25, k = 5 c = 9, k = 3 c = 64, k = 8

18 18 Topic 7.2.1 If the Coefficient of x 2 isn’t 1, Add a Number Completing the Square With an expression of the form ax 2 + bx, you can add a number to make an expression of the form a ( x + k ) 2.

19 19 Topic 7.2.1 Example 5 Solution follows… Completing the Square If 3 x 2 – 12 x + m is equal to 3( x + d ) 2, what is m ? Solution Multiply out the parentheses of 3( x + d ) 2 to get: 3( x + d ) 2 = 3 x 2 + 6 dx + 3 d 2 So 3 x 2 – 12 x + m = 3 x 2 + 6 dx + 3 d 2 Equate the coefficients of x, and the constants, to get: –12 = 6 d and m = 3 d 2 The first equation tells you that d = –2. And the second tells you that m = 3(–2) 2, or m = 12. So 3 x 2 – 12 x + 12 = 3( x – 2) 2.

20 20 Topic 7.2.1 Completing the Square Completing the square for ax 2 + bx The expression ax 2 + bx can be changed to a trinomial of the form a ( x + k ) 2. The resulting trinomial is a. b 2a2a 2 x + b 2 2 To do this, add =. 1 a b2b2 4a4a

21 21 Topic 7.2.1 Example 6 Solution follows… Completing the Square Convert 2 x 2 + 10 x to a perfect square trinomial. Solution 2 Here, a = 2 and b = 10, so you add = =. 10 2 1 2 1 2 100 4 1 2 25 2 This gives you 2 x 2 + 10 x +. 25 2

22 22 Topic 7.2.1 Guided Practice Solution follows… Completing the Square The quadratics below are of the form a ( x + d ) 2. Find the value of m and d in each equation. 31. 5 x 2 + 10 x + m = 5( x + d ) 2 32. 4 x 2 – 24 x + m = 4( x + d ) 2 33. 2 x 2 – 28 x + m = 2( x + d ) 2 34. 3 x 2 – 30 x + m = 3( x + d ) 2 35. 4 x 2 + 32 x + m = 4( x + d ) 2 m = 5, d = 1 m = 64, d = 4 m = 75, d = –5 m = 98, d = –7 m = 36, d = –3

23 23 Topic 7.2.1 Guided Practice Solution follows… Completing the Square The quadratics below are of the form a ( x + d ) 2. Find the value of m and d in each equation. 36. 20 x 2 + 60 x + m = 5(2 x + d ) 2 37. 20 x 2 – 20 x + m = 5(2 x + d ) 2 38. 27 x 2 + 18 x + m = 3(3 x + d ) 2 39. 27 x 2 + 36 x + m = 3(3 x + d ) 2 40. 16 x 2 – 80 x + m = 4(2 x + d ) 2 m = 45, d = 3 m = 100, d = –5 m = 12, d = 2 m = 3, d = 1 m = 5, d = –1

24 24 Topic 7.2.1 Guided Practice Solution follows… Completing the Square Add a term to convert each of the following into an expression of the form a ( x + k ) 2. 41. 2 x 2 – 12 x 42. 3 a 2 + 12 a 43. 6 y 2 – 60 y 44. 4 x 2 – 48 x 45. 5 x 2 + 245 46. 8 x 2 + 2 47. 36 x + 12 48. 120 x + 100 2 x 2 – 12 x + 18 = 2( x – 3) 2 5 x 2 + 70 x + 245 = 5( x + 7) 2 4 x 2 – 48 x + 144 = 4( x – 6) 2 6 y 2 – 60 y + 150 =6( y – 5) 2 3 a 2 + 12 a + 12 = 3( a + 2) 2 8 x 2 + 8 x + 2 = 8( x + ) 2 1 2 27 x 2 + 63 x + 12 = 27( x + ) 2 2 3 36 x 2 + 120 x + 100 = 36( x + ) 2 5 3

25 25 Topic 7.2.1 Independent Practice Solution follows… Completing the Square Find the value of c that will make each expression below a perfect square trinomial. 1. x 2 + 9 x + c 2. x 2 – 11 x + c 3. x 2 + 12 xy + c 4. x 2 – 10 xy + c 36 y 2 25 y 2 81 4 121 4

26 26 Topic 7.2.1 Independent Practice Solution follows… Completing the Square Complete the square for each quadratic expression below. 5. x 2 – 6 x 6. a 2 – 14 a 7. b 2 – 10 b 8. x 2 + 8 xy 9. c 2 – 12 bc 10. x 2 + 4 xy x 2 – 6 x + 9 x 2 + 4 xy + 4 y 2 c 2 – 12 bc + 36 b 2 x 2 + 8 xy + 16 y 2 b 2 – 10 b + 25 a 2 – 14 a + 49

27 27 Topic 7.2.1 Independent Practice Solution follows… Completing the Square Find the value of m and d in each of the following. 11. 5 x 2 – 40 x + m = 5( x + d ) 2 12. 2 x 2 + 20 x + m = 2( x + d ) 2 13. 3 x 2 – 6 x + m = 3( x + d ) 2 14. 3 x 2 – 30 x + m = 3( x + d ) 2 15. 4 x 2 + 24 x + m = 4( x + d ) 2 16. 7 x 2 – 28 x + m = 7( x + d ) 2 m = 80, d = –4 m = 28, d = –2 m = 36, d = 3 m = 75, d = –5 m = 3, d = –1 m = 50, d = 5

28 28 Topic 7.2.1 Independent Practice Solution follows… Completing the Square Add a term to convert each of the following into an expression of the form a ( x + k ) 2. 17. 3 x 2 – 30 x 18. 2 x 2 + 8 x 19. 18 x 2 – 48 x 20. 5 x 2 + 180 21. 27 x 2 + 12 22. 36 x 2 + 196 3( x – 5) 2 2( x + 2) 2 5( x + 6) 2 or 5( x – 6) 2 18( x – ) 2 4 3 27( x + ) 2 or 27( x – ) 2 2 3 2 3 36( x + ) 2 or 36( x – ) 2 7 3 7 3

29 29 Topic 7.2.1 Independent Practice Solution follows… Completing the Square 23. The length of a rectangle is twice its width. If the area can be found by completing the square for (18 x 2 + 60 x ) ft 2, find the width of the rectangle. (3 x + 5) ft

30 30 Topic 7.2.1 Round Up Completing the Square In the next Topic you’ll learn how to convert any quadratic expression into perfect square trinomial form — and then in Topic 7.2.3 you’ll use this to solve quadratic equations. OK, so now you know how to add a number to a binomial to make a perfect square trinomial.


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