Chapter 4 Quadratics 4.7 Solving Quadratic Equations: The Quadratic Formula

Chapter 4 Quadratics4.7 Solving Quadratic Equations:

The Quadratic Formula

Humour Break

4.7 Solving Quadratic Equations: The Quadratic Formula

Goals for Today:(1) We can use the quadratic formula to find the zeros

of any hard-to-factor quadratic equation in standard form

(2) In the quadratic formula, a portion of the formula called the discriminant can be used to tell us whether there are one, two, or no zeros

(3) We can find the zeros of a quadratic equation in vertex form using algebra instead of expanding and then factoring

4.7 Solving Quadratic Equations: The Quadratic Formula – Finding The Zeros

• When we have a quadratic equation in standard form, we learned how to find the zeros by factoring (Chapter 3)

• When we have a quadratic equation in vertex form, we could expand the equation and put it into standard form & then factor it to find the zeros


Recall from Chapter 3 that there are times where we couldn’t find the factors of a trinomial because they didn’t factor evenly to integers

This didn’t necessarily that there are no factors – but it did mean that there were no factors that we could discover using the butterfly method which is really an organized way of doing guess & check


At that time I said there is a way...


a

acbbx

2

42

Where we have a trinomial in standard form y = ax² + bx + c


Step 1: Identify what “a”, “b” & “c” are in the trinomial equation

Ex.1 x² - 4x - 1


Step 1: Identify what “a”, “b” & “c” are in the trinomial equation

Ex.1 x² - 4x – 1

a is 1b is - 4c is - 1


a

acbbx

2

42

Step 2: Plug a, b & c into the quadratic formulaEx.1 a = 1, b = – 4 & c = - 1


)1(2

)1)(1(4)4()4( 2 x

Step 2: Plug a, b & c into the quadratic formulaEx. 1 a = 1, b = – 4 & c = - 1


)1(2

)1)(1(4)4()4( 2 x

Step 3: Solve...Ex. 1


2

4164 x

Step 3: Solve...Ex. 1


2

204x

Step 3: (Cont’d... ) Solve...Ex. 1


2

47.44x



2

47.44......

2

47.44

xorx



2

47.0......

2

47.8 xorx



24.0......24.4 xorx


4.7 Solving Quadratic Equations: The Quadratic Formula - Discriminant

• We can use a portion of the quadratic formula to tell us whether or not a quadratic equation has 1, 2 or no zeros without solving for the zeros


acb 42


• If b² - 4ac is > 0... there are 2 zeros• Ex. y = -2x² + 8x - 1


• If b² - 4ac is > 0... there are 2 zeros• Ex. y = -2x² + 8x - 1


• If b² - 4ac is > 0... there are 2 zeros• Ex. y = -2x² + 8x – 1• a = -2, b = 8, c = -1 (Identify a, b & c)• b² - 4ac (discriminant form.)• = 8² - 4(-2)(-1)• = 64 – 8• = 56 (+ve so 2 zeros)


• If b² - 4ac is = 0... there is 1 zero• Ex. y = -2x² + 8x - 8


• If b² - 4ac is = 0... there is 1 zero• Ex. y = -2x² + 8x - 8


• If b² - 4ac is = 0... there is 1 zero• Ex. y = -2x² + 8x – 8• a = -2, b = 8, c = -8 (Identify a, b & c)• b² - 4ac (discriminant form.)• = 8² - 4(-2)(-8)• = 64 – 64• = 0 (so 1 zero)


• If b² - 4ac is < 0... there are no zeros• Ex. y = 2x² + 8x +15


• If b² - 4ac is < 0... there are no zeros• Ex. y = 2x² + 8x +15


• If b² - 4ac is < 0... there are no zeros• Ex. y = 2x² + 8x +15• a = 2, b = 8, c = 15 (Identify a, b & c)• b² - 4ac (discriminant form.)• = 8² - 4(2)(15)• = 64 – 120• = -56 (so no zeros)


Ex. y = 2(x – 3)² - 2Find the zeros...


y = 2(x – 3)² - 2y = 2(x² - 6x – 6x + 9) – 2 ... expand using FOILy = 2(x² - 6x + 9) – 2y = 2x² - 12x + 18 – 2y = 2x² - 12x + 16 .... now in standard formy = 2(x² - 6x + 8) ... factor our 2 (GCF)y = 2(x – 4)(x – 2) ... butterfly chartSo, the zeros are 4 and 2




Step 1: Set y = 0 & move the “k” or the “- 2” to the left hand side of the equation



Step 1: Set y = to 0 & move the “- 2” to the left hand side (where it becomes a positive 2)

2 = 2(x – 3)²


Step 2: Divide the left & right-hand side by 2

2 = 2(x – 3)²


Step 2: Divide the left & right-hand side by 2

2 = 2(x – 3)²-- ----------2 2


Step 3: Take the square root of both sides

1 = (x – 3)²


Step 3: Take the square root of both sides*

√1 = √(x – 3)²

*When you take the square root of the constant, you have to recognize that it could be either

+ 1 or -1 (i.e. 1 x 1 = 1 & -1 x -1 also = 1)


Step 4: Use algebra to solve for x

+1 = x – 3 or -1 = x – 3


Step 4: Use algebra to solve for x

+1 = x – 3 or -1 = x – 3

1 + 3 = x or -1 + 3 = x

x = 4 or x = 2

4.7 Solving Quadratic Equations: The Quadratic Formula – Movement Ex.

• A digital sensor records the path of Rachel’s soother after she throws it into the air. The equation that models the situation is

y = -4.9x² + 20.58x + 0.491 When does the soother hit the ground?

a

acbbx

2

42



• y = -4.9x² + 20.58x + 0.491 When does the soother hit the ground?a = -4.9, b = 20.58 & c = 0.491

)9.4(2

)491.0)(9.4(458.2058.20 2

x



8.9

6236.95364.42358.20

x



8.9

8125.2058.20

x



8.9

8125.205.20......

8.9

8125.2058.20

xorx



8.9

3125.41......

8.9

2325.0

xorx



22.4......02.0 xorxSo, the soother hits the ground after 4.22 seconds

Homework

• Tuesday, May 24th

• p.403, #4, 5, 6, 11 odd• Wednesday, May 25th

• p.404, #16, 17, 18 & 19

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Chapter 4 Quadratics 4.7 Solving Quadratic Equations: The Quadratic Formula