HW: Pg. 287 #47-67eoo, 73, 75, 79, 81

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5.6 The Quadratic Formula and the Discriminant

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EXAMPLE 1 Solve an equation with two real solutions

Solve x2 + 3x = 2.x2 + 3x = 2 Write original equation.

x2 + 3x – 2 = 0 Write in standard form.

x =– b + b2 – 4ac2a

Quadratic formula

x =– 3 + 32 – 4(1)(–2) 2(1)

a = 1, b = 3, c = –2

Simplify.x =– 3 + 172

The solutions arex = –3 + 172

0.56 andx = –3 – 17

2–3.56.

ANSWER

EXAMPLE 1 Solve an equation with two real solutions

CHECK

Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about –3.56.

EXAMPLE 2 Solve an equation with one real solutions

Solve 25x2 – 18x = 12x – 9.25x2 – 18x = 12x – 9. Write original equation.

Write in standard form.

x =30 + (–30)2– 4(25)(9)2(25)

a = 25, b = –30, c = 9

Simplify.

25x2 – 30x + 9 = 0.

x =30 + 0

50

x = 35

Simplify.

35The solution is

ANSWER

EXAMPLE 2 Solve an equation with one real solutions

CHECK

Graph y =25x2 – 30x + 9 and note that the only x-intercept is 0.6 = 3

5

EXAMPLE 3 Solve an equation with imaginary solutions

Solve –x2 + 4x = 5.–x2 + 4x = 5 Write original equation.

Write in standard form.

x =–4 + 42 – 4(–1)(–5)2(–1)

a = –1, b = 4, c = –5

Simplify.

–x2 + 4x – 5 = 0.

x =–4 + –4

–2–4 + 2i

x = –2Simplify.

Rewrite using the imaginary unit i.

x = 2 + i

The solution is 2 + i and 2 – i.

ANSWER

EXAMPLE 3 Solve an equation with imaginary solutions

CHECK

Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + i is shown.

–(2 + i)2 + 4(2 + i) = 5?

–3 – 4i + 8 + 4i = 5?

5 = 5

GUIDED PRACTICE for Examples 1, 2, and 3

Use the quadratic formula to solve the equation.

x2 = 6x – 41.

3 + 5

4x2 – 10x = 2x – 92.

12

1

7x – 5x2 – 4 = 2x + 33.

115 5 + i10

ANSWER

ANSWER

ANSWER

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

SOLUTION

x 2 – 8 x + 17 = 0

x 2 – 8x + 17 = 0 Two imaginary: 4 ± i

a x + b x + c = 0

EQUATION DISCRIMINANT SOLUTION(S)

b 2 – 4a c

(–8) 2 – 4(1)(17) = – 4

x = –b ± b 2 – 4a c

2a

Using the DiscriminantEXAMPLE 4

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

x 2 – 8 x + 17 = 0

x 2 – 8 x + 17 = 0 Two imaginary: 3 ± i

a x + b x + c = 0

EQUATION DISCRIMINANT SOLUTION(S)

b 2 – 4a c

(–8) 2 – 4(1)(17) = – 4

x 2 – 8 x + 16 = 0 (–8)

2 – 4(1)(16) = 0 One real: 4

x 2 – 8 x + 16 = 0

SOLUTION

x = –b ± b 2 – 4a c

2a

Using the DiscriminantEXAMPLE 4

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

x 2 – 8 x + 17 = 0

x 2 – 8 x + 17 = 0 Two imaginary: 3 ± i

a x + b x + c = 0

EQUATION DISCRIMINANT SOLUTION(S)

b 2 – 4a c

(–8) 2 – 4(1)(17) = – 4

x 2 – 8 x + 16 = 0 (–8)

2 – 4(1)(16) = 0 One real: 3

x 2 – 8 x + 16 = 0 x 2 – 8 x + 15 = 0

x 2 – 8x + 15 = 0 (–8)

2 – 4(1)(15) = 4 Two real: 3, 1

SOLUTION

x = –b ± b 2 – 4a c

2a

Using the DiscriminantEXAMPLE 4

In the previous example you may have noticed that the number of real solutions of x 2 – 6 x + c = 0 can be changed just by changing the value of c.

y = x 2 – 6x + 10

y = x 2 – 6 x + 9

y = x 2 – 6 x + 8

Graph is above x-axis (no x-intercept)

Graph touches x-axis (one x-intercept)

Graph crosses x-axis (two x-intercepts)

A graph can help you see why this occurs. By changing c, you can move the graph of y = x 2 – 6x + c up or down in the coordinate plane. If the graph is moved too high, it won’t have an x-intercept and the equation x 2 – 6x + c = 0 won’t have a real-number solution.

Using the DiscriminantEXAMPLE 4

GUIDED PRACTICE for Example 4

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

4. 2x2 + 4x – 4 = 048 ; Two real solutions

5.

0 ; One real solution

3x2 + 12x + 12 = 0

6. 8x2 = 9x – 11

–271 ; Two imaginary solutionsANSWER

ANSWER

ANSWER

Previously you studied the model h = –16t 2 + h 0 for the height of an object that is dropped. For an object that is launched or thrown, an extra term v 0 t must be added to the model to account for the object’s initial vertical velocity v 0.

Modelsh = –16 t

2 + h 0

h = –16 t 2 + v 0 t + h 0

Object is dropped.

Object is launched or thrown.

Labelsh = height

t = time in motionh 0 = initial height

v 0 = initial vertical velocity

(feet)

(seconds)

(feet)

(feet per second)

VOCABULARY Using the Quadratic Formula in Real Life

The initial velocity of a launched object can be positive, negative, or zero.

v 0 > 0 v 0 < 0 v 0 = 0

If the object is launched upward, its initial vertical velocity is positive (v 0 > 0).

If the object is launched downward, its initial vertical velocity is negative (v 0 < 0).

If the object is launched parallel to the ground, its initial vertical velocity is zero (v 0 = 0).

Using the Quadratic Formula in Real LifeVOCABULARY

A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet above the ground and has an initial vertical velocity of 45 feet per second. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air?

SOLUTION

h = –16 t 2 + v 0 t + h 0

h = 5, v 0 = 45, h 0=6

Since the baton is thrown (not dropped), use the model h = –16t

2 + v 0 t + h 0 with v 0 = 45 and h 0 = 6. To determine how long the baton is in the air, find the value of t for which h = 5.

5 = –16 t 2 + 45t + 6

0 = –16 t 2 + 45t + 1 a = –16, b = 45, c = 1

Solving a Vertical Motion Problem

Write height model.

EXAMPLE 5

Use quadratic formula

t is about -0.22 and 2.83, so the baton is in the air for about 2.83 sec.

t = – 45 ± 2089–32

7.

GUIDED PRACTICE for Example 5

ANSWER

In July of 1997, the first Cliff Diving World Championship were held in Brontallo, Switzerland. Participants performed acrobatic dives from heights of up to 92 feet. Suppose a cliff diver jumps from this height with an initial upward velocity of 5 feet per second. How much time does the diver have to perform acrobatic maneuvers before hitting the water?

The diver had about 2.56 seconds to perform acrobatic maneuvers before hitting the water.

HOMEWORK:

Pg. 295-297 #25-61eoo, 76, 78