Transcript 5.6 Lesson

HW: Pg. 287 #47-67eoo, 73, 75, 79, 81
5.6 The Quadratic Formula and the Discriminant
•
___________________ ,
____________ ,
_________ .
•
replacing the 0 with a "y" and then calculating
Check by:
"zeroes" on the graphing calculator
• The expression ______________ , where _____________ ,
are ________________ , of the quadratic equation
___________________ , is called the ____________________ .
EXAMPLE 1
Solve an equation with two real solutions
Solve x2 + 3x = 2.
x2 + 3x = 2
x2 + 3x – 2 = 0
x =– b + b2 – 4ac
2a
x = – 3 + 32 – 4(1)(–2)
2(1)
x =– 3 + 17
2
Write original equation.
Write in standard form.
Quadratic formula
a = 1, b = 3, c = –2
Simplify.
ANSWER
The solutions are x = –3 + 17
2
x = –3 – 17
–3.56.
2
0.56 and
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.
25x2 – 30x + 9 = 0.
x = 30 + (–30)2– 4(25)(9)
2(25)
30 + 0
x = 50
x = 53
ANSWER
The solution is 53
Write original equation.
Write in standard form.
a = 25, b = –30, c = 9
Simplify.
Simplify.
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
–x2 + 4x – 5 = 0.
x =–4 + 42 – 4(–1)(–5)
2(–1)
–4 + –4
x = –2
–4 + 2i
x = –2
x=2+i
ANSWER
The solution is 2 + i and 2 – i.
Write original equation.
Write in standard form.
a = –1, b = 4, c = –5
Simplify.
Rewrite using the
imaginary unit i.
Simplify.
EXAMPLE 3
Solve an equation with imaginary solutions
CHECK
Graph y = 2x2 + 4x – 5.
There are no xintercepts. So, the
original equation has no
real solutions. The
algebraic check for the
imaginary solution 2 + i
is shown.
?
2
–(2 + i) + 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.
1.
x2 = 6x – 4
ANSWER
2.
5
4x2 – 10x = 2x – 9
ANSWER
3.
3+
1 1
2
7x – 5x2 – 4 = 2x + 3
ANSWER
5 + i 115
10
EXAMPLE 4 Using the Discriminant
Find the discriminant of the quadratic equation and give
the number and type of solutions of the equation.
x 2 – 8x + 17 = 0
SOLUTION
EQUATION
ax + bx + c = 0
x 2 – 8x + 17 = 0
DISCRIMINANT
SOLUTION(S)
2 – 4ac
–b
±
b
x=
b 2 – 4ac
2a
(–8) 2 – 4(1)(17) = –4 Two imaginary: 4 ± i
EXAMPLE 4 Using the Discriminant
Find the discriminant of the quadratic equation and
give the number and type of solutions of the equation.
x 2 – 8x + 17 = 0
x 2 – 8x + 16 = 0
SOLUTION
EQUATION
DISCRIMINANT
SOLUTION(S)
–b ± b 2 – 4ac
x
=
ax + bx + c = 0
b 2 – 4ac
2a
x 2 – 8x + 17 = 0 (–8) 2 – 4(1)(17) = –4 Two imaginary: 3 ± i
x 2 – 8x + 16 = 0
(–8) 2 – 4(1)(16) = 0
One real: 4
EXAMPLE 4 Using the Discriminant
Find the discriminant of the quadratic equation and
give the number and type of solutions of the equation.
x 2 – 8x + 17 = 0
x 2 – 8x + 16 = 0
x 2 – 8x + 15 = 0
SOLUTION
EQUATION
ax + bx + c = 0
x 2 – 8x + 17 = 0
x 2 – 8x + 16 = 0
x 2 – 8x + 15 = 0
DISCRIMINANT
SOLUTION(S)
2 – 4ac
–b
±
b
x=
b 2 – 4ac
2a
(–8) 2 – 4(1)(17) = –4 Two imaginary: 3 ± i
(–8) 2 – 4(1)(16) = 0 One real: 3
(–8) 2 – 4(1)(15) = 4 Two real: 3, 1
EXAMPLE 4 Using the Discriminant
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.
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.
y = x 2 – 6x + 10 Graph is above x-axis (no x-intercept)
y = x 2 – 6x + 9
Graph touches x-axis (one x-intercept)
y = x 2 – 6x + 8
Graph crosses x-axis (two x-intercepts)
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 = 0
ANSWER 48 ; Two real solutions
5. 3x2 + 12x + 12 = 0
ANSWER 0 ; One real solution
6. 8x2 = 9x – 11
ANSWER –271 ; Two imaginary solutions
VOCABULARY
Using the Quadratic Formula in Real Life
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.
Models
Labels
h = –16t 2 + h 0
Object is dropped.
h = –16t 2 + v 0 t + h 0
Object is launched or thrown.
h = height
(feet)
t = time in motion
(seconds)
h 0 = initial height
(feet)
v 0 = initial vertical velocity (feet per second)
VOCABULARY
Using the Quadratic Formula in Real Life
The initial velocity of a launched object can be
positive, negative, or zero.
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).
v0 > 0
v0 < 0
v0 = 0
EXAMPLE 5 Solving a Vertical Motion Problem
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
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.
h = –16t 2 + v 0 t + h 0
Write height model.
5 = –16t 2 + 45t + 6
0 = –16t 2 + 45t + 1
t = –45 ± 2089
–32
h = 5, v 0 = 45, h 0=6
a = –16, b = 45, c = 1
Use quadratic formula
t is about -0.22 and 2.83, so the baton is in the air for about 2.83 sec.
GUIDED PRACTICE for Example 5
7.
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?
ANSWER
The diver had about 2.56 seconds to perform
acrobatic maneuvers before hitting the water.
HOMEWORK:
Pg. 295-297 #25-61eoo, 76, 78