22.3 Using the quadratic formula to Solve Equations

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Name
Class
Date
22.3 Using the q uadratic f ormula
to Solve Equations
Essential Question: What is the quadratic formula, and how can you use it to solve
quadratic equations?
Resource
Locker
Deriving the q uadratic f ormula
Explore
You can complete the square on the general form of a quadratic equation to derive a formula that can be used to solve
any quadratic equation.
A
Write the standard form of a quadratic equation.
ax 2 + bx + c =
B
0
Subtract c from both sides.
ax 2 + bx = -c
C
Multiply both sides by 4a to make the coefficient of x 2 a perfect square.
4a 2x 2 +
D
4abx
=
-4ac
Add b 2 to both sides of the equation to complete the square.
© Houghton Mifflin Harcourt Publishing Company
4a 2x 2 + 4abx + b 2 = -4ac + b 2
E
F
Factor the left side to write the trinomial as the square of a binomial.
(
2ax + b
2
Take the square roots
of both sides.
__
2ax + b
G
) = b - 4ac
2
√
=±
Subtract b from both
sides.
__
2ax = -b ±
H
b 2 - 4ac
√
b 2 - 4ac
Divide both sides
by 2a to solve for x.
__
√
b 2 - 4ac
-b ±
__
x=
Module 22
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__

-b ± √b 2 - 4ac
The formula you just derived, x = __, is called the quadratic formula.
2a
It gives you the values of x that solve any quadratic equation where a ≠ 0.
Reflect
1.
What If? If the derivation had begun by dividing each term by a, what would the resulting binomial of x
have been after completing the square? Does one derivation method appear to be simpler than the other?
Explain.
+ bx + c = 0
b
c
2
_
x
+_
ax + a = 0
b
c
2
_
x
+_
ax = - a 2
2
b
b
b
c+ _
2
_
_
x
+_
x
+
=
a 2 2a
a 2 2a
b
b
c
_
_
x+
= -a +
2a
2a
b
The resulting binomial is x +
. The previous derivation appears simpler because no
2a
fractions are involved in the derivation.
2
ax
(
( )
_)
( )
( )
(_)
Using the Discriminant to Determine
the Number of Real Solutions
Explain 1
Recall that a quadratic equation, ax 2 + bx + c, can have two, one, or no real solutions. By evaluating the part of
the quadratic formula under the radical sign, b 2 - 4ac, called the discriminant, you can determine the number of
real solutions.
Example 1 Determine how many real solutions each quadratic equation has.

x 2 - 4x + 3 = 0
Identify a, b, and c.
b 2 - 4ac
Use the discriminant.
(-4) 2 - 4(1)(3)
Substitute the identified values into the discriminant.
16 - 12 = 4
Simplify.
a = 1, b = -4, c = 3
Since b 2 - 4ac > 0, the equation has two real solutions.

x 2 - 2x + 2 = 0
a = 1 , b = -2 , c = 2
b 2 - 4ac
( -2 ) - 4(
2
4
- 8
1
)(
= -4
2
)
Simplify.
Since b 2 - 4ac < 0, the equation has no real solution(s).
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Reflect
2.
When the discriminant is positive, the quadratic equation has two real solutions. When the discriminant
is negative, there are no real solutions. How many real solutions does a quadratic equation have if its
discriminant equals 0? Explain.
One real solution; if the discriminant is 0, then you are adding or subtracting the square
root of 0 in the quadratic formula. Since the only square root of 0 is 0, the answer will be
the same whether it is added or subtracted.
Your Turn
Use the discriminant to determine the number of real solutions for each
quadratic equation.
3.
x 2 + 4x + 1 = 0
4.
2x 2 - 6x + 15 = 0
a = 1, b = 4, c = 1
a = 2, b = -6, c = 15
(4) - 4(1)(1)
(-6) - 4(2)(15)
b 2 - 4ac
b 2 - 4ac
2
2
16 - 4 = 12
two real solutions
36 - 120 = -84
no real solutions
5.
x 2 + 6x + 9 = 0
a = 1, b = 6, c = 9
b 2 - 4ac
(6) 2 - 4(1)(9)
36 - 36 = 0
one real solution
Solving Equations by Using the Quadratic Formula
Explain 2
To use the quadratic formula to solve a quadratic equation, check that the equation is in standard form. If not, rewrite
it in standard form. Then substitute the values of a, b, and c into the formula.
Example 2 Solve using the quadratic formula.

2x 2 + 3x - 5 = 0
a = 2, b = 3, c = -5
-b ± √b - 4ac
x = __
2a
Use the quadratic formula.
__
2
___
2
-3 ± √(3) - 4(2)(-5)
x = ___
2(2)
_
Substitute the identified values into the
quadratic formula.
-3 ± √ 49
x=_
4
-3 ± 7
x=_
4
-3 + 7
-3 - 7
_
x=
or
x=_
4
4
5
_
x = 1 or
x=2
5.
The solutions are 1 and
-_
2
Write as two equations.
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Simplify the radicand and the denominator.
Evaluate the square root.
Simplify both equations.
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Graph y =2 x 2 + 3x - 5 to verify your answers.
The graph does verify the solutions.

2x = x 2 - 4
x 2 - 2x - 4 = 0
Write in standard form.
a = 1 , b = -2 , c = -4
-b ± √b 2 - 4ac
x = __
2a
―――
―――――――――
( ) √( -2 ) - 4( 1 )( -4 )
x = ――――――――――――――
2( 1 )
――
- -2 ±
2
quadratic formula.
√
2±
20
x = __
2
―――
√
―
Simplify the radicand and the
denominator.
―
2±
2 ± 2 √5
4 ∙5
x = __ = _= 1 ± √5
2
2
Simplify.
x = 1 + √5 or x = 1 - √5
―
or x ≈ -1.236
The exact solutions are
3.236
―
―
Use a calculator to find approximate
solutions to three decimal places.
1 + √5 and 1 - √5 . The approximate solutions are
and -1.236 .
Graph y = x 2 - 2x - 4 and find the zeros using the graphing calculator. The calculator
will give approximate values.
The graph
does
confirm the solutions.
x ≈ 3.236
―
Reflect
6.
Discussion How can you use substitution to check your solutions?
Substitute each value into the given quadratic equation to see if it leads to a true equality.
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Your Turn
Solve using the quadratic formula.
7.
x 2 - 6x - 7 = 0
――――――
-(-6)± √(-6) - 4(1)(-7)
___
x=
2(1)
6 ± √―
64
x=_
2
6±8
_
x=
2
6+8
6-8
_
x=
or x = _
8.
2
2
x = 7 or
2x 2 = 8x - 7
――――――
-(-8)± √(-8) - 4(2)(7)
___
x=
2(2)
8 ± √―
8
x=_
4
8 ±2 √―
2
_
x=
4 ―
8 -2 √―
8 +2 √2
2
or
x=_
x=_
4
4
√―
√―
2
2
or x = 2 - _
x=2+_
2
2
√―
√―
2
2
The solutions are 2 + _and 2 - _
.
2
2
x = -1
The solutions are 7 and -1.
2
Explain 3
2
Using the Discriminant with Real-World Models
Given a real-world situation that can be modeled by a quadratic equation, you can find the number of real solutions
to the problem using the discriminant, and then apply the quadratic formula to obtain the solutions. After finding the
solutions, check to see if they make sense in the context of the problem.
In projectile motion problems where the projectile height h is modeled by the equation h = −16t 2 + vt + s, where
t is the time in seconds the object has been in the air, v is the initial vertical velocity in feet per second, and s is the
initial height in feet. The -16 coefficient in front of the t 2 term refers to the effect of gravity on the object. This
equation can be written using metric units as h = −4.9t 2 + vt + s, where the units are converted from feet to meters.
Time remains in units of seconds.
Example 3 For each problem, use the discriminant to determine the number of real
solutions for the equation. Then, find the solutions and check to see if
they make sense in the context of the problem.
 A diver jumps from a platform 10 meters above the surface of the water. The diver’s height is
given by the equation h = −4.9t 2 + 3.5t + 10, where t is the time in seconds after the diver
jumps. For what time t is the diver’s height 1 meter?
Substitute h = 1 into the height equation. Then, write the resulting quadratic equation in
standard form to solve for t.
1 = −4.9t 2 + 3.5t + 10 0
= −4.9t 2 + 3.5t + 9
First, use the discriminant to find the number of real solutions of the equation.
b 2 - 4ac
(3.5)2 - 4(-4.9)(9) = 188.65
Since
Module22
b 2 − 4ac > 0, the equation has two real solutions.
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Correcti
Next, use the quadratic formula to find the real number solutions.
a = −4.9, b = 3.5, c = 10
―――
-b ± √b - 4ac
2
t = __
2a
-3.5 ± √188.65
t = __
2(-4.9)
―――
Substitute the identified values into the quadratic
formula and the value of the discriminant.
-3.5 ± 13.73
t ≈ __
-9.8
-3.5 - 13.73
-3.5
+ 13.73 or t ≈ __
t ≈ __
-9.8
-9.8
t ≈ -1.04
or
t ≈ 1.76
Simplify.
Solutions
Disregard the negative solution because t represents the seconds after the diver jumps and a
negative value has no meaning in this context. So, the diver is at height 1 meter after a time
of t ≈ 1.76 seconds.
B
The height in meters of a model rocket on a particular launch can be modeled by the
equation h = −4.9t 2 + 102t + 100, where t is the time in seconds after its engine burns out
100 meters above the ground. When will the rocket reach a height of 600 meters?
Substitute h = 600 into the height equation. Then, write the resulting quadratic equation
in standard form to solve for t.
h = −4.9t 2 + 102t + 100
600 = −4.9t 2 + 102t + 100
0 = −4.9t 2 + 102t - 500
First, use the discriminant to find the number of real solutions of the equation.
a = −4.9, b = 102 , c = -500
b 2 - 4ac
10404
-
9800
-500
)
= 604
Simplify.
Since b 2 - 4ac > 0, the equation has 2 real solutions.
Module 22
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( 102 ) - 4(-4.9)(
2
Lesson 3
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Next, use the quadratic formula to find the real number solutions.
―――
√
102
604
- ±
t = __
2(-4.9)
quadratic formula and the value of the
discriminant.
-102
24.58
±
t = ___
-9.8
Simplify.
24.58
24.58
- 102 +
- 102 t ≈ __ or t ≈ __
-9.8
-9.8
t ≈ -7.90
Solutions
Disregard the
or
negative
launched and a negative
t ≈ 12.92
solution because t represents the seconds after the rocket has
value has no meaning in this context. So, the rocket is at
height 600 meters after a time of t ≈
12.92
seconds.
Your Turn
For each problem, use the discriminant to determine the number of real solutions
for the equation. Then, find the solutions and check to see if they make sense in the
context of the problem.
9.
A soccer player uses her head to hit a ball up in the air from a height of 2 meters with an initial vertical
velocity of 5 meters per second. The height h in meters of the ball is given by h = −4.9t 2 + 5t + 2, where t
is the time elapsed in seconds. How long will it take the ball to hit the ground if no other players touch it?
h = −4.9t 2 + 5t + 2
0 = −4.9t 2 + 5t + 2
Find the discriminant.
(5)2 − 4(-4.9)(2) = 64.2
Since b 2 - 4ac > 0, the equation has two real solutions.
Use the quadratic formula to find the solutions of the quadratic equation.
64.2
-5 ± √――
__
-9.8
-5 ± 8.01
t≈_
t=
-9.8
t ≈ -0.31 or t ≈ 1.33
Disregard the negative solution because there is no negative time in this problem context.
The soccer ball reached the ground after about t ≈ 1.33 seconds.
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10. The quarterback of a football team throws a pass to the
team’s receiver. The height h in meters of the football
can be modeled by h = −4.9t 2+ 3t + 1.75, where t is the
time elapsed in seconds. The receiver catches the football
at a height of 2 meters. How long does the ball remain in
the air until it is caught by the receiver?
The ball is caught by the receiver when h = 2.
h = −4.9t 2+ 3t + 1.75
0.25 = −4.9t 2+ 3t + 1.75
0 = −4.9t 2+ 3t + 1.5
(3) - 4(−4.9)( 1.5)= 38.4
2
Since b 2- 4ac > 0, the equation has two real solutions.
―― 
__
 
_ 
Use the quadratic formula to find the solutions of the quadratic equation.
-3 ±  √38.4 
 
 
t =  
  
-9.8
-3 ± 6.20
t ≈  
  
-9.8
t ≈ -0.33 or t ≈ 0.94
Disregard the negative solution because there is no negative time in this problem
context.The ball was in the air for t ≈ 0.94 second.
Elaborate intersect the x-axis at two points. If the discriminant is negative, the equation will have no
solutions: the graph will not intersect the x-axis at all.
12. What advantage does using the quadratic formula have over other methods of solving quadratic equations?
The quadratic formula works for all quadratic equations. Other methods only work in
certain situations.
13. Essential Question Check-In How can you derive the quadratic formula?
The quadratic formula is derived by completing the square for the standard form of a
quadratic equation.
Module 22
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© Houghton Mifflin Harcourt Publishing Company · Image Credits: © Corbis
Royalty Free
11. How can the discriminant of a quadratic equation be used to determine the number of zeros (x-intercepts)
that the graph of the equation will have?
If the discriminant is zero, the equation will have one solution: it will intersect the x-axis
at one point. If the discriminant is positive, the equation will have two solutions: it will
Lesson 3
3/23/14 6:30 AM

22.3 Using the quadratic formula to Solve Equations

Transcription

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