Solving Quadratics
Tutorial 11g
Relating to the Real World
Members of the science club launch a model
rocket from ground level with a starting
velocity of 96 feet per second. After how
many seconds will the rocket have an altitude
of 128 ft?
The quadratic formula is important in physics
when finding vertical motion. When an object
is dropped, thrown, or launched either straight
up or down, you can use the vertical motion
formula to find the height of the object.
Vertical motion formula: h = -16t2 + vt +s
Introduction
 The vertical motion formula is a quadratic
equation.
 A quadratic equation is any equation that can
be written in the form ax2 + bx + c = 0, where
a, b, and c are real numbers and a  0.
 In this lesson you will learn how to solve any
quadratic by using the quadratic formula.
 Later in this lesson we will take a look back at
the previous vertical motion problem and learn
how to solve it.
The Quadratic Formula:
If ax2 + bx +c = 0 and a  0, then:
 b  b  4ac
x
2a
2
Remember: The solution of ax2 + bx +c = 0 are
numbers we call roots of the equation.
Also, the solution set of y = ax2 + bx +c is a set of
ordered pairs of numbers that satisfy the function.
The Quadratic Formula:
cont. . .
If ax2 + bx +c = 0 and a  0, then:
 b  b  4ac
x
2a
2
As you can see above, the coefficients of a
quadratic equation (values of a, b, & c) are used
to solve the equation.
Example 1
Use the Quadratic Formula to solve:
x2 + 5x = -6= -6
x2 + 5x = -6.
 First we must put the
equation in standard form.
 Identify the values of a, b, & c:
Remember
ax2
+ bx + c = 0
 Substitute 1 for a, 5 for b, and 6
for c in the Quadratic Formula.
 Then simplify the resulting
expression. CLICK HERE!
+6 =+6
5x + 6
6=0
1x2 + 5
a = 1,
1
b=5
5,
c = 66
 b  b  4ac
x
2a
2
 5  5  4 1 6
x
2 1
2
Example 1
Simplify:
Use the Quadratic Formula to solve:
x2 + 5x = -6.
Cont . . .
 5  5  4 1 6
x
2 1
2
 5  25  24
x
2 1
5 1
x
2 1
 5 1
x
2
or
 5 1
x
2
x  2
or
x  3
Relating to the Real World
Members of the science club launch a model
rocket from ground level with a starting
velocity of 96 feet per second. After how
many seconds will the rocket have an altitude
of 128 ft?
We will need to use the vertical motion
formula to find the height of the object.
Vertical motion formula: h = -16t2 + vt + s
h is the height of the object in feet.
t is the time it takes an object to rise or fall to a given height.
v is the starting velocity in feet per second.
s is the starting height.
Relating to the Real World
Members of the science club launch a model rocket from
ground level with a starting velocity of 96 feet per second.
After how many seconds will the rocket have an altitude of
128 ft?
Vertical motion formula: h = -16t2 + vt + s
h is the height of the object in feet.
t is the time it takes an object to rise or fall
to a given height.
v is the starting velocity in feet per second.
s is the starting height.
h = -16t2 + vt + s
128 = -16t2 + 96t + 0
 Use the Vertical motion formula.
 Substitute 128 for h, 96 for v, and 0 for s.
0 = -16t2 + 96t –128  Subtract 128 from each side.
Members of the science club launch a model rocket from ground level with a starting velocity
of 96 feet per second. After how many seconds will the rocket have an altitude of 128 ft?
Relating to the Real World
0 = -16t2 + 96t –128
h = -16t2 + vt + s
128 = -16t2 + 96t + 0
 Use the Vertical motion formula.
 Substitute 128 for h, 96 for v, and 0 for s.
0 = -16t2 + 96t –128  Subtract 128 from each side.
Members of the science club launch a model rocket from ground level with a starting velocity
of 96 feet per second. After how many seconds will the rocket have an altitude of 128 ft?
Relating to the Real World
0 = -16t2 + 96t –128
 b  b 2  4ac
t
2a
 96  962  4(16)(128)
t
2(16)
 96  9216  8192
t
 32
 96  1024
t
 32
 96  1024
t
 32
 96  1024
or t 
 32
 To solve, use the quadratic
formula.
 Substitute –16 for a, 96 for b,
and –128 for c.
 Simplify
 Write two solutions and then
simplify.
Members of the science club launch a model rocket from ground level with a starting velocity
of 96 feet per second. After how many seconds will the rocket have an altitude of 128 ft?
Relating to the Real World
0 = -16t2 + 96t –128
 b  b 2  4ac
t
2a
 96  962  4(16)(128)
t
2(16)
 96  9216  8192
t
 32
 96  1024
t
 32
t
 96  1024
 32
or t 
 96  1024
 32
 96  1024
 32
 96  32
t
 32
t
 96  1024
 32
 96  32
or t 
 32
or t 
t  2 or t  4
The rocket is 128 ft. off the ground
after 2 seconds & after 4 seconds.
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Tutorial 11g - C on T ech Math : : An application

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