Geometry problems Projectile problems

1. 1.5 Applications of Quadratic
Equations
Chapter 1 Equations and Inequalities

2. Concepts & Objectives
⚫ Applications of Quadratic Equations
⚫ Geometry problems
⚫ Height of a projected object
⚫ Modeling with Quadratic Equations

3. Geometry Problems
⚫ Problems involving the area or volume of a geometric
object or problems involving the Pythagorean Theorem
often lead to quadratic equations.
⚫ As we did on applying linear equations, it’s important to
⚫ Read the problem. Make a diagram if necessary, so
you understand what you are trying to find.
⚫ Assign a variable – remember, you can often express
one quantity in terms of another.
⚫ Write an equation and solve it.
⚫ State your answer – are you answering the question?
⚫ Check – does your answer make sense?

4. Geometry Problems (cont.)
Example: A piece of machinery is capable of producing
rectangular sheets of metal such that the length is three
times the width. Furthermore, equal-sized squares
measuring 5 in. on a side can be cut from the corners so
that the resulting piece of metal can be shaped into an open
box by folding up the flaps.
If specifications call for the volume of the box to be
1435 in3, what should the dimensions of the original piece
of metal be?

5. Geometry Problems (cont.)
⚫ Read the problem: We need to find the dimensions of
the original piece of metal. Draw a rectangle.
⚫ Assign a variable: Let x be the width; this means that the
length would be represented as 3x. Each corner has a
5×5 square cut out. We then subtract that from the
length and the width.
3x
x
5
5
3x – 10
x – 10

6. Geometry Problems (cont.)
⚫ Write an equation. The volume of a rectangular solid is
V = lwh. This means our equation is:
5
3x – 10
x – 10
( )( )( )1435 3 10 10 5x x= − −
2
1435 15 200 500x x= − +
2
15 200 935 0x x− − =
2
3 40 187 0x x− − =
( )( )
( )
2
40 40 4 3 187 40 3844 40 62
2 3 6 6
x
 − −  
= = =
11
or 17
3
x = −
X

7. Geometry Problems (cont.)
⚫ State the answer.
The width is 17 in., so the length is 3(17) = 51 in.
⚫ Check:
( )( )( )
( )( )( ) 3
51 10 17 10 5
41 7 5 1435 in
V lwh=
= − −
= =

8. Pythagorean Theorem
⚫ The Pythagorean Theorem sets up the relationship
between the sides of a right triangle. It states that
⚫ The hypotenuse will always be the longest side.
In a right triangle, the sum of the squares
of the lengths of the legs is equal to the
square of the hypotenuse.
hypotenuse (c)
right
angle
a
b
2 2 2
a b c+ =

9. Pythagorean Theorem (cont.)
⚫ Example: Erik finds a piece of property in the shape of a
right triangle. He finds that the longer leg is 20 m longer
than twice the length of the shorter leg. The hypotenuse
is 10 m longer than the length of the longer leg. Find
the lengths of the sides of the triangular lot.
x
2x + 20
2x + 30
( ) ( )
2 22
2 20 2 30x x x+ + = +

10. Pythagorean Theorem (cont.)
⚫ The lengths of the sides of the property are 50 m, 120 m,
and 130 m.
( ) ( )
2 22
2 20 2 30x x x+ + = +
2 2 2
4 80 400 4 120 900x x x x x+ + + = + +
2
40 500 0x x− − =
–500
–50 10
–40
( )( )50 10 0x x− + =
50, 10x = −X
( )2 50 20 120 m+ =
( )2 50 30 130 m+ =

11. Height of a Projected Object
⚫ If air resistance is neglected, the height s (in feet) of an
object projected directly upward from an initial height of
s0 feet, with initial velocity v0 feet per second is
where t is the number of seconds after the object is
projected.
⚫ The coefficient of t2, –16, is a constant based on the
gravitational force of Earth. This constant varies on
other surfaces, such as the moon and other planets.
2
0 016 ,s t v t s= − + +

12. Height of a Projected Object
⚫ Example: If a projectile is shot vertically upward from
the ground with an initial velocity of 100 ft per sec,
neglecting air resistance, its height s (in feet) above the
ground t seconds after projection is given by
a) After how many seconds will it be 50 ft above the
ground?
b) How long will it take for the projectile to return to the
ground?
2
16 100 .s t t= − +

13. Height of a Projected Object
a) We must find value(s) of t so that height s is 50 ft.
2
50 16 100t t= − +
2
16 100 50 0t t− + =
2
8 50 25 0t t− + =
( ) ( ) ( )( )
( )
2
50 50 4 8 25
2 8
t
− −  − −
=
50 1700
16
t

=
.55 or 5.70t t 
Both values are acceptable,
since the projectile reaches
50 ft on the way up and the
way down.

14. Height of a Projected Object
b) When the projectile returns to the ground, its height
will be 0 ft, so let s = 0 in the given equation.
t = 0 represents when the projectile is launched, and
6.25 sec is when it returns to the ground.
2
0 16 100t t= − +
2
0 4 25t t= −
( )0 4 25t t= −
25
0 or 6.25
4
t t= = =

15. Height of a Project Object

16. Classwork
⚫ 10/14 Attendance Check/Progress Check
⚫ 1.5 Assignment (College Algebra)
⚫ Page 127: 6-18 (even); page 119: 32-56 (×4; omit
44); page 110: 70-90 (even; omit 82)
⚫ 1.5 Classwork Check
⚫ Quiz 1.4