Lesson 10: Quadratic Equations - Duke TIP · Lesson 10: Quadratic Equations Problem Solving...
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Lesson 10: Quadratic Equations Problem Solving Assignment solutions Algebra 1 © 2009 Duke University Talent Identification Program Page 1 of 5 Solve. 1. The length of a rectangle is 3 less than twice the width. If the area of the rectangle is 2 25 in , find the length of each side to the nearest hundredth. ( ) 2 3 25 A l w w w = ⋅ = − = ( ) 2 2 2 3 25 2 3 25 2 3 25 0 w w w w w w − = − = − − = This won’t factor so we solve by the quadratic formula. ( ) ( ) ( )( ) ( ) 2 3 3 42 25 22 3 9 200 4 3 209 4 w −− ± − − − = ± + = ± = Since w cannot be negative, we get: 3 209 4.36 in 4 w + = = And the length is ( ) 2 3 2 4.36 3 5.72 in l w = − = − = . Check: ( ) ( ) ? ? 25 5.72 4.36 25 24.94 25 l w ⋅ = ⋅ = ≈ w 2w-3
Transcript of Lesson 10: Quadratic Equations - Duke TIP · Lesson 10: Quadratic Equations Problem Solving...
Lesson 10: Quadratic Equations Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 1 of 5
Solve.
1. The length of a rectangle is 3 less than twice the width. If the area of the
rectangle is 225 in , find the length of each side to the nearest hundredth.
( )2 3 25A l w w w= ⋅ = − =
( )2
2
2 3 25
2 3 25
2 3 25 0
w w
w w
w w
− =
− =
− − =
This won’t factor so we solve by the quadratic formula.
( ) ( ) ( ) ( )( )
23 3 4 2 25
2 2
3 9 200
4
3 209
4
w− − ± − − −
=
± +=
±=
Since w cannot be negative, we get:
3 209
4.36 in4
w+
= =
And the length is ( )2 3 2 4.36 3 5.72 inl w= − = − = .
Check:
( ) ( )
?
?
25
5.72 4.36 25
24.94 25
l w⋅ =
⋅ =
≈ �
w
2w-3
Lesson 10: Quadratic Equations Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 2 of 5
Note: Because of rounding errors, the check does not come out to be exactly the correct answer. If we used the exact answer
(and not the decimal approximation), the length would be:
2 3
3 2092 3
4
3 209 6
2 2
3 209
2
l w= −
+= −
+= −
− +=
Then the product of the length and the width is:
3 209 3 209
2 4
9 209
8
200
8
25
l w − + +
⋅ = − +
=
=
=
2. Find three consecutive even integers with the property that the sum of
the product of the first two integers, the product of the first and last integers, and the product of the second and third integers is 296.
Let x be the first integer. Then 2x + and 4x + are the next consecutive even integers.
( ) ( ) ( ) ( ) ( )2 2 2
2
2
2 4 2 4 296
2 4 6 8 296
3 12 288 0
4 96 0
x x x x x x
x x x x x x
x x
x x
+ + + + + + =
+ + + + + + =
+ − =
+ − =
We could solve by the quadratic formula, but it is easy to
complete the square for this one.
Lesson 10: Quadratic Equations Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 3 of 5
( )
2
2
2
4 96 0
4 4 96 4
2 100
2 10
2 10
8 or 12
x x
x x
x
x
x
x x
+ − =
+ + = +
+ =
+ = ±
= − ±
= = −
The numbers are 8, 10, and 12, or -12, -10, and -8.
Check:
8, 10, 12
?
?
?
8 10 8 12 10 12 296
80 96 120 296
296 296
⋅ + ⋅ + ⋅ =
+ + =
= �
-12, -10, -8
?
?
?
12 10 12 8 10 8 296
120 96 80 296
296 296
− ⋅ − + − ⋅ − + − ⋅ − =
+ + =
= �
3. The legs of a right triangle are the same length. If the hypotenuse is 32 mm, find the length of the legs.
2 2 2
2
2
2
32
2 1024
512
512
16 2
16 2
x x
x
x
x
x
x
+ =
=
=
= ±
= ± ⋅
= ±
Lesson 10: Quadratic Equations Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 4 of 5
The length of a side cannot be negative, so 16 2 mmx = .
4. Two cars start at the same point, one traveling north and one traveling
west. When the car traveling west had gone 15 miles, the distance between the two cars was 5 less than twice the distance the other car
had traveled. How far apart are the cars? Round your answer to the nearest hundredth.
( )22 2
2 2
2
15 2 5
225 4 20 25
0 3 20 200
d d
d d d
d d
+ = −
+ = − +
= − −
( ) ( ) ( ) ( )( )
220 20 4 3 200
2 3
20 400 2400
6
20 2800
6
20 400 7
6
20 20 7
6
10 10 7
3
d− − ± − − −
=
± +=
±=
± ⋅=
±=
±=
Since distance can’t be negative, we get 10 10 7
12.153
d+
= ≈ .
The distance between the two cars is
( )2 5 2 12.15 5
19.3 miles
d − = −
=
Check:
d
2d-5
15
Lesson 10: Quadratic Equations Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 5 of 5
( ) ( )?
2 22
?
12.15 15 19.3
147.62 225 372.49
372.62 372.49
+ =
+ =
≈ �
Note: Because of rounding errors, the check does not come out
to be exactly the correct answer. If we used the exact answer (and not the decimal approximation), the length would be:
10 10 72 5 2 5
3
20 20 7 15
3 3
5 20 7
3
d +
− = −
+= −
+=
Using the Pythagorean theorem gives:
2 2?
2
?
?
?
10 10 7 5 20 715
3 3
100 200 7 700 25 200 7 2800225
9 9
100 200 7 700 2025 2825 200 7
9 9 9
2825 200 7 2825 200 7
9 9
+ ++ =
+ + + ++ =
+ + ++ =
+ += �
