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Lesson 10: Quadratic Equations Quiz Solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 1 of 5
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1. Use the discriminant to determine the nature of the solutions to the given
quadratic equations.
a) 24 4 1q q− +
In this quadratic expression, 4a = , 4b = − , and 1c = .
( ) ( ) ( )22 4 4 4 4 1
16 16
0
b ac− = − −
= −
=
Since the discriminant is 0, there are 2 real solutions. These
solutions are actually the same number.
b) 2 3 5 0f f+ + =
In this quadratic expression, 1a = , 3b = , and 5c = .
( ) ( ) ( )22 4 3 4 1 5
9 20
11
b ac− = −
= −
= −
Since the discriminant is negative, there are 2 imaginary solutions.
c) 24 5 3r r− =
Writing the expression in standard form gives:
2
2
4 5 3
4 3 5 0
r r
r r
− =
− − =
In this quadratic expression, 4a = , 3b = − , and 5c = − .
( ) ( ) ( )22 4 3 4 4 5
9 80
89
b ac− = − − −
= +
=
Lesson 10: Quadratic Equations Quiz Solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 2 of 5
Since the discriminant is positive, there are 2 real solutions.
2. Solve by factoring: 23 10 8x x+ =
( ) ( )
2
2
3 10 8
3 10 8 0
3 2 4 0
x x
x x
x x
+ =
+ − =
− + =
3 2 0 or 4 0
3 2 or 4
2 or 4
3
x x
x x
x x
− = + =
= = −
= = −
3. Solve by completing the square: 22 6 6 0x x− + =
2
2
2
2
2
2
9 9
4
2 6 6 0
2 6 6 0
2 2
3 3 0
3 3
3 3
3 3
2 4
3 3
2 4
3 3
2 2
3 3
2
4
x x
x x
x x
x x
x x
x
x
ix
ix
− + =
− +=
− + =
− = −
− = −
− − =
−− = ±
− = ±
±=
+ +
Lesson 10: Quadratic Equations Quiz Solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 3 of 5
4. Solve by using the quadratic formula: 24 5 6s s+ =
2
2
4 5 6
4 5 6 0
s s
s s
+ =
+ − =
( ) ( )( )
25 5 4 4 6
2 4
5 25 96
8
5 121
8
5 11
8
s
s
s
s
− ± − −=
− ± +=
− ±=
− ±=
5 11 5 11 or
8 8
6 16 or
8 8
3 or 2
4
s s
s s
s s
− + − −= =
−= =
= = −
5. Solve by any method.
a) 24 37 0q − =
2
2
2
4 37 0
4 37
37
4
37
4
37
2
q
q
q
q
q
− =
=
=
= ±
= ±
Lesson 10: Quadratic Equations Quiz Solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 4 of 5
b) 2 4 12a a+ =
( ) ( )
2
2
4 12
4 12 0
6 2 0
a a
a a
a a
+ =
+ − =
+ − =
6 0 or 2 0
6 or 2
a a
a a
+ = − =
= − =
c) 28 5x x=
( )
2
2
8 5
8 5 0
8 5 0
x x
x x
x x
=
− =
− =
0 or 8 5 0
0 or 8 5
50 or
8
x x
x x
x x
= − =
= =
= =
d) 22 3 5 0y y− + =
( ) ( ) ( ) ( )( )
23 3 4 2 5
2 2
3 9 40
4
3 31
4
3 31
4
y
y
y
iy
− − ± − −=
± −=
± −=
±=
6. Perform the indicated operations and simplify.
a) 35i
Since 35 4 8 3r÷ = , 35 3i i i= = − .
Lesson 10: Quadratic Equations Quiz Solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 5 of 5
b) ( ) ( ) ( )6 7 8 3 11 2i i i− + + − −
( ) ( ) ( )6 7 8 3 11 2 6 7 8 3 11 2
3 2
i i i i i i
i
− + + − − = − + + − +
= −
c) ( ) ( )4 2 3 2i i+ −
( ) ( )( )
24 2 3 2 12 8 6 4
12 2 4 1
12 2 4
16 2
i i i i i
i
i
i
+ − = − + −
= − − −
= − +
= −
d) ( )( )3 5
1 2
i
i
+
+
( )( )
( )( )
( )( )
( )( )
2
2
3 5 3 5 1 2
1 2 1 2 1 2
3 6 5 10
1 4
3 10 1
1 4 1
3 10
1 4
13
5
i i i
i i i
i i i
i
i
i
i
+ + −= ⋅
+ + −
− + −=
−− − −
=− −
− +=
+−
=