Liberty Union High
School District
Algebra 1
Study Guide
Benchmark 4 Assessment
LUHSD Algebra I
1
Benchmark 4 Study Guide
1´
Simplify: 125n
Simplify:
98k 2
= 125n
= 5i5i5in
= 52 i5in
= 52 ! 5n
= 5 5n
2
Simplify: 2 6 − 2 24
2´
Simplify:
a) 2 6 + 3 4
= 2 6 ! 2 2i2i2i3
= 2 6 ! 2 22 i2i3
= 2 6 ! 2i 22 i 2i3
b) 2 5 + 3 20
= 2 6 ! 2i2 6
=2 6 !4 6
= !2 6
3
3´
Multiply: 15n 2 i 10n 3
Multiply: 18a 2 i 45a 2
= 15n 2 i10n 3
= 2i3i5i5in5
= 2i3i52 in 4 in
= 52 i n 4 i 2i3in
= 5in 2 i 6n
= 5n 2 6n
Page 1 of 6
MDC@ACOE (NEBMC) 05/13/11
LUHSD Algebra I
4
Rationalize :
=
==
=
5
4
3
i
Benchmark 4 Study Guide
4
3
6
.
5
4´
Rationalize:
5´
Solve: 4 n + 8 = 36
6
Solve using the quadratic formula:
2 x2 − 7 = 5x
3
3
4 3
9
4 3
3
Solve:
m
=3
5
m
=3
5
! 5$ m ! 5$
#" 1 &% i 5 = #" 1 &% i3
m = 15
∴ m = 15 or m = −15
6
Solve using the quadratic formula:
2 x 2 + 9 x = −7
2x 2 + 9x + 7 = 0 (Rewrite in standard form)
a = 2, b = 9, c = 7
x=
x=
−b ± b 2 − 4ac
2a
−9 ± 92 − 4 ( 2 )( 7 )
2 (2)
−9 ± 81 − 56
4
−9 ± 25
x=
4
−9 ± 5
x=
4
7
x = −1, −
2
x=
Page 2 of 6
MDC@ACOE (NEBMC) 05/13/11
LUHSD Algebra I
7
Benchmark 4 Study Guide
Solve by completing the square:
x 2 + 16 x − 22 = 0
Solve: (3a − 4) = 25
then add 82 to each
side of the equation.
x 2 + 16 x + 82 = 22 + 82
( x + 8)
8´
2
x + 16 x = 22
±
Solve by completing the square:
x 2 + 2 x − 33 = 0
Remember 1 (16 ) = 8
2
( x + 8)
7´
2
= 86
2
= ± 86
x + 8 = ± 86
x = −8 ± 86
8
Solve: ( 6 x −1) = 9
2
( 6 x − 1)
2
2
=± 9
6 x − 1 = ±3
6x −1 = 3
6x −1 = 3
6x −1+1 = 3 + 1
6x = 4
6x 4
=
6 6
2•2
x=
2•3
2
x=
3
or
6 x − 1 = −3
6 x − 1 = −3
6 x − 1 + 1 = −3 + 1
6 x = −2
6 x −2
=
6
6
2 •1
x=−
2•3
1
x=−
3
Page 3 of 6
MDC@ACOE (NEBMC) 05/13/11
LUHSD Algebra I
9
Benchmark 4 Study Guide
Solve: −20 ≤ −6m − 2 ≤ 58
9´
Solve : −53 < 9 x + 1 < −26
−20 + 2 ≤ −6m − 2 + 2 ≤ 58 + 2
−18
≤ − 6m
≤ 60
−18
−6
3
≥
≥
−6m
−6
m
60
−6
≥ − 10
≥
10´ Determine if the relation is a function.
10 Determine if the relation is a function.
Since the domain (input)
of -2 is mapped to both
13 and 15 in the range,
the relation is not a
function.
11
Determine if the following graph represents a
function.
11´ Determine if the following graph represents a
function.
Apply the
vertical line test.
A vertical line
through any x
value will
intersect the
graph only once
if the graph is a
function.
Answer: The graph is a function
Page 4 of 6
MDC@ACOE (NEBMC) 05/13/11
LUHSD Algebra I
Benchmark 4 Study Guide
Answer Key
1. = 98k 2
5.
4 n + 8 = 36
1
1
• 4 n + 8 = • 36
4
4
n+8 = 9
= 7•7•2•k •k
= 7k 2
2. a)
n + 8 = ±9
= 2 6 +3 4
= 2 6 + 3• 2
= 2 6 +6
b)
= 2 5 + 3 20
n+8 = 9
n + 8 = −9
n +8−8 = 9−8
n + 8 − 8 = −9 − 8
n =1
= 2 5 + 3 4•5
6.
= 2 5 + 3• 4 • 5
2 x2 − 7 = 5x
2 x2 − 7 − 5x = 5x − 5x
= 2 5 + 3• 2 • 5
2 x2 − 5x − 7 = 0
= 2 5+6 5
a = 2, b = −5, c = −7
=8 5
x=
3. = 18a 2 • 45a 2
= 18 • 45 • a 2 • a 2
x=
= 9 • 2 • 9 • 5 • a2 • a2
= 9 • 9 • 2 • 5 • a2 • a2
x=
= 9a 2 • 2 • 5
4.
= 9a 2 10
x=
=
6
5
x=
=
6
5
•
5
5
=
6• 5
5• 5
=
n = −17
− ( −5 ) ±
5±
5±
( −5 ) − 4 ( 2 )( −7 )
2 (2)
25 − 4 ( 2 )( −7 )
2 (2)
25 − ( −56 )
2 (2)
2
5 ± 81
2 (2)
5+9 5−9
,
4
4
14 −4
x= ,
4 4
7
x = , −1
2
6 5
5
Page 5 of 6
MDC@ACOE (NEBMC) 05/13/11
LUHSD Algebra I
7.
Benchmark 4 Study Guide
x 2 + 2 x − 33 = 0
x 2 + 2 x − 33 + 33 = 0 + 33
x 2 + 2 x = 33
x 2 + 2 x + 1 = 33 + 1
( x + 1)
( x + 1)
2
= 34
2
= ± 34
x + 1 = ± 34
x + 1 − 1 = −1 ± 34
x = −1 ± 34
(3a − 4 )
8.
2
= 25
(3a − 4 ) = ± 25
(3a − 4 ) = ±5
2
3a − 4 = 5
or
3a − 4 = 5
3a − 4 + 4 = 5 + 4
3a − 4 = −5
3a − 4 + 4 = −5 + 4
3a = 9
3a 9
=
3 3
a=3
9.
3a = −1
−53 ≤ 9 x + 1 ≤ −26
−53 − 1 ≤ 9 x + 1 − 1 ≤ −26 − 1
−54 ≤
9x
3a − 4 = −5
3a −1
=
3
3
1
a=−
3
≤ −27
−54
9x
−27
≤
≤
9
9
9
−6 ≤ x
≤ −3
10. Each value for the domain ( x ) has only
one value for y therefore it is a Function
11. It fails the vertical line test therefore it is
NOT a function
Page 6 of 6
MDC@ACOE (NEBMC) 05/13/11