Determine whether the following systems always, sometimes, or never have solutions. (Assume that different letters refer to unequal constants.) Explain.
-
{
y
=
x
2
+
c
y
=
x
2
+
d
left brace . table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus c , row2 column 1 , y equals , x squared , plus d , end table
-
{
y
=
a
x
2
+
c
y
=
b
x
2
+
c
left brace . table with 2 rows and 1 column , row1 column 1 , y equals eh , x squared , plus c , row2 column 1 , y equals b , x squared , plus c , end table
-
{
y
=
(
x
+
a
)
2
y
=
(
x
+
b
)
2
left brace . table with 2 rows and 1 column , row1 column 1 , y equals . open , x plus eh , close squared , row2 column 1 , y equals . open , x plus b , close squared , end table
-
{
y
=
a
(
x
+
m
)
2
+
c
y
=
b
(
x
+
n
)
2
+
d
left brace . table with 2 rows and 1 column , row1 column 1 , y equals eh . open , x plus m , close squared . plus c , row2 column 1 , y equals b . open , x plus n , close squared . plus d , end table
- Find the side of the square with vertical and horizontal sides inscribed in the region representing the solution of the system
{
y
≤
−
x
2
+
1
y
≥
x
2
−
1
.
left brace . table with 2 rows and 1 column , row1 column 1 , y less than or equal to negative , x squared , plus 1 , row2 column 1 , y greater than or equal to , x squared , minus 1 , end table . .
Standardized Test Prep
SAT/ACT
-
How many solutions does the system have?
{
y
=
−
1
4
x
2
−
2
x
y
=
x
2
+
3
4
left brace . table with 2 rows and 1 column , row1 column 1 , y equals negative , 1 fourth , x squared , minus 2 x , row2 column 1 , y equals , x squared , plus , 3 fourths , end table
- 0
- 1
- 2
- 3
- Which expression is equivalent to
(
−
3
+
2
i
)
(
2
−
3
i
)
?
open negative 3 plus 2 i close open 2 minus 3 i close question mark
- 13 i
- 12
- 12 + 13 i
-
−
12
negative 12
- Which expression is equivalent to
(
2
−
7
i
)
÷
(
2
i
)
3
?
open 2 minus 7 i close divides open 2 i , close cubed , question mark
-
7
8
−
1
4
i
7 eighths , minus , 1 fourth , i
-
1
4
−
7
8
i
1 fourth , minus , 7 eighths , i
-
7
8
+
1
4
i
7 eighths , plus , 1 fourth , i
-
1
4
+
7
8
i
1 fourth , plus , 7 eighths , i
Short Response
- Solve the equation
−
3
x
2
+
5
x
+
4
=
0
.
negative 3 , x squared , plus 5 x plus 4 equals 0 . Show your work.
Mixed Review
See Lesson 4-8.
Find the sum or difference.
-
(
1
−
i
)
+
(
−
5
+
4
i
)
open 1 minus i close plus open negative 5 plus 4 i close
-
(
3
+
4
i
)
−
(
−
4
−
3
i
)
open 3 plus 4 i close minus open negative 4 minus 3 i close
-
(
1
+
i
)
+
(
2
+
2
i
)
open 1 plus i close plus open 2 plus 2 i close
See Lesson 4-7.
Solve each equation using the Quadratic Formula.
-
2
m
2
+
5
m
+
3
=
0
2 m squared , plus 5 m plus 3 equals 0
-
p
2
−
4
p
+
3
=
0
p squared , minus 4 p plus 3 equals 0
-
25
x
2
−
30
x
+
9
=
0
25 x squared , minus 30 x plus 9 equals 0
See Lesson 4-6.
Rewrite each equation in vertex form.
-
y
=
−
k
2
+
4
k
+
6
y equals negative , k squared , plus 4 k plus 6
-
y
=
x
2
+
6
x
+
1
y equals , x squared , plus 6 x plus 1
-
y
=
2
n
2
−
8
n
−
3
y equals , 2 n squared , minus 8 n minus 3
Get Ready! To prepare for Lesson 5-1, do Exercises 75–77.
See Lesson 1-3.
Simplify by combining like terms.
-
3
q
+
9
q
−
q
3 q plus 9 q minus q
-
−
2
a
b
2
+
2
a
2
b
+
3
a
b
2
negative 2 eh , b squared , plus , 2 eh squared , b plus . 3 eh b squared
-
−
4
y
2
+
2
y
+
3
y
2
negative 4 , y squared , plus 2 y plus , 3 y squared
