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Let’s check out your problem:
−
81
x
2
=
−
11
-81 x^{2}=-11
−
81
x
2
=
−
11
\newline
What are the solutions to the given equation?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
=
11
81
x=\frac{\sqrt{11}}{81}
x
=
81
11
\newline
(B)
x
=
−
11
81
x=-\frac{\sqrt{11}}{81}
x
=
−
81
11
and
x
=
11
81
x=\frac{\sqrt{11}}{81}
x
=
81
11
\newline
(C)
x
=
−
11
9
x=-\frac{\sqrt{11}}{9}
x
=
−
9
11
and
x
=
11
9
x=\frac{\sqrt{11}}{9}
x
=
9
11
\newline
(D)
x
=
11
9
x=\frac{\sqrt{11}}{9}
x
=
9
11
View step-by-step help
Home
Math Problems
Algebra 2
Solve a quadratic equation using the zero product property
Full solution
Q.
−
81
x
2
=
−
11
-81 x^{2}=-11
−
81
x
2
=
−
11
\newline
What are the solutions to the given equation?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
=
11
81
x=\frac{\sqrt{11}}{81}
x
=
81
11
\newline
(B)
x
=
−
11
81
x=-\frac{\sqrt{11}}{81}
x
=
−
81
11
and
x
=
11
81
x=\frac{\sqrt{11}}{81}
x
=
81
11
\newline
(C)
x
=
−
11
9
x=-\frac{\sqrt{11}}{9}
x
=
−
9
11
and
x
=
11
9
x=\frac{\sqrt{11}}{9}
x
=
9
11
\newline
(D)
x
=
11
9
x=\frac{\sqrt{11}}{9}
x
=
9
11
Question Prompt:
question_prompt: "What are the solutions to the given equation
−
81
x
2
=
−
11
-81x^{2}=-11
−
81
x
2
=
−
11
?"
Step
1
1
1
:
Step
1
1
1
: Divide both sides of the equation by
−
81
-81
−
81
to isolate
x
2
x^2
x
2
. So,
x
2
=
−
11
/
−
81
x^2 = -11/-81
x
2
=
−
11/
−
81
.
Step
2
2
2
:
Step
2
2
2
: Simplify the right side of the equation.
x
2
=
11
81
x^2 = \frac{11}{81}
x
2
=
81
11
.
Step
3
3
3
:
Step
3
3
3
: Take the
square root
of both sides to solve for
x
x
x
. Remember, there are two solutions:
x
=
11
81
x = \sqrt{\frac{11}{81}}
x
=
81
11
and
x
=
−
11
81
x = -\sqrt{\frac{11}{81}}
x
=
−
81
11
.
Step
4
4
4
:
Step
4
4
4
: Simplify the square root.
11
81
\sqrt{\frac{11}{81}}
81
11
is the same as
11
/
81
\sqrt{11}/\sqrt{81}
11
/
81
.
Step
5
5
5
:
Step
5
5
5
: Since
81
\sqrt{81}
81
is
9
9
9
, we get
x
=
11
9
x = \frac{\sqrt{11}}{9}
x
=
9
11
and
x
=
−
11
9
x = -\frac{\sqrt{11}}{9}
x
=
−
9
11
.
More problems from Solve a quadratic equation using the zero product property
Question
The function
h
h
h
is defined over the real numbers. This table gives a few values of
h
h
h
.
\newline
\begin{tabular}{lllll}
\newline
x
x
x
&
−
6
-6
−
6
.
1
1
1
&
−
6
-6
−
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.
01
01
01
&
−
6
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−
6
.
001
001
001
&
−
5
-5
−
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.
9
9
9
\\
\newline
\hline
h
(
x
)
h(x)
h
(
x
)
&
−
0
-0
−
0
.
25
25
25
&
−
0
-0
−
0
.
74
74
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&
−
0
-0
−
0
.
98
98
98
&
−
1
-1
−
1
.
0
0
0
\newline
\end{tabular}
\newline
What is a reasonable estimate for
lim
x
→
−
6
h
(
x
)
\lim _{x \rightarrow-6} h(x)
lim
x
→
−
6
h
(
x
)
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
6
-6
−
6
\newline
(B)
−
2
-2
−
2
\newline
(C)
−
1
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−
1
\newline
(D) The limit doesn't exist
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)
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6
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\newline
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Question
What is the amplitude of
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−
3
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h(x)=5 \sin (4 x-2)-3 \text { ? }
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4
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−
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−
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\newline
Give an exact value.
\newline
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What is the period of the function
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Give an exact value.
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What is the period of the function
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What is the period of the function
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