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Find the exact solutions to f(x)equals0 in the complex numbers and confirm that the solutions are not real by showing that the...Question
Find the exact solutions to f(x)equals0 in the complex numbers and confirm that the solutions are not real by showing that the graph of yequalsf(x) does not cross the x-axis.
x squared plus 10 x plus 34 equals 0 nbsp
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Part 1
The solution(s) is/are xequals
negative 5 plus 3 i.
(Simplify your answer. Type your answer in the form aplusbi. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)
x squared plus 10 x plus 34 equals 0 nbsp
Question content area bottom
Part 1
The solution(s) is/are xequals
negative 5 plus 3 i.
(Simplify your answer. Type your answer in the form aplusbi. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the equation \(x^2 + 10x + 34 = 0\) in the complex numbers, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this equation, \(a = 1\), \(b = 10\), and \(c = 34\).
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 10^2 - 4 \cdot 1 \cdot 34 = 100 - 136 = -36
\]
Since the discriminant is negative, we will have complex solutions.
Now, we will find the solutions using the quadratic formula:
\[
x = \frac{-10 \pm \sqrt{-36}}{2 \cdot 1}
\]
Calculating the square root of \(-36\):
\[
\sqrt{-36} = 6i
\]
Substituting this back into the quadratic formula gives:
\[
x = \frac{-10 \pm 6i}{2}
\]
Now we can simplify this:
\[
x = \frac{-10}{2} \pm \frac{6i}{2}
\]
\[
x = -5 \pm 3i
\]
Thus, the exact solutions are:
\[
x = -5 + 3i, \quad x = -5 - 3i
\]
### Part 2: Confirming the solutions are not real by showing the graph does not cross the x-axis.
To confirm that the solutions are not real, we can analyze the quadratic function \(f(x) = x^2 + 10x + 34\).
The vertex of the parabola represented by this equation can be found using the vertex formula \(x = -\frac{b}{2a}\):
\[
x = -\frac{10}{2 \cdot 1} = -5
\]
Now, we substitute \(x = -5\) back into \(f(x)\) to find the value of the function at the vertex:
\[
f(-5) = (-5)^2 + 10(-5) + 34
\]
\[
= 25 - 50 + 34
\]
\[
= 9
\]
Since \(f(-5) = 9\) and is positive, the vertex is above the x-axis, indicating that the parabola does not cross the x-axis.
Overall, the solutions are confirmed as:
\[
x = -5 + 3i, \quad x = -5 - 3i
\]
Therefore, the final answer is:
\[
\text{x equals } -5 + 3i, -5 - 3i
\]
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this equation, \(a = 1\), \(b = 10\), and \(c = 34\).
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 10^2 - 4 \cdot 1 \cdot 34 = 100 - 136 = -36
\]
Since the discriminant is negative, we will have complex solutions.
Now, we will find the solutions using the quadratic formula:
\[
x = \frac{-10 \pm \sqrt{-36}}{2 \cdot 1}
\]
Calculating the square root of \(-36\):
\[
\sqrt{-36} = 6i
\]
Substituting this back into the quadratic formula gives:
\[
x = \frac{-10 \pm 6i}{2}
\]
Now we can simplify this:
\[
x = \frac{-10}{2} \pm \frac{6i}{2}
\]
\[
x = -5 \pm 3i
\]
Thus, the exact solutions are:
\[
x = -5 + 3i, \quad x = -5 - 3i
\]
### Part 2: Confirming the solutions are not real by showing the graph does not cross the x-axis.
To confirm that the solutions are not real, we can analyze the quadratic function \(f(x) = x^2 + 10x + 34\).
The vertex of the parabola represented by this equation can be found using the vertex formula \(x = -\frac{b}{2a}\):
\[
x = -\frac{10}{2 \cdot 1} = -5
\]
Now, we substitute \(x = -5\) back into \(f(x)\) to find the value of the function at the vertex:
\[
f(-5) = (-5)^2 + 10(-5) + 34
\]
\[
= 25 - 50 + 34
\]
\[
= 9
\]
Since \(f(-5) = 9\) and is positive, the vertex is above the x-axis, indicating that the parabola does not cross the x-axis.
Overall, the solutions are confirmed as:
\[
x = -5 + 3i, \quad x = -5 - 3i
\]
Therefore, the final answer is:
\[
\text{x equals } -5 + 3i, -5 - 3i
\]
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