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 yequals​f(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
  
enter your response here.

Answer 1

To solve the equation \( f(x) = x^2 + 10x + 34 = 0 \) in the complex numbers, we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1 \), \( b = 10 \), 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 substitute into the quadratic formula:

\[
x = \frac{-10 \pm \sqrt{-36}}{2 \cdot 1} = \frac{-10 \pm 6i}{2}
\]

This simplifies to:

\[
x = \frac{-10}{2} \pm \frac{6i}{2} = -5 \pm 3i
\]

So the solutions are:

\[
x = -5 + 3i \quad \text{and} \quad x = -5 - 3i
\]

### Part 2: Confirming that the solutions are not real

To show that the graph \( y = f(x) = x^2 + 10x + 34 \) does not cross the x-axis, we can analyze the vertex of the parabola represented by the quadratic function.

The vertex \( x \) coordinate of the parabola can be found using:

\[
x = -\frac{b}{2a} = -\frac{10}{2} = -5
\]

Now, we substitute \( x = -5 \) back into \( f(x) \) to find the corresponding \( y \) value:

\[
f(-5) = (-5)^2 + 10(-5) + 34 = 25 - 50 + 34 = 9
\]

Since \( f(-5) = 9 \), which is greater than zero, the vertex lies above the x-axis. Because the parabola opens upwards (as the coefficient of \( x^2 \) is positive), the function does not cross the x-axis.

### Conclusion

The exact solutions to \( f(x) = 0 \) in the complex numbers are:

\[
x = -5 + 3i \quad \text{and} \quad x = -5 - 3i
\]

And since the graph does not cross the x-axis, we confirm that the solutions are not real.