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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 graph of yequals​f(x) does not cross the​ x-axis.

x squared plus 8 x plus 52 equals 0 nbsp
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Part 1
The​ solution(s) is/are xequals
  
enter your response here.
​(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.)

1 answer

To find the exact solutions to the equation \( f(x) = x^2 + 8x + 52 = 0 \) in the complex numbers, we can use the quadratic formula:

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

In this case, \( a = 1 \), \( b = 8 \), and \( c = 52 \).

First, we calculate the discriminant:

\[ b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot 52 = 64 - 208 = -144 \]

Since the discriminant is negative, we will have complex solutions.

Now we can find the solutions using the quadratic formula:

\[ x = \frac{-8 \pm \sqrt{-144}}{2 \cdot 1} \]

We can simplify \( \sqrt{-144} \):

\[ \sqrt{-144} = \sqrt{144} \cdot i = 12i \]

Now substituting back into the quadratic formula:

\[ x = \frac{-8 \pm 12i}{2} \]

Breaking this down, we can simplify:

\[ x = \frac{-8}{2} \pm \frac{12i}{2} = -4 \pm 6i \]

Thus, the exact solutions are:

\[ x = -4 + 6i, \quad x = -4 - 6i \]

So, the solutions are:

\[ \boxed{-4 + 6i, -4 - 6i} \]

Part 2: Confirm that the solutions are not real

To show that the graph of \( y = f(x) \) does not cross the x-axis, we can analyze the quadratic function \( f(x) = x^2 + 8x + 52 \).

  1. Finding the vertex: The vertex form can help us locate the minimum point of the quadratic function, given that the coefficient of \( x^2 \) is positive.

    The x-coordinate of the vertex is given by:

    \[ x = -\frac{b}{2a} = -\frac{8}{2 \cdot 1} = -4 \]

    Now, we can find the corresponding y-coordinate by substituting \( x = -4 \) into the function:

    \[ f(-4) = (-4)^2 + 8(-4) + 52 = 16 - 32 + 52 = 36 \]

  2. Conclusion: The vertex of the parabola is at the point \( (-4, 36) \), which is above the x-axis. Since the parabola opens upwards and the vertex is above the x-axis, the graph does not cross the x-axis. Therefore, there are no real solutions, confirming that the solutions we found are indeed complex.

Thus, we conclusively show that the graph of \( y = f(x) \) does not cross the x-axis.