Answer
The values of \( p \) for which the roots of the equation are non-real are between \( -4 - 4\sqrt{3} \) and \( -4 + 4\sqrt{3} \).
Solution
To solve the problem, we will follow these steps:
1. **Identify the conditions for non-real roots**: The roots of a quadratic equation \( ax^2 + bx + c = 0 \) are non-real if the discriminant \( D = b^2 - 4ac < 0 \).
2. **Analyze the function \( f(x) = x^2 - px + 8 + 2p \)**: We need to find the conditions under which this function has two equal roots, which occurs when the discriminant is zero.
3. **Determine the turning point of \( h(x) = f(x) - 3 \)**: The turning point of a quadratic function \( ax^2 + bx + c \) can be found using the vertex formula \( x = -\frac{b}{2a} \).
4. **Solve the equation \( (x + 5)^2 = 1 - p^2 \)** for \( p \).
Let's start with step 1 and 2.
### Step 1: Conditions for Non-Real Roots
For the function \( f(x) = x^2 - px + (8 + 2p) \), the discriminant \( D \) is given by:
\[
D = (-p)^2 - 4 \cdot 1 \cdot (8 + 2p) = p^2 - 4(8 + 2p)
\]
Setting the discriminant less than zero for non-real roots:
\[
p^2 - 4(8 + 2p) < 0
\]
### Step 2: Equal Roots Condition
For equal roots, we set the discriminant to zero:
\[
p^2 - 4(8 + 2p) = 0
\]
Now, let's solve this equation for \( p \):
\[
p^2 - 32 - 8p = 0
\]
Rearranging gives:
\[
p^2 + 8p - 32 = 0
\]
Now, we will solve this quadratic equation using the quadratic formula:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1}
\]
Calculating the discriminant:
\[
D = 64 + 128 = 192
\]
Now substituting back into the formula:
\[
p = \frac{-8 \pm \sqrt{192}}{2}
\]
Calculating \( \sqrt{192} = 8\sqrt{3} \):
\[
p = \frac{-8 \pm 8\sqrt{3}}{2} = -4 \pm 4\sqrt{3}
\]
### Step 3: Values of \( p \) for Non-Real Roots
Now we need to find the values of \( p \) for which the discriminant is less than zero:
\[
p^2 - 32 - 8p < 0
\]
This is a quadratic inequality. We can find the roots of the equation \( p^2 + 8p - 32 = 0 \) which we already calculated as \( p = -4 \pm 4\sqrt{3} \).
The roots are:
\[
p_1 = -4 - 4\sqrt{3}, \quad p_2 = -4 + 4\sqrt{3}
\]
Now we can analyze the intervals:
- The quadratic opens upwards (since the coefficient of \( p^2 \) is positive).
- The inequality \( p^2 + 8p - 32 < 0 \) holds between the roots.
Thus, the values of \( p \) for which the roots of the equation are non-real are:
\[
-4 - 4\sqrt{3} < p < -4 + 4\sqrt{3}
\]
### Step 4: Turning Point of \( h(x) = f(x) - 3 \)
To find the turning point of \( h(x) \), we first need to find the vertex of \( f(x) \):
\[
x = -\frac{-p}{2 \cdot 1} = \frac{p}{2}
\]
Now substituting \( x = \frac{p}{2} \) into \( f(x) \):
\[
f\left(\frac{p}{2}\right) = \left(\frac{p}{2}\right)^2 - p\left(\frac{p}{2}\right) + 8 + 2p
\]
Calculating:
\[
= \frac{p^2}{4} - \frac{p^2}{2} + 8 + 2p = \frac{p^2}{4} - \frac{2p^2}{4} + 8 + 2p = -\frac{p^2}{4} + 8 + 2p
\]
Now, the turning point of \( h(x) \) is:
\[
h\left(\frac{p}{2}\right) = f\left(\frac{p}{2}\right) - 3 = -\frac{p^2}{4} + 8 + 2p - 3 = -\frac{p^2}{4} + 2p + 5
\]
### Summary
1. The values of \( p \) for which the roots of the equation are non-real are:
\[
-4 - 4\sqrt{3} < p < -4 + 4\sqrt{3}
\]
2. The coordinates of the turning point of \( h(x) = f(x) - 3 \) are:
\[
\left(\frac{p}{2}, -\frac{p^2}{4} + 2p + 5\right)
\]
Reviewed and approved by the UpStudy tutoring team
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