Determine the value(s) of in for which the roots of the equation are non-real. Given that $$ f(x)=x^{2}-p x+8+2 p $$ has two equal root and $$ p

Question

Determine the value(s) of in for which the roots of the equation are non-real. Given that $$ f(x)=x^{2}-p x+8+2 p $$ has two equal root and $$ p<0 $$, determine the coodinates of the turning point, $$ h(x)=f(x)-3 $$. Given: $$ (x+5)^{2}=1-p^{2} $$ Calculate the values of $$ p $$ for which the roots of the equation are non-real.

UpStudy AI · Accepted Answer

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:
   $$
   \left(\frac{p}{2}, -\frac{p^2}{4} + 2p + 5\right)
   $$
The values of $$ p $$ for which the roots of the equation are non-real are between $$ -4 - 4\sqrt{3} $$ and $$ -4 + 4\sqrt{3} $$.

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