Determine the value(s) of $$ m $$ 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 $$ m $$ 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, $$ h(x)=f(x)-3 $$.

UpStudy AI · Accepted Answer

To solve the problem, we'll break it down into two parts:

1. **Determining the Value(s) of $$ m $$ for Non-Real Roots**
2. **Finding the Coordinates of the Turning Point of $$ h(x) = f(x) - 3 $$**

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### 1. Determining the Value(s) of $$ m $$ for Non-Real Roots

**Given Equation:**
$$ f(x) = x^2 - p x + 8 + 2p $$

To determine when the roots of this quadratic equation are non-real, we need to examine the discriminant ($$ D $$) of the quadratic equation. The discriminant is given by:
$$ D = b^2 - 4ac $$

For the quadratic equation $$ ax^2 + bx + c = 0 $$:
- $$ a = 1 $$
- $$ b = -p $$
- $$ c = 8 + 2p $$

**Calculating the Discriminant:**
$$ D = (-p)^2 - 4 \times 1 \times (8 + 2p) $$
$$ D = p^2 - 32 - 8p $$
$$ D = p^2 - 8p - 32 $$

**Condition for Non-Real Roots:**
$$ D < 0 $$
$$ p^2 - 8p - 32 < 0 $$

**Solving the Inequality:**
$$ p^2 - 8p - 32 < 0 $$

First, find the roots of the equation $$ p^2 - 8p - 32 = 0 $$ using the quadratic formula:
$$ p = \frac{8 \pm \sqrt{64 + 128}}{2} $$
$$ p = \frac{8 \pm \sqrt{192}}{2} $$
$$ p = \frac{8 \pm 8\sqrt{3}}{2} $$
$$ p = 4 \pm 4\sqrt{3} $$

Thus, the roots are:
$$ p = 4 + 4\sqrt{3} \quad \text{and} \quad p = 4 - 4\sqrt{3} $$

Since $$ p < 0 $$, we consider the relevant root:
$$ p = 4 - 4\sqrt{3} \approx -2.928 $$

**Conclusion:**
For $$ p < 0 $$ and the roots to be non-real, $$ p $$ must lie in the interval:
$$ 4 - 4\sqrt{3} < p < 0 $$

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### 2. Finding the Coordinates of the Turning Point of $$ h(x) = f(x) - 3 $$

**Given Function:**
$$ f(x) = x^2 - p x + 8 + 2p $$
$$ h(x) = f(x) - 3 = x^2 - p x + 5 + 2p $$

**Vertex (Turning Point) of a Quadratic Function:**
The vertex $$ (h_x, h_y) $$ of a quadratic function $$ ax^2 + bx + c $$ is given by:
$$ h_x = -\frac{b}{2a} $$
$$ h_y = f(h_x) $$

**Calculating $$ h_x $$:**
$$ h_x = \frac{p}{2} $$

**Calculating $$ h_y $$:**
$$ h_y = \left(\frac{p}{2}\right)^2 - p\left(\frac{p}{2}\right) + 5 + 2p $$
$$ h_y = \frac{p^2}{4} - \frac{p^2}{2} + 5 + 2p $$
$$ h_y = -\frac{p^2}{4} + 5 + 2p $$

**Substituting $$ p = 4 - 4\sqrt{3} $$:**
$$ h_x = \frac{4 - 4\sqrt{3}}{2} = 2 - 2\sqrt{3} $$

$$ h_y = -\frac{(4 - 4\sqrt{3})^2}{4} + 5 + 2(4 - 4\sqrt{3}) $$
$$ h_y = -\frac{16 - 32\sqrt{3} + 48}{4} + 5 + 8 - 8\sqrt{3} $$
$$ h_y = -\frac{64 - 32\sqrt{3}}{4} + 13 - 8\sqrt{3} $$
$$ h_y = -16 + 8\sqrt{3} + 13 - 8\sqrt{3} $$
$$ h_y = -3 $$

**Conclusion:**
The coordinates of the turning point are:
$$ \boxed{\,\left(2 - 2\sqrt{3},\; -3\,\right)} $$
For the equation $$ f(x) = x^2 - p x + 8 + 2p $$ with $$ p < 0 $$, the roots are non-real when $$ p $$ is between $$ 4 - 4\sqrt{3} $$ and 0. The turning point of $$ h(x) = f(x) - 3 $$ is at $$ \left(2 - 2\sqrt{3},\; -3\right) $$.

Determine the value(s) of \( m \) 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, \( h(x)=f(x)-3 \).

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