The least integer x for wich the inequality `((x-3)^(2))/(x^(2)+8x-22)lt0` is satisfied, is

Options

A. `-8`
B. `-9`
C. `-10`
D. `-11`

Explanation

To solve the inequality \(\frac{(x-3)^2}{x^2 + 8x - 22} < 0\), we need to analyze both the numerator and the denominator.

### Step 1: Analyze the Numerator
The numerator is \((x-3)^2\). Since this is a square term, it is always non-negative (i.e., \((x-3)^2 \geq 0\)) for all real \(x\). The numerator is equal to zero when \(x = 3\).

### Step 2: Analyze the Denominator
Next, we need to analyze the denominator \(x^2 + 8x - 22\). We will find the roots of the quadratic equation by using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = 8\), and \(c = -22\).

Calculating the discriminant:
\[ b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot (-22) = 64 + 88 = 152 \]

Now, finding the roots:
\[ x = \frac{-8 \pm \sqrt{152}}{2} = \frac{-8 \pm 2\sqrt{38}}{2} = -4 \pm \sqrt{38} \]
Thus, the roots are:
\[ x_1 = -4 + \sqrt{38}, \quad x_2 = -4 - \sqrt{38} \]

### Step 3: Determine the Sign of the Denominator
The quadratic \(x^2 + 8x - 22\) opens upwards (since the coefficient of \(x^2\) is positive). Therefore, it is negative between its roots and positive outside:
- The interval where the denominator is negative is \((-4 - \sqrt{38}, -4 + \sqrt{38})\).

### Step 4: Find the Intervals
We need to find the intervals where the entire expression \(\frac{(x-3)^2}{x^2 + 8x - 22}\) is less than zero. Since the numerator is non-negative, the fraction can only be negative when the denominator is negative.

### Step 5: Identify the Least Integer \(x\)
We need to find the least integer \(x\) in the interval \((-4 - \sqrt{38}, -4 + \sqrt{38})\).

Calculating \(-4 + \sqrt{38}\):
\[ \sqrt{38} \approx 6.16 \quad \Rightarrow \quad -4 + \sqrt{38} \approx -4 + 6.16 \approx 2.16 \]
Calculating \(-4 - \sqrt{38}\):
\[ -4 - \sqrt{38} \approx -4 - 6.16 \approx -10.16 \]
Thus, the interval is approximately \((-10.16, 2.16)\).

The least integer satisfying this inequality is:
\[ \lfloor -10.16 \rfloor = -10 \]

### Final Answer
The least integer \(x\) for which the inequality \(\frac{(x-3)^2}{x^2 + 8x - 22} < 0\) is satisfied is:
\[ \boxed{-10} \]