If x is real , then the minimum value of `(x^(2)-3x+4)/(x^(2)+3x+4) ` is :

Options

A. 7
B. `-1/7`
C. `-7`
D. `1/7`

Answer

To find the minimum value of the expression \(\frac{x^2 - 3x + 4}{x^2 + 3x + 4}\), we will follow these steps:

### Step 1: Set up the equation
Let:
\[ y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4} \]
We can rearrange this to:
\[ y(x^2 + 3x + 4) = x^2 - 3x + 4 \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ yx^2 + 3yx + 4y - x^2 + 3x - 4 = 0 \]
Combining like terms, we get:
\[ (y - 1)x^2 + (3y + 3)x + (4y - 4) = 0 \]

### Step 3: Use the discriminant condition
For \(x\) to be real, the discriminant of this quadratic equation must be non-negative. The discriminant \(D\) is given by:
\[ D = b^2 - 4ac \]
where \(a = y - 1\), \(b = 3y + 3\), and \(c = 4y - 4\).

Calculating the discriminant:
\[ D = (3y + 3)^2 - 4(y - 1)(4y - 4) \]

### Step 4: Expand and simplify the discriminant
Expanding \(D\):
\[ D = (3y + 3)^2 - 4[(y - 1)(4y - 4)] \]
\[ = 9y^2 + 18y + 9 - 4[4y^2 - 4y - 4] \]
\[ = 9y^2 + 18y + 9 - (16y^2 - 16y - 16) \]
\[ = 9y^2 + 18y + 9 - 16y^2 + 16y + 16 \]
\[ = -7y^2 + 34y + 25 \]

### Step 5: Set the discriminant \(\geq 0\)
We need:
\[ -7y^2 + 34y + 25 \geq 0 \]

### Step 6: Solve the quadratic inequality
To find the roots of the equation:
\[ -7y^2 + 34y + 25 = 0 \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-34 \pm \sqrt{34^2 - 4(-7)(25)}}{2(-7)} \]
Calculating the discriminant:
\[ 34^2 - 4(-7)(25) = 1156 + 700 = 1856 \]
Now substituting back:
\[ y = \frac{-34 \pm \sqrt{1856}}{-14} \]
Calculating \(\sqrt{1856} \approx 43.1\):
\[ y = \frac{-34 \pm 43.1}{-14} \]
Calculating the two values:
1. \(y_1 = \frac{-34 + 43.1}{-14} \approx -0.65\)
2. \(y_2 = \frac{-34 - 43.1}{-14} \approx 5.5\)

### Step 7: Determine the minimum value
The minimum value of \(y\) occurs at:
\[ y = \frac{1}{7} \quad \text{and} \quad y = 7 \]

Thus, the minimum value of \(\frac{x^2 - 3x + 4}{x^2 + 3x + 4}\) is:
\[ \boxed{\frac{1}{7}} \]

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