The exhausiive Intervals of real values of x such that `sqrt(12-4x)gt1+sqrt(4x+4)` is

Options

A. `[1-(sqrt(31))/8,1+(sqrt(31))/8)`
B. `[-1,1+(sqrt(31))/8)`
C. `[-1,3]`
D. `[-1,1-(sqrt(31))/8)`

Answer

To solve the inequality \( \sqrt{12 - 4x} > 1 + \sqrt{4x + 4} \), we will follow these steps:

### Step 1: Rearranging the Inequality
First, we will rearrange the inequality to isolate the square roots:
\[ \sqrt{12 - 4x} - \sqrt{4x + 4} > 1 \]

### Step 2: Squaring Both Sides
Next, we will square both sides to eliminate the square roots. However, we must be cautious because squaring can introduce extraneous solutions:
\[ (\sqrt{12 - 4x} - \sqrt{4x + 4})^2 > 1^2 \]
Expanding the left side:
\[ (12 - 4x) + (4x + 4) - 2\sqrt{(12 - 4x)(4x + 4)} > 1 \]
This simplifies to:
\[ 16 - 2\sqrt{(12 - 4x)(4x + 4)} > 1 \]
Rearranging gives:
\[ 15 > 2\sqrt{(12 - 4x)(4x + 4)} \]
Dividing by 2:
\[ \frac{15}{2} > \sqrt{(12 - 4x)(4x + 4)} \]

### Step 3: Squaring Again
Now we square both sides again:
\[ \left(\frac{15}{2}\right)^2 > (12 - 4x)(4x + 4) \]
Calculating the left side:
\[ \frac{225}{4} > (12 - 4x)(4x + 4) \]

### Step 4: Expanding the Right Side
Expanding the right side:
\[ (12 - 4x)(4x + 4) = 48 + 48x - 16x - 16x^2 = -16x^2 + 32x + 48 \]
So we have:
\[ \frac{225}{4} > -16x^2 + 32x + 48 \]

### Step 5: Bringing All Terms to One Side
Multiplying through by 4 to eliminate the fraction:
\[ 225 > -64x^2 + 128x + 192 \]
Rearranging gives:
\[ 64x^2 - 128x + (192 - 225) > 0 \]
This simplifies to:
\[ 64x^2 - 128x - 33 > 0 \]

### Step 6: Finding Roots of the Quadratic
Now we will find the roots of the quadratic equation \( 64x^2 - 128x - 33 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 64, b = -128, c = -33 \):
\[ x = \frac{128 \pm \sqrt{(-128)^2 - 4 \cdot 64 \cdot (-33)}}{2 \cdot 64} \]
Calculating the discriminant:
\[ x = \frac{128 \pm \sqrt{16384 + 8448}}{128} = \frac{128 \pm \sqrt{24832}}{128} \]
Calculating the square root:
\[ \sqrt{24832} = 157.5 \quad (\text{approx.}) \]
So:
\[ x = \frac{128 \pm 157.5}{128} \]
Calculating the two roots:
\[ x_1 \approx 2.0, \quad x_2 \approx -0.23 \]

### Step 7: Testing Intervals
We will test the intervals determined by the roots \( x_1 \) and \( x_2 \):
1. \( (-\infty, -0.23) \)
2. \( (-0.23, 2.0) \)
3. \( (2.0, \infty) \)

### Step 8: Conclusion
After testing these intervals, we find that the inequality holds for:
\[ x \in (-1, 1 - \frac{\sqrt{31}}{8}) \cup (1 + \frac{\sqrt{31}}{8}, 3) \]