Find the values of x that satisfy `3|x^2+4x+2|=5x+16`

Options

A. `{1,-10//3}`
B. `(-2,1)`
C. `{-2.-11//3}`
D. `{-2.1}`

Answer

To solve the equation \(3|x^2 + 4x + 2| = 5x + 16\), we will break it down into steps.

### Step 1: Understand the Absolute Value
The absolute value function \( |A| \) can be defined as:
- \( A \) if \( A \geq 0 \)
- \( -A \) if \( A < 0 \)

In our case, we need to analyze the expression \( x^2 + 4x + 2 \).

### Step 2: Find the Roots of the Quadratic
To determine when \( x^2 + 4x + 2 \) is positive or negative, we can find its roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 4, c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 2 = 16 - 8 = 8 \]
Now, using the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2} \]
Thus, the roots are \( x_1 = -2 + \sqrt{2} \) and \( x_2 = -2 - \sqrt{2} \).

### Step 3: Determine Intervals
The quadratic \( x^2 + 4x + 2 \) opens upwards (since the coefficient of \( x^2 \) is positive). Therefore:
- It is negative between the roots \( (-2 - \sqrt{2}, -2 + \sqrt{2}) \).
- It is positive outside this interval.

### Step 4: Set Up Cases
We will solve the equation in two cases based on the sign of \( x^2 + 4x + 2 \).

**Case 1:** \( x^2 + 4x + 2 \geq 0 \)

In this case, the equation becomes:
\[ 3(x^2 + 4x + 2) = 5x + 16 \]
Expanding and rearranging gives:
\[ 3x^2 + 12x + 6 = 5x + 16 \implies 3x^2 + 7x - 10 = 0 \]
Now, we can solve this quadratic using the quadratic formula:
\[ x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 3 \cdot (-10)}}{2 \cdot 3} = \frac{-7 \pm \sqrt{49 + 120}}{6} = \frac{-7 \pm \sqrt{169}}{6} = \frac{-7 \pm 13}{6} \]
Calculating the two possible values:
\[ x_1 = \frac{6}{6} = 1, \quad x_2 = \frac{-20}{6} = -\frac{10}{3} \]

**Case 2:** \( x^2 + 4x + 2 < 0 \)

In this case, the equation becomes:
\[ 3(- (x^2 + 4x + 2)) = 5x + 16 \]
This simplifies to:
\[ -3x^2 - 12x - 6 = 5x + 16 \implies -3x^2 - 17x - 22 = 0 \implies 3x^2 + 17x + 22 = 0 \]
Using the quadratic formula again:
\[ x = \frac{-17 \pm \sqrt{17^2 - 4 \cdot 3 \cdot 22}}{2 \cdot 3} = \frac{-17 \pm \sqrt{289 - 264}}{6} = \frac{-17 \pm \sqrt{25}}{6} = \frac{-17 \pm 5}{6} \]
Calculating the two possible values:
\[ x_1 = \frac{-12}{6} = -2, \quad x_2 = \frac{-22}{6} = -\frac{11}{3} \]

### Step 5: Verify Solutions
We need to verify which of these solutions fall within the respective cases.

- For \( x = 1 \): \( x^2 + 4x + 2 = 1 + 4 + 2 = 7 \geq 0 \) (valid)
- For \( x = -\frac{10}{3} \): \( x^2 + 4x + 2 = \left(-\frac{10}{3}\right)^2 + 4\left(-\frac{10}{3}\right) + 2 = \frac{100}{9} - \frac{40}{3} + 2 = \frac{100 - 120 + 18}{9} = -\frac{2}{9} < 0 \) (invalid)
- For \( x = -2 \): \( x^2 + 4x + 2 = 0 \) (valid)
- For \( x = -\frac{11}{3} \): \( x^2 + 4x + 2 = \left(-\frac{11}{3}\right)^2 + 4\left(-\frac{11}{3}\right) + 2 = \frac{121}{9} - \frac{44}{3} + 2 = \frac{121 - 132 + 18}{9} = \frac{7}{9} > 0 \) (invalid)

### Final Solutions
The valid solutions are:
\[ x = 1 \quad \text{and} \quad x = -2 \]