What are the real solutions of `2absx^2 + 4absx-6=0` ?

Options

A. {1}
B. {-1,1}
C. {1,-3}
D. {1,3}

Answer

To solve the equation \( 2|x|^2 + 4|x| - 6 = 0 \), we will follow these steps:

### Step 1: Substitute \( m \) for \( |x| \)
Let \( m = |x| \). Then, we can rewrite the equation as:
\[ 2m^2 + 4m - 6 = 0 \]

### Step 2: Simplify the equation
To simplify the equation, we can divide the entire equation by 2:
\[ m^2 + 2m - 3 = 0 \]

### Step 3: Factor the quadratic equation
Next, we need to factor the quadratic equation. We look for two numbers that multiply to \(-3\) (the constant term) and add up to \(2\) (the coefficient of \(m\)). The numbers \(3\) and \(-1\) work:
\[ (m + 3)(m - 1) = 0 \]

### Step 4: Set each factor to zero
Now, we set each factor equal to zero:
1. \( m + 3 = 0 \)
2. \( m - 1 = 0 \)

### Step 5: Solve for \( m \)
From the first equation:
\[ m + 3 = 0 \implies m = -3 \]
From the second equation:
\[ m - 1 = 0 \implies m = 1 \]

### Step 6: Consider the modulus definition
Since \( m = |x| \), we know that \( |x| \) cannot be negative. Therefore, \( m = -3 \) is not a valid solution. We only consider \( m = 1 \).

### Step 7: Solve for \( x \)
Now we convert back to \( x \):
\[ |x| = 1 \]
This gives us two possible solutions:
\[ x = 1 \quad \text{or} \quad x = -1 \]

### Final Answer
The real solutions of the equation \( 2|x|^2 + 4|x| - 6 = 0 \) are:
\[ x = 1 \quad \text{and} \quad x = -1 \]

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