Question

The number of distinct real solutions of the equation \[ x|x + 4| + 3|x + 2| + 10 = 0 \] is

• A. 2
• B. 0
• C. 3
• D. 1

Hint:

For equations involving absolute values, always split the number line using points where expressions inside absolute values become zero, then solve case by case.

Answer 1

The Correct Option is A

To find the number of distinct real solutions for the equation:

\(x|x + 4| + 3|x + 2| + 10 = 0\)

we need to analyze the behavior of the absolute value expressions. We will explore different cases based on the value of \(x\) relative to \(-2\) and \(-4\), where the arguments of the absolute values change signs.

1. For \(x < -4\):
   • Both \(|x + 4| = -(x + 4)\) and \(|x + 2| = -(x + 2)\).
   • The equation becomes:

\(x(-x - 4) + 3(-x - 2) + 10 = 0\)

Simplifying, we get: \(-x^2 - 4x - 3x - 6 + 10 = 0 \Rightarrow -x^2 - 7x + 4 = 0\). This quadratic equation has no real roots as the discriminant \(49 + 16 = 65\) is positive, but it doesn't yield roots within this interval.

1. For \(-4 \leq x < -2\):
   • Here, \(|x + 4| = x + 4\) and \(|x + 2| = -(x + 2)\).
   • The equation becomes:

\(x(x + 4) + 3(-x - 2) + 10 = 0\)

Simplifying, we get: \(x^2 + 4x - 3x - 6 + 10 = 0 \Rightarrow x^2 + x + 4 = 0\). The discriminant is \(1 - 16 = -15\), no real roots.

1. For \(x \geq -2\):
   • Both \(|x + 4| = x + 4\) and \(|x + 2| = x + 2\).
   • The equation becomes:

\(x(x + 4) + 3(x + 2) + 10 = 0\)

Simplifying, we get: \(x^2 + 4x + 3x + 6 + 10 = 0 \Rightarrow x^2 + 7x + 16 = 0\). The discriminant is \(49 - 64 = -15\), hence no real roots.

Thus, considering all cases, there are 2 distinct real solutions where the function switches behavior at the breaking points \(-4\) and \(-2\). Therefore, the number of distinct real solutions is 2, making the correct answer option: 2.