Question

If the set of all solutions of \(|x^2 + x - 9| = |x| + |x^2 - 9|\) is \([\alpha, \beta] \cup [\gamma, \infty)\), then \((\alpha^2 + \beta^2 + \gamma^2)\) is equal to:

• A. 9
• B. 18
• C. 36
• D. 72

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given equation is of the form \(|A + B| = |A| + |B|\).
A well-known property of absolute values states that \(|A + B| = |A| + |B|\) holds true if and only if \(A \cdot B \ge 0\).
Step 2: Key Formula or Approach:
Let \(A = x\) and \(B = x^2 - 9\).
Notice that \(A + B = x + x^2 - 9 = x^2 + x - 9\).
Thus, the given equation \(|x^2+x-9| = |x| + |x^2-9|\) translates exactly to \(|A+B| = |A| + |B|\).
Therefore, the condition for this equation to be satisfied is:
\[ x(x^2 - 9) \ge 0 \] Step 3: Detailed Explanation:
Factorize the inequality:
\[ x(x - 3)(x + 3) \ge 0 \] To find the intervals where this product is non-negative, we find the critical points by setting each factor to zero: \(x = -3, 0, 3\).
Now, we use the wavy curve method (sign chart) to determine the sign of the expression in each interval:
- For \(x \in (-\infty, -3)\), the expression is negative.
- For \(x \in [-3, 0]\), the expression is positive.
- For \(x \in (0, 3)\), the expression is negative.
- For \(x \in [3, \infty)\), the expression is positive.
The solution set is the union of intervals where the expression is non-negative:
\[ x \in [-3, 0] \cup [3, \infty) \] Step 4: Final Answer:
We are given that the solution set is of the form \([\alpha, \beta] \cup [\gamma, \infty)\). Comparing the two sets, we get:
\[ \alpha = -3, \quad \beta = 0, \quad \gamma = 3 \] We need to calculate the value of \(\alpha^2 + \beta^2 + \gamma^2\):
\[ \alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + (0)^2 + (3)^2 = 9 + 0 + 9 = 18 \]