The problem asks for the value of $(\alpha^2 + \beta^2 + \gamma^2)$ derived from the solution set of the equation $|x^2 + x - 9| = |x| + |x^2 - 9|$, which is given as $[\alpha, \beta] \cup [\gamma, \infty)$.
Understanding the Absolute Value Equation
We utilize a key property of absolute values: $|a| = |b| + |c|$ holds if and only if $a = b + c$ and the product $b \cdot c \ge 0$.
In this equation, let $a = x^2 + x - 9$, $b = x$, and $c = x^2 - 9$. First, we verify if $a = b + c$: $b + c = x + (x^2 - 9) = x^2 + x - 9$ This matches $a$. Therefore, the original equation $|x^2 + x - 9| = |x| + |x^2 - 9|$ is equivalent to the condition that $b$ and $c$ have the same sign or one/both are zero, meaning $b \cdot c \ge 0$.
Solving the Inequality Condition
The condition simplifies to:
$x \cdot (x^2 - 9) \ge 0$Factorizing the quadratic term gives:
$x(x - 3)(x + 3) \ge 0$The critical points where the expression equals zero are $x = -3$, $x = 0$, and $x = 3$. We test the sign of $x(x - 3)(x + 3)$ in the intervals determined by these points:
- Interval $x < -3$: Negative (e.g., $(-4)(-7)(-1) < 0$)
- Interval $-3 \le x \le 0$: Non-negative (e.g., $(-1)(-4)(2) > 0$)
- Interval $0 < x < 3$: Negative (e.g., $(1)(-2)(4) < 0$)
- Interval $x \ge 3$: Non-negative (e.g., $(4)(1)(7) > 0$)
The inequality $x(x - 3)(x + 3) \ge 0$ is satisfied when $-3 \le x \le 0$ or $x \ge 3$.
Identifying Solution Set Parameters
The derived solution set is $[-3, 0] \cup [3, \infty)$.
Comparing this with the given format $[\alpha, \beta] \cup [\gamma, \infty)$, we find:
- $\alpha = -3$
- $\beta = 0$
- $\gamma = 3$
Final Calculation
We now calculate the required value $(\alpha^2 + \beta^2 + \gamma^2)$: $ \alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + (0)^2 + (3)^2 $ $ = 9 + 0 + 9 $ $ = 18 $
Similar Questions
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
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Important Questions from Algebra
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
