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 A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- 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 A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- 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 ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
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