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** 12. Solve the following inequality \( \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq-2 \). Hence without the use of graphic calculator, solve the inequality \( \frac{2^{x}\left(2^{x}+3\right)\left(2^{x}-3\right)+16}{2^{2 x}+2^{x}-6} \geq-2 \) where \( x \neq 1 \).

Ask by Munoz Osborne. in Singapore
Jan 22,2025

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To solve the inequality \[ \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq -2, \] we first rearrange it to: \[ \frac{x(x+3)(x-3)+16 + 2(x+3)(x-2)}{(x+3)(x-2)} \geq 0. \] Simplifying the numerator: \[ x^3 + 2x^2 - 7x + 4 \geq 0. \] Factoring the numerator: \[ (x - 1)^2(x + 4) \geq 0. \] The critical points are \(x = -4\), \(x = -3\), and \(x = 1\). Testing intervals: - \(x < -4\): Negative - \(-4 < x < -3\): Positive - \(-3 < x < 1\): Positive - \(x > 1\): Positive Thus, the solution is: \[ x \in (-4, -3) \cup (-3, 1] \cup (1, \infty). \] For the second inequality: \[ \frac{2^{x}(2^{x}+3)(2^{x}-3)+16}{2^{2x}+2^{x}-6} \geq -2, \] let \(y = 2^x\), then: \[ \frac{y(y+3)(y-3)+16}{y^2 + y - 6} \geq -2. \] Solving this inequality similarly will yield the solution for \(x\).

Solución

To solve the inequality \[ \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq -2, \] we will first manipulate the inequality step by step. ### Step 1: Rearranging the Inequality We start by moving \(-2\) to the left side: \[ \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} + 2 \geq 0. \] Next, we can express \(2\) with a common denominator: \[ \frac{x(x+3)(x-3)+16 + 2(x+3)(x-2)}{(x+3)(x-2)} \geq 0. \] ### Step 2: Simplifying the Numerator Now, we need to simplify the numerator: \[ x(x+3)(x-3) + 16 + 2(x+3)(x-2). \] Calculating \(2(x+3)(x-2)\): \[ 2(x+3)(x-2) = 2(x^2 + x - 6) = 2x^2 + 2x - 12. \] Now, we can expand \(x(x+3)(x-3)\): \[ x(x^2 - 9) = x^3 - 9x. \] Combining everything, we have: \[ x^3 - 9x + 16 + 2x^2 + 2x - 12 = x^3 + 2x^2 - 7x + 4. \] ### Step 3: Final Inequality Now we can rewrite the inequality: \[ \frac{x^3 + 2x^2 - 7x + 4}{(x+3)(x-2)} \geq 0. \] ### Step 4: Finding Roots of the Numerator Next, we need to find the roots of the polynomial \(x^3 + 2x^2 - 7x + 4\). We can use the Rational Root Theorem to test possible rational roots. Testing \(x = 1\): \[ 1^3 + 2(1^2) - 7(1) + 4 = 1 + 2 - 7 + 4 = 0. \] Thus, \(x = 1\) is a root. We can factor the polynomial using synthetic division: \[ x^3 + 2x^2 - 7x + 4 = (x - 1)(x^2 + 3x - 4). \] Now we can factor \(x^2 + 3x - 4\): \[ x^2 + 3x - 4 = (x + 4)(x - 1). \] Thus, the complete factorization is: \[ (x - 1)^2(x + 4). \] ### Step 5: Analyzing the Inequality Now we have: \[ \frac{(x - 1)^2(x + 4)}{(x + 3)(x - 2)} \geq 0. \] The critical points are \(x = -4\), \(x = -3\), \(x = 1\), and \(x = 2\). We will test intervals around these points: 1. **Interval \((- \infty, -4)\)**: Choose \(x = -5\): \[ \frac{(-5 - 1)^2(-5 + 4)}{(-5 + 3)(-5 - 2)} = \frac{36(-1)}{(-2)(-7)} < 0. \] 2. **Interval \((-4, -3)\)**: Choose \(x = -3.5\): \[ \frac{(-3.5 - 1)^2(-3.5 + 4)}{(-3.5 + 3)(-3.5 - 2)} > 0. \] 3. **Interval \((-3, 1)\)**: Choose \(x = 0\): \[ \frac{(0 - 1)^2(0 + 4)}{(0 + 3)(0 - 2)} > 0. \] 4. **Interval \((1, 2)\)**: Choose \(x = 1.5\): \[ \frac{(1.5 - 1)^2(1.5 + 4)}{(1.5 + 3)(1.5 - 2)} < 0. \] 5. **Interval \((2, \infty)\)**: Choose \(x = 3\): \[ \frac{(3 - 1)^2(3 + 4)}{(3 + 3)(3 - 2)} > 0. \] ### Step 6: Conclusion for the First Inequality The solution to the inequality is: \[ (-4, -3) \cup (-3, 1] \cup (2, \infty). \] ### Step 7: Solving the Second Inequality Now we need to solve the inequality \[ \frac{2^{x}(2^{x}+3)(2^{x}-3)+16}{2^{2x}+2^{x}-6} \geq -2. \] Let \(y = 2^x\

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Bonus Knowledge

To tackle the first inequality \( \frac{x(x+3)(x-3)+16}{(x+3)(x-2)} \geq-2 \), we first simplify the left-hand side. Rewrite it as \( \frac{x(x+3)(x-3)+16 + 2(x+3)(x-2)}{(x+3)(x-2)} \geq 0 \). This will combine terms to allow us to investigate when the numerator is non-negative. By expanding and simplifying the numerator, we find \( x^3 - 9x + 12 \). You can then find the roots of this polynomial using synthetic division or the Rational Root Theorem, which means testing factors of the constant term over the leading coefficient. After solving for \( x \), you would test intervals based on these roots against the inequality \( x^3 - 9x + 12 \geq 0 \). For the second inequality \( \frac{2^{x}\left(2^{x}+3\right)\left(2^{x}-3\right)+16}{2^{2 x}+2^{x}-6} \geq-2 \), we recognize the terms \( 2^x = y \). Hence, rewrite it as \( \frac{y(y+3)(y-3)+16}{y^2+y-6} \geq-2 \). The same method applies: simplify the numerator, find roots, and check the sign of the expression across intervals identified by those roots. Remember, when dealing with rational inequalities, always consider where the denominator may be zero and exclude those values from your solution set. Lastly, double-check your intervals; it's easy to flip signs if you miss a negative factor! Good luck!

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