Math Problem Statement

Consider the equation x^2 - 9/x^2 = 10. Solve for x such that the inequality x^3 - 3x > 0 is also satisfied. Additionally, show that y = x^3 - 3x is increasing at x = 1.

Solution

The problem you uploaded asks for two things:

  1. Solve the equation: x29x2=10x^2 - \frac{9}{x^2} = 10 for xx such that the inequality x33x>0x^3 - 3x > 0 is also satisfied.

  2. Show that the function: y=x33xy = x^3 - 3x is increasing at x=1x = 1.

Step 1: Solving the equation

The equation is: x29x2=10x^2 - \frac{9}{x^2} = 10 Multiplying both sides by x2x^2 to eliminate the fraction gives: x49=10x2x^4 - 9 = 10x^2 Rearranging this, we get a quadratic equation in terms of x2x^2: x410x29=0x^4 - 10x^2 - 9 = 0 Let z=x2z = x^2. This simplifies to: z210z9=0z^2 - 10z - 9 = 0 We can solve this quadratic using the quadratic formula: z=(10)±(10)24(1)(9)2(1)z = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(-9)}}{2(1)} z=10±100+362=10±1362z = \frac{10 \pm \sqrt{100 + 36}}{2} = \frac{10 \pm \sqrt{136}}{2} z=10±2342z = \frac{10 \pm 2\sqrt{34}}{2} z=5±34z = 5 \pm \sqrt{34} Thus, x2=5+34x^2 = 5 + \sqrt{34} or x2=534x^2 = 5 - \sqrt{34}.

Since 34\sqrt{34} is approximately 5.83, the second solution 5345 - \sqrt{34} is negative, which is not possible for x2x^2. Hence, the only valid solution is: x2=5+34x^2 = 5 + \sqrt{34} Taking the square root of both sides: x=±5+34x = \pm \sqrt{5 + \sqrt{34}}

Step 2: Checking the inequality x33x>0x^3 - 3x > 0

The inequality to check is x33x>0x^3 - 3x > 0. Factorizing this, we get: x(x23)>0x(x^2 - 3) > 0 The critical points occur when x=0x = 0 or x=±3x = \pm \sqrt{3}.

Testing intervals:

  • For x(,3)x \in (-\infty, -\sqrt{3}), the expression is positive.
  • For x(3,0)x \in (-\sqrt{3}, 0), the expression is negative.
  • For x(0,3)x \in (0, \sqrt{3}), the expression is negative.
  • For x(3,)x \in (\sqrt{3}, \infty), the expression is positive.

Thus, x33x>0x^3 - 3x > 0 when x>3x > \sqrt{3} or x<3x < -\sqrt{3}.

Step 3: Verifying at x=1x = 1 that the function is increasing

To show that y=x33xy = x^3 - 3x is increasing at x=1x = 1, we take the derivative: y=ddx(x33x)=3x23y' = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3 At x=1x = 1: y(1)=3(1)23=33=0y'(1) = 3(1)^2 - 3 = 3 - 3 = 0 This means the function is not increasing or decreasing at x=1x = 1, as the derivative equals zero (it has a critical point).

Would you like more details or have any further questions?

Here are 5 related questions to explore further:

  1. How do we solve quadratic equations in general?
  2. What does the sign of the derivative tell us about a function's behavior?
  3. What is the graphical interpretation of the function y=x33xy = x^3 - 3x?
  4. How can we determine intervals where a function is increasing or decreasing?
  5. How do we solve equations involving higher powers (quartic equations)?

Tip: Always verify the domain of solutions when dealing with square roots or fractions in equations!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Inequalities
Derivatives
Functions

Formulas

x^2 - 9/x^2 = 10
x^3 - 3x > 0
y' = 3x^2 - 3 (derivative of y = x^3 - 3x)

Theorems

Quadratic formula
First derivative test (for increasing or decreasing functions)

Suitable Grade Level

Grades 10-12