Math Problem Statement

Jika x_1 dan x_2 adalah penyelesaian persamaan \( \left( \frac{4}{9} \right)^{x^2 - 3} \left( \frac{8}{27} \right)^{1 - x} = \frac{3}{2} \), maka \( (x_1 - x_2)^2 = \) ... Pilihan jawaban: A) \( \frac{9}{4} \), B) \( \frac{25}{4} \), C) \( \frac{41}{4} \), D) \( \frac{25}{2} \), E) 25

Solution

To solve this problem, let's go through it step-by-step.

We are given the equation:

(49)x23(827)1x=32\left( \frac{4}{9} \right)^{x^2 - 3} \left( \frac{8}{27} \right)^{1 - x} = \frac{3}{2}

and we are told that x1x_1 and x2x_2 are the solutions to this equation. We need to find the value of (x1x2)2(x_1 - x_2)^2.

Step 1: Simplify the Equation

The left-hand side of the equation can be simplified using properties of exponents. Let's express 49\frac{4}{9} and 827\frac{8}{27} in terms of base powers:

49=(23)2and827=(23)3.\frac{4}{9} = \left( \frac{2}{3} \right)^2 \quad \text{and} \quad \frac{8}{27} = \left( \frac{2}{3} \right)^3.

Thus, we can rewrite the equation as:

((23)2)x23((23)3)1x=32.\left( \left( \frac{2}{3} \right)^2 \right)^{x^2 - 3} \left( \left( \frac{2}{3} \right)^3 \right)^{1 - x} = \frac{3}{2}.

Using the power of a power rule (am)n=amn(a^m)^n = a^{m \cdot n}, we get:

(23)2(x23)(23)3(1x)=32.\left( \frac{2}{3} \right)^{2(x^2 - 3)} \cdot \left( \frac{2}{3} \right)^{3(1 - x)} = \frac{3}{2}.

Now, we can combine the exponents:

(23)2x26+33x=32.\left( \frac{2}{3} \right)^{2x^2 - 6 + 3 - 3x} = \frac{3}{2}.

Simplify the exponent:

(23)2x23x3=32.\left( \frac{2}{3} \right)^{2x^2 - 3x - 3} = \frac{3}{2}.

Step 2: Equate Exponents

Since 23\frac{2}{3} and 32\frac{3}{2} are reciprocal values, we can set the exponent equal to 1-1 (since (2/3)1=3/2(2/3)^{-1} = 3/2):

2x23x3=1.2x^2 - 3x - 3 = -1.

Add 1 to both sides:

2x23x2=0.2x^2 - 3x - 2 = 0.

Step 3: Solve the Quadratic Equation

Now we have a quadratic equation:

2x23x2=0.2x^2 - 3x - 2 = 0.

Using the quadratic formula x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=2a = 2, b=3b = -3, and c=2c = -2:

x=(3)±(3)242(2)22.x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}.

x=3±9+164.x = \frac{3 \pm \sqrt{9 + 16}}{4}.

x=3±254.x = \frac{3 \pm \sqrt{25}}{4}.

x=3±54.x = \frac{3 \pm 5}{4}.

This gives us two solutions:

x1=3+54=2andx2=354=12.x_1 = \frac{3 + 5}{4} = 2 \quad \text{and} \quad x_2 = \frac{3 - 5}{4} = -\frac{1}{2}.

Step 4: Find (x1x2)2(x_1 - x_2)^2

Now that we have x1=2x_1 = 2 and x2=12x_2 = -\frac{1}{2}, let's find (x1x2)2(x_1 - x_2)^2:

x1x2=2(12)=2+12=52.x_1 - x_2 = 2 - \left(-\frac{1}{2}\right) = 2 + \frac{1}{2} = \frac{5}{2}.

Then,

(x1x2)2=(52)2=254.(x_1 - x_2)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}.

Answer

The answer is:

254\boxed{\frac{25}{4}}


Do you have any questions or need further details on any step?


Related Questions

  1. How would the solution change if the base in the equation was different?
  2. Can we solve this equation using logarithms instead of simplifying the bases?
  3. What if we were asked to find x1x2x_1 \cdot x_2 instead of (x1x2)2(x_1 - x_2)^2?
  4. How does the discriminant in the quadratic formula help us determine the nature of the roots?
  5. What other methods can be used to solve quadratic equations?

Tip

When working with exponentials, try expressing terms with common bases to simplify the equation.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Exponentials

Formulas

Quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Theorems

Exponent rules
Quadratic formula theorem

Suitable Grade Level

Grades 10-12