2(a + b) = (3 + \sqrt{5})(a - b)
Mastering the Equation: Solving 2(a + b) = (3 + √5)(a - b)
Understanding algebraic equations is a fundamental skill in mathematics, and equations involving radicals like √5 often appear in advanced algebra, trigonometry, and mathematical modeling. One such equation that learners frequently encounter is:
> 2(a + b) = (3 + √5)(a - b)
Whether you’re solving for one variable in terms of the other or exploring deeper algebraic properties, mastering this equation strengthens your problem-solving abilities. In this article, we’ll guide you step-by-step through simplifying, solving, and interpreting the equation — all optimized for clarity and SEO-friendly content.
What Does the Equation Represent?
The equation 2(a + b) = (3 + √5)(a − b) is a linear relationship linking two expressions involving variables a and b. The term (3 + √5) is an irrational coefficient, making this equation ideal for practicing simplification and algebraic manipulation, especially when working with radicals.
Step-by-Step Solution
Step 1: Expand both sides
Start by expanding both sides to eliminate parentheses:
Left-hand side: 2(a + b) = 2a + 2b
Right-hand side: (3 + √5)(a − b) = 3a − 3b + a√5 − b√5
So, the equation becomes: 2a + 2b = 3a − 3b + a√5 − b√5
Step 2: Move all terms to one side
Collect every term to the left to group like terms:
2a + 2b − 3a + 3b − a√5 + b√5 = 0
Combine like terms:
- a-terms: 2a − 3a = −a
- b-terms: 2b + 3b = 5b
- radical terms: −a√5 + b√5 = √5(b − a)
Resulting equation: −a + 5b + √5(b − a) = 0
Step 3: Factor intelligent grouping
Rewriting: √5(b − a) − a + 5b = 0
Group terms strategically: √5(b − a) + (5b − a) = 0
Now, isolate the radical term: √5(b − a) = a − 5b
Step 4: Rationalize or substitute (optional)
Since √5 is irrational, solving explicitly for one variable is often more useful. Let’s solve for a in terms of b, or vice versa.
From above: √5(b − a) = a − 5b
Square both sides to eliminate the square root (valid since both sides are real, assuming valid domain):
(√5(b − a))² = (a − 5b)² 5(b² − 2ab + a²) = a² − 10ab + 25b²
Expand: 5b² − 10ab + 5a² = a² − 10ab + 25b²
Subtract −10ab from both sides: 5b² + 5a² = a² + 25b²
Bring all terms to one side: 5a² − a² + 5b² − 25b² = 0 4a² − 20b² = 0
Divide by 4: a² − 5b² = 0
So: a² = 5b² ⇒ a = ±√5 b
Interpretation of the Solution
From the simplification, the general solution to 2(a + b) = (3 + √5)(a − b) is: a = √5 b or a = −√5 b
This reveals a linear relationship between a and b defined by the irrational coefficient 3 + √5. These proportional solutions are important in vector geometry, similarity transformations, and eigenvalues in linear algebra.
Practical Applications
- Geometry & Trigonometry: Equations with √5 appear when dealing with golden ratios, pentagonal symmetry, and trigonometric identities involving golden angles.
- Physics Problems: Modeling wave interference or resonance systems often leads to radical-based equations.
- Algebraic通用性: Understanding how to isolate variables when irrational terms are present helps build foundational algebra skills applicable across disciplines.
Key Takeaways
- Always expand both sides before rearranging terms.
- Isolating radicals and squaring both sides eliminates irrational numbers safely.
- Practice solving for variables either explicitly or in terms of a proportional relationship.
- The irrational coefficient 3 + √5 defines a linear functional dependence, useful in modeling and advanced math.
Frequently Asked Questions (FAQ)
Q: Can I leave the equation unsolved? A: Yes, expressing a relationship like a = √5 b is sufficient in many contexts, especially when analyzing variable dependencies.
Q: Does squaring both sides introduce extraneous solutions? A: Yes, always check solutions by substituting back, especially when radicals are involved.
Q: Is there a geometric meaning? A: Works with vectors and proportions—especially relevant when dealing with the golden ratio (≈1.618), where √5 emerges naturally.
Conclusion
The equation 2(a + b) = (3 + √5)(a − b) may seem complex due to the radical, but through careful expansion, grouping, and squaring, we reveal a clean proportional relationship: a = ±√5 b. Mastering such equations builds strong algebraic intuition and equips you with tools for advanced mathematics.
For deeper understanding, practice applying similar techniques with other radical-based equations and explore their real-world applications in engineering, physics, and geometry.
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If you want more detailed worked examples or interactive exercises, explore online algebra tools or textbooks focusing on equations with irrational coefficients!
