Solving (5+2√3)/(7+√3) Find A And B Rational Numbers

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In the captivating realm of mathematics, irrational equations present a unique challenge that requires a blend of algebraic manipulation and a deep understanding of number systems. These equations, often involving radicals and irrational numbers, demand meticulous attention to detail and a strategic approach to unravel their solutions. This article delves into the intricacies of solving such equations, specifically focusing on an example that involves finding rational numbers 'a' and 'b' in an expression containing square roots.

This exploration is not just an academic exercise; it's a journey into the heart of mathematical problem-solving. By tackling irrational equations, we sharpen our algebraic skills, enhance our understanding of number properties, and cultivate the ability to think critically and strategically. The problem at hand, 5+237+3=a−3b3{ \frac{5+2\sqrt{3}}{7+\sqrt{3}} = a - 3b\sqrt{3} }, serves as an excellent illustration of the techniques involved in simplifying expressions with radicals and isolating variables to arrive at a solution. Through this detailed analysis, we aim to demystify irrational equations and empower you with the tools to conquer them.

The beauty of mathematics lies in its ability to connect seemingly disparate concepts. In this case, we'll be drawing upon our knowledge of algebraic operations, rationalization techniques, and the properties of irrational numbers. Each step in the solution process is a testament to the interconnectedness of mathematical ideas, revealing how a firm grasp of fundamental principles can unlock the answers to complex problems. So, let's embark on this mathematical adventure and uncover the values of 'a' and 'b' that satisfy the given equation.

The equation 5+237+3=a−3b3{ \frac{5+2\sqrt{3}}{7+\sqrt{3}} = a - 3b\sqrt{3} } presents a fascinating challenge. Our mission is to determine the values of the rational numbers 'a' and 'b' that make this equation a mathematical truth. The presence of the square root of 3 (3{ \sqrt{3} }) in both the numerator and denominator of the fraction, as well as in the term −3b3{ -3b\sqrt{3} }, adds a layer of complexity that requires a strategic approach.

The key to unraveling this equation lies in a technique known as rationalizing the denominator. This process involves eliminating the radical from the denominator of the fraction, thereby simplifying the expression and making it easier to compare with the right-hand side of the equation. Rationalizing the denominator is achieved by multiplying both the numerator and denominator of the fraction by the conjugate of the denominator. The conjugate of a binomial expression of the form x+y{ x + y } is x−y{ x - y }, and vice versa. This clever manipulation leverages the difference of squares identity, (x+y)(x−y)=x2−y2{ (x + y)(x - y) = x^2 - y^2 }, to eliminate the radical.

In this specific problem, the denominator is 7+3{ 7 + \sqrt{3} }, so its conjugate is 7−3{ 7 - \sqrt{3} }. By multiplying both the numerator and denominator of the fraction by this conjugate, we'll transform the left-hand side of the equation into a form that allows us to isolate the terms with and without the square root of 3. This will pave the way for us to equate the corresponding parts of the equation and solve for 'a' and 'b'. The elegance of this technique lies in its ability to convert a seemingly complex expression into a more manageable form, highlighting the power of algebraic manipulation in simplifying mathematical problems.

To embark on the journey of solving for 'a' and 'b', our first crucial step involves rationalizing the denominator of the fraction 5+237+3{ \frac{5+2\sqrt{3}}{7+\sqrt{3}} }. This process, as mentioned earlier, is the key to simplifying the expression and making it amenable to further analysis. We achieve this by multiplying both the numerator and the denominator by the conjugate of the denominator. In this case, the conjugate of 7+3{ 7 + \sqrt{3} } is 7−3{ 7 - \sqrt{3} }. Let's break down this process step-by-step:

  1. Identify the Conjugate: The conjugate of the denominator 7+3{ 7 + \sqrt{3} } is 7−3{ 7 - \sqrt{3} }.
  2. Multiply Numerator and Denominator: We multiply both the numerator and the denominator of the fraction by this conjugate: 5+237+3×7−37−3{ \frac{5+2\sqrt{3}}{7+\sqrt{3}} \times \frac{7-\sqrt{3}}{7-\sqrt{3}} }
  3. Expand the Numerator: We carefully expand the product in the numerator using the distributive property: (5+23)(7−3)=5(7)+5(−3)+23(7)+23(−3){ (5 + 2\sqrt{3})(7 - \sqrt{3}) = 5(7) + 5(-\sqrt{3}) + 2\sqrt{3}(7) + 2\sqrt{3}(-\sqrt{3}) } =35−53+143−6{ = 35 - 5\sqrt{3} + 14\sqrt{3} - 6 }
  4. Simplify the Numerator: We combine the like terms in the numerator: 35−6−53+143=29+93{ 35 - 6 - 5\sqrt{3} + 14\sqrt{3} = 29 + 9\sqrt{3} }
  5. Expand the Denominator: We expand the product in the denominator using the difference of squares identity, (a+b)(a−b)=a2−b2{ (a + b)(a - b) = a^2 - b^2 }: (7+3)(7−3)=72−(3)2{ (7 + \sqrt{3})(7 - \sqrt{3}) = 7^2 - (\sqrt{3})^2 } =49−3{ = 49 - 3 }
  6. Simplify the Denominator: We simplify the denominator: 49−3=46{ 49 - 3 = 46 }
  7. The Rationalized Fraction: Now we have the rationalized fraction: 29+9346{ \frac{29 + 9\sqrt{3}}{46} }

By meticulously following these steps, we have successfully rationalized the denominator of the fraction. This simplified form is much easier to work with and will be instrumental in our quest to find the values of 'a' and 'b'. The careful application of algebraic principles and the strategic use of the conjugate have transformed a complex expression into a more manageable one, showcasing the power of these techniques in solving mathematical problems.

Now that we've skillfully rationalized the denominator, our equation stands as 29+9346=a−3b3{ \frac{29 + 9\sqrt{3}}{46} = a - 3b\sqrt{3} }. The next strategic move is to separate the rational and irrational parts of the fraction on the left-hand side. This separation will allow us to equate the corresponding terms on both sides of the equation, paving the way for us to solve for 'a' and 'b'.

To separate the rational and irrational parts, we can rewrite the fraction as a sum of two fractions, each with the denominator 46: 29+9346=2946+9346{ \frac{29 + 9\sqrt{3}}{46} = \frac{29}{46} + \frac{9\sqrt{3}}{46} }

Now, our equation looks like this: 2946+9346=a−3b3{ \frac{29}{46} + \frac{9\sqrt{3}}{46} = a - 3b\sqrt{3} }

The left-hand side now clearly exhibits a rational part (2946{ \frac{29}{46} }) and an irrational part (9346{ \frac{9\sqrt{3}}{46} }). The right-hand side, a−3b3{ a - 3b\sqrt{3} }, also has a rational part ('a') and an irrational part (−3b3{ -3b\sqrt{3} }). The fundamental principle we'll employ here is that for two expressions of the form p+qr=x+yr{ p + q\sqrt{r} = x + y\sqrt{r} } to be equal, where p, q, x, and y are rational numbers and r{ \sqrt{r} } is irrational, then p=x{ p = x } and q=y{ q = y }.

Applying this principle to our equation, we can equate the rational parts and the irrational parts separately:

  • Equating Rational Parts: a=2946{ a = \frac{29}{46} }
  • Equating Irrational Parts: 9346=−3b3{ \frac{9\sqrt{3}}{46} = -3b\sqrt{3} }

By separating the equation into these two simpler equations, we've made significant progress in isolating the unknowns 'a' and 'b'. We now have a direct expression for 'a' and a simple equation involving 'b'. The elegance of this approach lies in its ability to break down a complex equation into manageable components, highlighting the power of careful observation and strategic manipulation in solving mathematical problems.

With the equation neatly separated into its rational and irrational components, we are now poised to solve for the values of 'a' and 'b'. This is the culmination of our efforts, where the strategic steps we've taken so far will lead us to the solution.

From equating the rational parts, we have: a=2946{ a = \frac{29}{46} } This elegantly provides us with the value of 'a'. It's a straightforward result, a testament to the power of rationalizing the denominator and separating terms.

Now, let's focus on the irrational parts. We have the equation: 9346=−3b3{ \frac{9\sqrt{3}}{46} = -3b\sqrt{3} } To solve for 'b', we need to isolate it. The presence of 3{ \sqrt{3} } on both sides of the equation allows us to simplify by dividing both sides by 3{ \sqrt{3} }: 946=−3b{ \frac{9}{46} = -3b } Next, we divide both sides of the equation by -3 to isolate 'b': b=946÷(−3){ b = \frac{9}{46} \div (-3) } To divide by -3, we multiply by its reciprocal, which is −13{ -\frac{1}{3} }: b=946×(−13){ b = \frac{9}{46} \times \left(-\frac{1}{3}\right) } Now, we perform the multiplication: b=−946×3{ b = -\frac{9}{46 \times 3} } We can simplify this fraction by canceling the common factor of 3 between 9 and 3: b=−346{ b = -\frac{3}{46} } And there we have it! We've successfully found the value of 'b'.

In this mathematical journey, we set out to find the values of the rational numbers 'a' and 'b' in the equation 5+237+3=a−3b3{ \frac{5+2\sqrt{3}}{7+\sqrt{3}} = a - 3b\sqrt{3} }. Through a series of strategic steps, including rationalizing the denominator, separating rational and irrational parts, and equating like terms, we have triumphantly arrived at the solution.

Our meticulous efforts have revealed that:

  • a=2946{ a = \frac{29}{46} }
  • b=−346{ b = -\frac{3}{46} }

These values of 'a' and 'b' satisfy the given equation, confirming the validity of our solution process. This exploration not only provided us with the answers but also reinforced the importance of fundamental algebraic techniques and strategic problem-solving in mathematics.

The beauty of this problem lies in its ability to illustrate the interconnectedness of various mathematical concepts. We utilized the properties of irrational numbers, the technique of rationalizing the denominator, and the principle of equating like terms to dissect the equation and extract the desired information. Each step was a testament to the power of mathematical reasoning and the elegance of algebraic manipulation.

By successfully navigating this irrational equation, we've honed our mathematical skills and deepened our understanding of the intricacies of number systems. This experience serves as a valuable stepping stone for tackling more complex mathematical challenges in the future, empowering us with the confidence and tools to unravel any equation that comes our way. The world of mathematics is full of such intriguing puzzles, waiting to be solved with a combination of knowledge, strategy, and a dash of mathematical curiosity.

  • Irrational Equations
  • Rationalizing the Denominator
  • Solving for Variables
  • Algebraic Manipulation
  • Square Roots
  • Conjugate of a Binomial
  • Equating Like Terms
  • Rational Numbers
  • Mathematical Problem-Solving
  • Simplifying Expressions