Proof If X = (1/2) + (1/2)√5, Show (x² + X⁻²)/(x - X⁻¹) = 3

by Kenji Nakamura 60 views

Hey there, math enthusiasts! Today, we're diving deep into an intriguing algebraic problem that involves manipulating expressions with radicals and fractions. Our mission, should we choose to accept it, is to prove that if $x=\frac{1}{2}+\frac{1}{2} \sqrt{5}$, then $\frac{x2+x{-2}}{x-x^{-1}}=3$. Buckle up, because we're about to embark on a mathematical journey filled with twists, turns, and ultimately, a satisfying solution. Let's get started, guys!

Understanding the Problem

Before we jump into the solution, let's take a moment to truly understand the problem at hand. We're given a value for x, which is expressed as a combination of a fraction and a radical. This value, $x=\frac{1}{2}+\frac{1}{2} \sqrt{5}$, might seem a bit intimidating at first glance, but fear not! We're going to break it down step by step. What we need to show, our ultimate goal, is that the expression $ rac{x2+x{-2}}{x-x^{-1}}$ simplifies to the number 3. This expression involves x squared, the reciprocal of x squared, x itself, and the reciprocal of x. To tackle this, we'll need to carefully manipulate these terms and see how they interact. The beauty of mathematics lies in its precision, and by following the rules of algebra, we can transform complex expressions into simpler forms. So, before we start crunching numbers, let's make sure we have a clear roadmap in our minds. We know our starting point (the value of x), and we know our destination (showing that the given expression equals 3). Now, let's figure out the best route to get there. Think of it like solving a puzzle – each step we take should bring us closer to the final picture. We'll be using algebraic identities, simplification techniques, and a bit of mathematical intuition to unravel this mystery. Are you ready to put on your thinking caps? Great! Let's move on to the next step and start exploring the properties of x that will help us in our quest. Remember, the key to success in math is persistence and a willingness to explore different approaches. So, let's dive in and see what we can discover about our friend, x.

Finding the Reciprocal of x

The first step in simplifying our target expression is to find the reciprocal of x, denoted as x⁻¹. Remember, the reciprocal of a number is simply 1 divided by that number. So, if $x=\frac{1}{2}+\frac{1}{2} \sqrt{5}$, then $x^{-1} = \frac{1}{x} = \frac{1}{\frac{1}{2}+\frac{1}{2} \sqrt{5}}$. Now, this looks a bit messy, doesn't it? We have a fraction within a fraction, and we want to simplify it. The trick here is to rationalize the denominator. This means we want to get rid of the square root in the denominator. To do this, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of $\frac{1}{2}+\frac{1}{2} \sqrt{5}$ is $\frac{1}{2}-\frac{1}{2} \sqrt{5}$. Why the conjugate? Because when we multiply a binomial by its conjugate, we get the difference of squares, which eliminates the square root. So, let's do it! We have:

x−1=112+125⋅12−12512−125x^{-1} = \frac{1}{\frac{1}{2}+\frac{1}{2} \sqrt{5}} \cdot \frac{\frac{1}{2}-\frac{1}{2} \sqrt{5}}{\frac{1}{2}-\frac{1}{2} \sqrt{5}}

Now, we multiply the numerators and the denominators:

x−1=12−125(12)2−(125)2x^{-1} = \frac{\frac{1}{2}-\frac{1}{2} \sqrt{5}}{(\frac{1}{2})^2 - (\frac{1}{2} \sqrt{5})^2}

Let's simplify the denominator:

(12)2=14(\frac{1}{2})^2 = \frac{1}{4}

(125)2=14⋅5=54(\frac{1}{2} \sqrt{5})^2 = \frac{1}{4} \cdot 5 = \frac{5}{4}

So, the denominator becomes:

14−54=−44=−1\frac{1}{4} - \frac{5}{4} = -\frac{4}{4} = -1

Now we have:

x−1=12−125−1x^{-1} = \frac{\frac{1}{2}-\frac{1}{2} \sqrt{5}}{-1}

Multiplying the numerator by -1, we get:

x−1=−12+125x^{-1} = -\frac{1}{2} + \frac{1}{2} \sqrt{5}

Great job, guys! We've successfully found the reciprocal of x. This was a crucial step, as we'll need this value to evaluate the expression we're trying to simplify. Now that we have x⁻¹, we can move on to calculating x² and x⁻². Remember, each step we take is building upon the previous one, so it's important to be meticulous and accurate. Let's keep up the momentum and tackle the next part of the problem!

Calculating x² and x⁻²

Now that we have both x and x⁻¹, our next task is to calculate x² and x⁻². This means we need to square both x and its reciprocal. Let's start with x². We know that $x=\frac{1}{2}+\frac{1}{2} \sqrt{5}$, so:

x2=(12+125)2x^2 = (\frac{1}{2}+\frac{1}{2} \sqrt{5})^2

To square this binomial, we can use the formula $(a+b)^2 = a^2 + 2ab + b^2$. In our case, $a = \frac{1}{2}$ and $b = \frac{1}{2} \sqrt{5}$. Let's plug these values into the formula:

x2=(12)2+2(12)(125)+(125)2x^2 = (\frac{1}{2})^2 + 2(\frac{1}{2})(\frac{1}{2} \sqrt{5}) + (\frac{1}{2} \sqrt{5})^2

Now, let's simplify each term:

(12)2=14(\frac{1}{2})^2 = \frac{1}{4}

2(12)(125)=1252(\frac{1}{2})(\frac{1}{2} \sqrt{5}) = \frac{1}{2} \sqrt{5}

(125)2=14⋅5=54(\frac{1}{2} \sqrt{5})^2 = \frac{1}{4} \cdot 5 = \frac{5}{4}

Putting it all together, we get:

x2=14+125+54x^2 = \frac{1}{4} + \frac{1}{2} \sqrt{5} + \frac{5}{4}

Combining the fractions, we have:

x2=64+125=32+125x^2 = \frac{6}{4} + \frac{1}{2} \sqrt{5} = \frac{3}{2} + \frac{1}{2} \sqrt{5}

Awesome! We've found x². Now, let's tackle x⁻². We know that $x^{-1} = -\frac{1}{2} + \frac{1}{2} \sqrt{5}$, so:

x−2=(−12+125)2x^{-2} = (-\frac{1}{2} + \frac{1}{2} \sqrt{5})^2

Again, we can use the formula $(a+b)^2 = a^2 + 2ab + b^2$, but this time, $a = -\frac{1}{2}$ and $b = \frac{1}{2} \sqrt{5}$. Plugging in the values:

x−2=(−12)2+2(−12)(125)+(125)2x^{-2} = (-\frac{1}{2})^2 + 2(-\frac{1}{2})(\frac{1}{2} \sqrt{5}) + (\frac{1}{2} \sqrt{5})^2

Simplifying each term:

(−12)2=14(-\frac{1}{2})^2 = \frac{1}{4}

2(−12)(125)=−1252(-\frac{1}{2})(\frac{1}{2} \sqrt{5}) = -\frac{1}{2} \sqrt{5}

(125)2=14⋅5=54(\frac{1}{2} \sqrt{5})^2 = \frac{1}{4} \cdot 5 = \frac{5}{4}

Putting it all together:

x−2=14−125+54x^{-2} = \frac{1}{4} - \frac{1}{2} \sqrt{5} + \frac{5}{4}

Combining the fractions:

x−2=64−125=32−125x^{-2} = \frac{6}{4} - \frac{1}{2} \sqrt{5} = \frac{3}{2} - \frac{1}{2} \sqrt{5}

Excellent work! We've successfully calculated both x² and x⁻². This was another key step in our journey. Now that we have these values, we're well-equipped to simplify the numerator and denominator of our target expression. We're making great progress, guys! Let's keep moving forward and see how these pieces fit together.

Simplifying the Numerator: x² + x⁻²

Now that we have calculated x² and x⁻², we can move on to simplifying the numerator of our expression, which is x² + x⁻². We found that:

x2=32+125x^2 = \frac{3}{2} + \frac{1}{2} \sqrt{5}

x−2=32−125x^{-2} = \frac{3}{2} - \frac{1}{2} \sqrt{5}

So, to find x² + x⁻², we simply add these two expressions together:

x2+x−2=(32+125)+(32−125)x^2 + x^{-2} = (\frac{3}{2} + \frac{1}{2} \sqrt{5}) + (\frac{3}{2} - \frac{1}{2} \sqrt{5})

Notice something beautiful here? The terms involving the square root of 5 cancel each other out! This is a common occurrence in these types of problems, and it's always a satisfying sight. So, let's simplify:

x2+x−2=32+32=62=3x^2 + x^{-2} = \frac{3}{2} + \frac{3}{2} = \frac{6}{2} = 3

Wow! That was surprisingly simple, wasn't it? We've found that the numerator, x² + x⁻², simplifies to 3. This is a significant step towards our goal. Now, we need to simplify the denominator, x - x⁻¹, and then we can see how the entire expression behaves. We're on a roll, guys! Let's keep the momentum going and tackle the denominator.

Simplifying the Denominator: x - x⁻¹

Alright, let's move on to simplifying the denominator of our expression, which is x - x⁻¹. We know that:

x=12+125x = \frac{1}{2} + \frac{1}{2} \sqrt{5}

x−1=−12+125x^{-1} = -\frac{1}{2} + \frac{1}{2} \sqrt{5}

So, to find x - x⁻¹, we subtract the second expression from the first:

x−x−1=(12+125)−(−12+125)x - x^{-1} = (\frac{1}{2} + \frac{1}{2} \sqrt{5}) - (-\frac{1}{2} + \frac{1}{2} \sqrt{5})

Be careful with the signs here! Subtracting a negative is the same as adding, so we have:

x−x−1=12+125+12−125x - x^{-1} = \frac{1}{2} + \frac{1}{2} \sqrt{5} + \frac{1}{2} - \frac{1}{2} \sqrt{5}

Again, we see some nice cancellation happening. This time, the terms involving the square root of 5 cancel each other out:

x−x−1=12+12=1x - x^{-1} = \frac{1}{2} + \frac{1}{2} = 1

Fantastic! We've found that the denominator, x - x⁻¹, simplifies to 1. This is another significant piece of the puzzle. Now that we've simplified both the numerator and the denominator, we're ready to put it all together and see if we've reached our destination. We're almost there, guys! Let's take the final step and evaluate the entire expression.

Evaluating the Expression (x² + x⁻²)/(x - x⁻¹)

We've reached the final stage of our mathematical journey! We've successfully simplified both the numerator and the denominator of our expression. We found that:

x2+x−2=3x^2 + x^{-2} = 3

x−x−1=1x - x^{-1} = 1

Now, we can substitute these values back into the original expression:

x2+x−2x−x−1=31=3\frac{x^2 + x^{-2}}{x - x^{-1}} = \frac{3}{1} = 3

Eureka! We've done it! We've shown that if $x=\frac{1}{2}+\frac{1}{2} \sqrt{5}$, then $\frac{x2+x{-2}}{x-x^{-1}}=3$. This was quite a journey, but we navigated it successfully by breaking the problem down into smaller, manageable steps. We found the reciprocal of x, calculated x² and x⁻², simplified the numerator and denominator separately, and then put it all together to arrive at our final answer. Give yourselves a pat on the back, guys! You've demonstrated excellent problem-solving skills and a deep understanding of algebraic manipulation. This is the beauty of mathematics – taking a seemingly complex problem and unraveling it through logical steps and careful calculations. Remember, the key to success in math is practice and persistence. The more you practice, the more comfortable you'll become with these types of problems, and the more confident you'll feel in your abilities. So, keep exploring, keep learning, and keep challenging yourselves. And remember, math can be fun!

Conclusion

In conclusion, we have successfully shown that if $x=\frac{1}{2}+\frac{1}{2} \sqrt{5}$, then $\frac{x2+x{-2}}{x-x^{-1}}=3$. This problem highlights the power of algebraic manipulation and the importance of breaking down complex problems into smaller, more manageable steps. We used techniques such as rationalizing the denominator, squaring binomials, and simplifying expressions to arrive at our final answer. This exercise not only demonstrates a specific mathematical result but also reinforces fundamental algebraic skills that are crucial for further mathematical studies. Keep practicing and exploring, guys, and you'll continue to unlock the fascinating world of mathematics!