Five Times the Sum of a Number and Its Reciprocal: A Step‑by‑Step Guide to Solving the Equation
When you first encounter the expression five times the sum of a number and its reciprocal, it can feel like a puzzle. That said, by breaking the problem into clear, manageable steps, you can solve it quickly and confidently. This article walks you through the entire process—from understanding the terminology to applying algebraic techniques—so you can tackle similar problems with ease That's the whole idea..
Not the most exciting part, but easily the most useful Worth keeping that in mind..
Introduction
In algebra, many problems revolve around manipulating expressions that involve a number and its reciprocal. A reciprocal is simply the inverse of a number: if (x) is a non‑zero real number, then its reciprocal is (\frac{1}{x}). When the problem statement says “five times the sum of a number and its reciprocal,” it is asking you to evaluate the expression
[ 5\left(x + \frac{1}{x}\right) ]
where (x) is the unknown number. Often, the problem will set this expression equal to a given value, and your task is to find the possible values of (x) But it adds up..
Below, we’ll cover:
- Understanding the Problem – What the expression really means.
- Setting Up the Equation – Translating the verbal description into algebra.
- Solving the Equation – Step‑by‑step algebraic manipulation.
- Checking for Extraneous Solutions – Why we must verify our answers.
- Common Variations – How to adapt the method to similar problems.
- Frequently Asked Questions (FAQ) – Quick answers to typical doubts.
- Conclusion – Recap and final thoughts.
1. Understanding the Problem
Let’s dissect the phrase:
- “Five times”: Multiply the entire following expression by 5.
- “The sum of a number and its reciprocal”: Add the number (x) to (\frac{1}{x}).
Putting it together, we get the algebraic form:
[ 5\left(x + \frac{1}{x}\right) ]
If the problem states that this expression equals, say, 12, the full equation becomes:
[ 5\left(x + \frac{1}{x}\right) = 12 ]
Our goal is to solve for (x).
2. Setting Up the Equation
2.1 Write the Expression Clearly
Start by writing the expression in a single line:
[ 5x + \frac{5}{x} ]
2.2 Isolate the Variable
If the problem gives a specific value on the right side, move it to the left (or vice versa) so that the equation equals zero. To give you an idea, with 12 on the right:
[ 5x + \frac{5}{x} - 12 = 0 ]
2.3 Clear the Denominator
Multiplying both sides by (x) (the common denominator) eliminates the fraction:
[ x(5x) + x\left(\frac{5}{x}\right) - 12x = 0 ]
Simplify:
[ 5x^2 + 5 - 12x = 0 ]
2.4 Arrange in Standard Quadratic Form
Reorder terms to match the familiar quadratic template (ax^2 + bx + c = 0):
[ 5x^2 - 12x + 5 = 0 ]
Now we have a standard quadratic equation ready to solve Practical, not theoretical..
3. Solving the Equation
3.1 Use the Quadratic Formula
For a quadratic equation (ax^2 + bx + c = 0), the solutions are:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Plugging in (a = 5), (b = -12), (c = 5):
-
Compute the discriminant: [ \Delta = b^2 - 4ac = (-12)^2 - 4(5)(5) = 144 - 100 = 44 ]
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Take the square root: [ \sqrt{\Delta} = \sqrt{44} = 2\sqrt{11} ]
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Apply the formula: [ x = \frac{-(-12) \pm 2\sqrt{11}}{2(5)} = \frac{12 \pm 2\sqrt{11}}{10} ]
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Simplify: [ x = \frac{6 \pm \sqrt{11}}{5} ]
Thus, the two possible solutions are:
[ x_1 = \frac{6 + \sqrt{11}}{5}, \qquad x_2 = \frac{6 - \sqrt{11}}{5} ]
3.2 Verify the Solutions
Always substitute back into the original equation to confirm they satisfy it Easy to understand, harder to ignore..
For (x_1):
[ 5\left(\frac{6 + \sqrt{11}}{5} + \frac{5}{6 + \sqrt{11}}\right) \overset{?}{=} 12 ]
After simplifying (rationalizing the denominator if needed), you’ll find the equality holds. Because of that, repeat for (x_2). If both satisfy the equation, you’re done.
4. Checking for Extraneous Solutions
When dealing with reciprocals, you must see to it that the denominator is never zero. In our case, (x) cannot be 0 because (\frac{1}{x}) would be undefined. Neither (x_1) nor (x_2) equals zero, so both are valid.
5. Common Variations
The method above applies broadly. Here are a few twists you might encounter:
| Variation | Expression | Key Adjustment |
|---|---|---|
| Different constant multiplier | (k\left(x + \frac{1}{x}\right) = N) | Keep (k) outside; multiply both sides by (x) to clear the fraction. |
| Additional linear term | (5\left(x + \frac{1}{x}\right) + 3x = 12) | Expand, combine like terms, then clear denominators. |
| Reciprocal of reciprocal | (5\left(x + \frac{1}{x}\right) = \frac{1}{x}) | Multiply both sides by (x) to eliminate the fraction on the right. |
| Higher degree | (5\left(x^2 + \frac{1}{x^2}\right) = 7) | Use the identity (\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2). |
By recognizing the structure, you can adapt the same systematic approach.
6. FAQ
| Question | Answer |
|---|---|
| **Why must I multiply by the denominator?Now, | |
| **What if the discriminant is negative? ** | Yes. |
| What if the problem says “find all real numbers”? | For simple coefficients, factoring may work, but the quadratic formula is reliable for all cases. ** |
| **Can the solutions be negative?That's why ** | It eliminates fractions, turning the equation into a polynomial that is easier to solve. Also, |
| **Is there a shortcut for solving such equations? Here's the thing — always check for extraneous solutions, especially when the reciprocal is involved. The sign depends on the specific constants. ** | Verify that the discriminant is non‑negative and that no denominator becomes zero. |
7. Conclusion
Solving an equation that involves five times the sum of a number and its reciprocal is a matter of clear translation from words to algebra, careful manipulation to eliminate fractions, and systematic application of the quadratic formula. By following these steps—understanding the problem, setting up the equation, clearing denominators, solving the quadratic, and checking for extraneous solutions—you can confidently tackle this and many similar algebraic challenges. Remember, practice is key: the more you work through diverse variations, the more intuitive the process becomes.
Quick note before moving on.
