If x 2+ 1 = 2x, then find x – \((\frac{1}{x})\)
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Understanding the Algebra Problem
The question asks us to find the value of an expression, \(x - (\frac{1}{x})\), given an algebraic equation, \(x^2 + 1 = 2x\). To solve this problem, we first need to find the value of \(x\) from the given equation. Once we know the value of \(x\), we can substitute it into the expression \(x - (\frac{1}{x})\) to find its value.
Solving the Given Equation: \(x^2 + 1 = 2x\)
The given equation is a quadratic equation. Let's rearrange it into the standard quadratic form, which is \(ax^2 + bx + c = 0\).
We have:
\[x^2 + 1 = 2x\]Subtract \(2x\) from both sides of the equation to move all terms to one side:
\[x^2 - 2x + 1 = 0\]This equation is a perfect square trinomial. It can be factored as \((x - 1)^2\).
So, the equation becomes:
\[(x - 1)^2 = 0\]To solve for \(x\), take the square root of both sides:
\[\sqrt{(x - 1)^2} = \sqrt{0}\] \[x - 1 = 0\]Now, add 1 to both sides to isolate \(x\):
\[x = 1\]So, the value of \(x\) that satisfies the given equation \(x^2 + 1 = 2x\) is \(1\).
Finding the Value of the Expression \(x - (\frac{1}{x})\)
Now that we have found \(x = 1\), we can substitute this value into the expression \(x - (\frac{1}{x})\).
Substitute \(x = 1\) into the expression:
\[x - \left(\frac{1}{x}\right) = 1 - \left(\frac{1}{1}\right)\]Calculate the value:
\[1 - \left(\frac{1}{1}\right) = 1 - 1 = 0\]Therefore, the value of \(x - (\frac{1}{x})\) when \(x^2 + 1 = 2x\) is \(0\).
Step-by-Step Solution Summary
Here's a quick recap of the steps taken to solve this algebra problem:
- Start with the given equation: \(x^2 + 1 = 2x\).
- Rearrange the equation into standard form: \(x^2 - 2x + 1 = 0\).
- Factor the quadratic equation: \((x - 1)^2 = 0\).
- Solve for \(x\): \(x - 1 = 0\), which gives \(x = 1\).
- Substitute the value of \(x\) into the required expression: \(x - (\frac{1}{x}) = 1 - (\frac{1}{1})\).
- Calculate the final result: \(1 - 1 = 0\).
Comparing with Options
Let's compare our calculated value with the given options:
| Option | Value | Matches our result? |
|---|---|---|
| 1 | 4 | No |
| 2 | 12 | No |
| 3 | 0 | Yes |
| 4 | 2 | No |
Our calculated value, \(0\), matches Option 3.
Revision Table: Key Concepts for Solving Algebraic Equations
| Concept | Description | Relevance to this problem |
|---|---|---|
| Quadratic Equation | An equation of the form \(ax^2 + bx + c = 0\), where \(a \neq 0\). | The given equation \(x^2 + 1 = 2x\) is a quadratic equation that can be rearranged to \(x^2 - 2x + 1 = 0\). |
| Factoring Quadratics | Breaking down a quadratic expression into a product of linear factors (e.g., \((x-r_1)(x-r_2)\)). | The equation \(x^2 - 2x + 1 = 0\) is factored as \((x-1)^2 = 0\). |
| Perfect Square Trinomial | A trinomial that results from squaring a binomial, like \((a+b)^2 = a^2 + 2ab + b^2\) or \((a-b)^2 = a^2 - 2ab + b^2\). | \(x^2 - 2x + 1\) is a perfect square trinomial, specifically \((x-1)^2\). |
| Solving Equations | Finding the value(s) of the variable that make the equation true. | We solved \((x-1)^2 = 0\) to find \(x=1\). |
| Substitution | Replacing a variable with its numerical value or another expression. | We substituted \(x=1\) into the expression \(x - (\frac{1}{x})\). |
Additional Information: Alternative Approaches
While factoring was the easiest way to solve \(x^2 - 2x + 1 = 0\), other methods could also be used for solving quadratic equations:
- Quadratic Formula: For \(ax^2 + bx + c = 0\), the solutions are given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our case, \(a=1\), \(b=-2\), \(c=1\).
- Completing the Square: This method involves manipulating the equation to create a perfect square trinomial on one side.
Also, for the expression \(x - (\frac{1}{x})\), we could have potentially manipulated the original equation differently. If we divide the original equation \(x^2 - 2x + 1 = 0\) by \(x\) (assuming \(x \neq 0\)), we get:
\[\frac{x^2}{x} - \frac{2x}{x} + \frac{1}{x} = 0\] \[x - 2 + \frac{1}{x} = 0\] \[x + \frac{1}{x} = 2\]This gives us the value of \(x + (\frac{1}{x})\), which is 2. However, the question asks for \(x - (\frac{1}{x})\). We know \((x - \frac{1}{x})^2 = x^2 - 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}\). Also, \((x + \frac{1}{x})^2 = x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}\). So, \((x - \frac{1}{x})^2 = (x + \frac{1}{x})^2 - 4\). Using \(x + (\frac{1}{x}) = 2\):
\[\left(x - \frac{1}{x}\right)^2 = (2)^2 - 4 = 4 - 4 = 0\]Taking the square root of both sides:
\[x - \frac{1}{x} = \sqrt{0} = 0\]This confirms our earlier result using a different algebraic manipulation. Note that \(x=1\) confirms \(x \neq 0\), so dividing by \(x\) was valid.
Important Questions from Quadratic Equation
If the equations x 2+ ax + b = 0 and x 2+ bx + a = 0 have a common root, then find the value of a + b (where a is not equal to b)
- Roots of the following equation are 6x 2+ 4x - 2 = 0
Solve : (x + 2y) (2x – y)
A. 2x 2+ 5xy – 2y 2
B. 2x 2+ 3xy – 2y 2
C. x 2+ 4xy + y 2
D. x 2+ 4xy – y 2
Find the factors of (x 2– x – 132)?
Which of the following is NOT a quadratic equation?
