If a - 1/a = 1, then a 2+ 1/a 2= ?
3
Solving Algebraic Expressions: Finding \(a^2 + \frac{1}{a^2}\)
The question asks us to find the value of the expression \(a^2 + \frac{1}{a^2}\), given the relationship \(a - \frac{1}{a} = 1\).
This type of problem is common in algebra and involves using algebraic identities to simplify or transform expressions. We are given a relationship involving the difference of a variable and its reciprocal and need to find the sum of the squares of the variable and its reciprocal.
Step-by-Step Solution to Find \(a^2 + \frac{1}{a^2}\)
We are given the equation:
\[ a - \frac{1}{a} = 1 \]
To find \(a^2 + \frac{1}{a^2}\), we can square both sides of the given equation. Squaring a binomial like \((x-y)\) gives \(x^2 - 2xy + y^2\).
Let \(x = a\) and \(y = \frac{1}{a}\). Then the expression \(a - \frac{1}{a}\) is in the form \((x-y)\).
Square both sides of the equation \(a - \frac{1}{a} = 1\):
\[ \left(a - \frac{1}{a}\right)^2 = (1)^2 \]
Apply the identity \((x-y)^2 = x^2 - 2xy + y^2\) on the left side:
\[ a^2 - 2 \times a \times \frac{1}{a} + \left(\frac{1}{a}\right)^2 = 1^2 \]
Simplify the middle term \(2 \times a \times \frac{1}{a}\). Since \(a \times \frac{1}{a} = 1\) (assuming \(a \neq 0\)), this term becomes \(2 \times 1 = 2\).
Also, \(\left(\frac{1}{a}\right)^2 = \frac{1^2}{a^2} = \frac{1}{a^2}\) and \(1^2 = 1\).
Substitute these simplified terms back into the equation:
\[ a^2 - 2 + \frac{1}{a^2} = 1 \]
Now, we want to isolate the term \(a^2 + \frac{1}{a^2}\). To do this, add 2 to both sides of the equation:
\[ a^2 + \frac{1}{a^2} = 1 + 2 \]
Calculate the sum on the right side:
\[ a^2 + \frac{1}{a^2} = 3 \]
So, the value of \(a^2 + \frac{1}{a^2}\) is 3.
Understanding the Options
Let's look at the given options:
- Option 1: 1
- Option 2: 3
- Option 3: 2
- Option 4: 4
Our calculated value for \(a^2 + \frac{1}{a^2}\) is 3, which corresponds to Option 2.
Common Algebraic Identities Used
The key identity used in this problem is the square of a difference:
- \((x-y)^2 = x^2 - 2xy + y^2\)
Another related identity is the square of a sum:
- \((x+y)^2 = x^2 + 2xy + y^2\)
These identities are fundamental in simplifying and manipulating algebraic expressions.
Revision Table: Key Concepts
| Concept | Description | Example/Application |
|---|---|---|
| Algebraic Identity | An equation that is true for all values of the variables involved. | \((x-y)^2 = x^2 - 2xy + y^2\) |
| Squaring an Equation | Applying the squaring operation to both sides of an equation maintains equality. Useful for transforming expressions, e.g., from \((a - \frac{1}{a})\) to terms involving \(a^2\) and \(\frac{1}{a^2}\). | If \(x = y\), then \(x^2 = y^2\). |
| Reciprocal | For a non-zero number \(a\), its reciprocal is \(\frac{1}{a}\). The product of a number and its reciprocal is 1. | \(a \times \frac{1}{a} = 1\) (for \(a \neq 0\)). |
Additional Information: Related Problems
Similar problems might ask you to find \(a^2 + \frac{1}{a^2}\) given \(a + \frac{1}{a} = k\). In that case, you would square \((a + \frac{1}{a})\):
\[ \left(a + \frac{1}{a}\right)^2 = k^2 \]
\[ a^2 + 2 \times a \times \frac{1}{a} + \left(\frac{1}{a}\right)^2 = k^2 \]
\[ a^2 + 2 + \frac{1}{a^2} = k^2 \]
\[ a^2 + \frac{1}{a^2} = k^2 - 2 \]
Also, you might be asked to find \(a - \frac{1}{a}\) given \(a^2 + \frac{1}{a^2}\), or even higher powers like \(a^3 + \frac{1}{a^3}\) or \(a^3 - \frac{1}{a^3}\), which involve other algebraic identities like \( (x^3+y^3) = (x+y)(x^2-xy+y^2) \) and \( (x^3-y^3) = (x-y)(x^2+xy+y^2) \). Understanding the relationship between \(a \pm \frac{1}{a}\) and \(a^2 + \frac{1}{a^2}\) is a key step in solving these more complex problems.
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