If (X - \(\rm \frac{1}{x}\)) = 6, and x > 0, find the value of (x2 - \(\rm \frac{1}{x^2}\)).
12\(\sqrt{10}\)
Understanding the Algebra Problem: Solving for x<sup>2</sup> - 1/x<sup>2</sup>
The question asks us to find the value of \(x^2 - \frac{1}{x^2}\), given that \(x - \frac{1}{x} = 6\) and \(x > 0\).
This is a common type of algebra problem that can be solved using algebraic identities. The expression we need to find, \(x^2 - \frac{1}{x^2}\), is a difference of squares. Recall the difference of squares identity:
\[a^2 - b^2 = (a - b)(a + b)\]Applying this identity to \(x^2 - \frac{1}{x^2}\), where \(a = x\) and \(b = \frac{1}{x}\), we get:
\[x^2 - \frac{1}{x^2} = \left(x - \frac{1}{x}\right)\left(x + \frac{1}{x}\right)\]We are already given the value of the first factor, \(x - \frac{1}{x} = 6\). So, to find \(x^2 - \frac{1}{x^2}\), we need to find the value of the second factor, \(x + \frac{1}{x}\).
Finding the Value of x + 1/x
We know the value of \(x - \frac{1}{x}\). There's a useful identity that relates the square of the sum and the square of the difference of two terms:
\[(a + b)^2 = (a - b)^2 + 4ab\]Using this identity with \(a = x\) and \(b = \frac{1}{x}\):
\[\left(x + \frac{1}{x}\right)^2 = \left(x - \frac{1}{x}\right)^2 + 4 \cdot x \cdot \frac{1}{x}\] \[\left(x + \frac{1}{x}\right)^2 = \left(x - \frac{1}{x}\right)^2 + 4 \cdot 1\] \[\left(x + \frac{1}{x}\right)^2 = \left(x - \frac{1}{x}\right)^2 + 4\]Now, substitute the given value \(x - \frac{1}{x} = 6\) into this equation:
\[\left(x + \frac{1}{x}\right)^2 = (6)^2 + 4\] \[\left(x + \frac{1}{x}\right)^2 = 36 + 4\] \[\left(x + \frac{1}{x}\right)^2 = 40\]To find \(x + \frac{1}{x}\), we take the square root of both sides:
\[x + \frac{1}{x} = \pm \sqrt{40}\] \[x + \frac{1}{x} = \pm \sqrt{4 \times 10}\] \[x + \frac{1}{x} = \pm 2\sqrt{10}\]The problem states that \(x > 0\). If \(x > 0\), then \(\frac{1}{x}\) is also positive. The sum of two positive numbers (\(x\) and \(\frac{1}{x}\)) must be positive. Therefore, we take the positive square root:
\[x + \frac{1}{x} = 2\sqrt{10}\]Calculating the Final Value
Now we have the values for both factors in the expression \(x^2 - \frac{1}{x^2} = \left(x - \frac{1}{x}\right)\left(x + \frac{1}{x}\right)\):
- \(x - \frac{1}{x} = 6\)
- \(x + \frac{1}{x} = 2\sqrt{10}\)
Substitute these values:
\[x^2 - \frac{1}{x^2} = (6) \times (2\sqrt{10})\] \[x^2 - \frac{1}{x^2} = 12\sqrt{10}\]Thus, the value of \(x^2 - \frac{1}{x^2}\) is \(12\sqrt{10}\).
Comparing with Options
Let's check the given options:
- \(12\sqrt{10}\)
- \(18\sqrt{10}\)
- \(24\sqrt{2}\)
- \(24\sqrt{10}\)
Our calculated value \(12\sqrt{10}\) matches Option 1.
| Given | To Find | Identity Used 1 | Identity Used 2 | Calculated \(x + \frac{1}{x}\) | Final Result |
|---|---|---|---|---|---|
| \(x - \frac{1}{x} = 6\), \(x > 0\) | \(x^2 - \frac{1}{x^2}\) | \(a^2 - b^2 = (a-b)(a+b)\) | \((a+b)^2 = (a-b)^2 + 4ab\) | \(2\sqrt{10}\) | \(12\sqrt{10}\) |
Algebraic Identities Revision Table
Here are some fundamental algebraic identities often useful in solving such problems:
| Identity Name | Formula |
|---|---|
| Difference of Squares | \(a^2 - b^2 = (a - b)(a + b)\) |
| Perfect Square (Sum) | \((a + b)^2 = a^2 + 2ab + b^2\) |
| Perfect Square (Difference) | \((a - b)^2 = a^2 - 2ab + b^2\) |
| Relationship between squares of sum and difference | \((a + b)^2 = (a - b)^2 + 4ab\) |
| Relationship between squares of sum and difference | \((a - b)^2 = (a + b)^2 - 4ab\) |
Additional Information on Solving Algebra Problems
Solving algebraic equations and expressions often involves recognizing patterns and applying appropriate identities or formulas. For problems involving expressions like \(x \pm \frac{1}{x}\) and \(x^2 \pm \frac{1}{x^2}\) or \(x^3 \pm \frac{1}{x^3}\), the key is usually to:
- Identify the target expression and try to factor it or relate it to the given expression.
- Use standard algebraic identities to manipulate the given expression or the target expression.
- Square or cube the given expression if needed to generate terms required for the target expression.
- Pay attention to conditions on variables, like \(x > 0\), as they can help determine the sign of square roots or other values.
- Practice recognizing the common forms and the identities associated with them.
These types of problems are common in competitive exams and help build a strong foundation in algebraic manipulation.
Similar Questions
What is the value of 64x3 + 38x2y + 20xy2 + y3, when x = 3 and y = - 4?
If a2 + b2 + c2 = ab + bc + ac, then the value of \(\rm \frac{11 a^4+13 b^4+17 c^4}{17 a^2 b^2+9 b^2 c^2+15 c^2 a^2}\) is ?
If \((x+\frac{1}{x})\) = 5, and x > 1, what is the value of \((x^8-\frac{1}{x^8} )\)?
If a + b + c = 6 and a2 + b2 + c2 = 14, then what is the value of (a - b)2 + (b - c)2 + (c - a)2 ?
If \(\rm (x+\frac{1}{x})=2\), then \(\rm x^7+\frac{1}{x^{117}}=\) ___________.
If x > 0 and \(\rm x^4 + \frac{1}{x^4} = 142\), what is the value of \(\rm x^7 + \frac{1}{x^7} \)?
If a + b + c = 7, ab + bc + ca = 11 and abc = −1, then a3 + b3 + c3 is equal to:
If (x + \(\rm \frac{1}{x}\)) = 3\(\sqrt2\), and x > 1, what is the value of (x8 - \(\frac{1}{x^8}\))?
If (a - b) = 1, then what is the value of (a3 - b3)?
If (a3 + b3 + c3 - 3abc) = 405, and (a - b)2 + (b - c)2 + (c -a)2 = 54, find the value of (a + b + c).
Important Questions from Algebra
The difference between the two positive numbers x and y where x > y, is 25% of x. If the value of y is 15, then the value of x is:
If p2 + q2 - r2 = 0, then the value of (p6 + q6 - r6) ÷ p2q2r2 is:
If √2 + √x = √3, then the value of x is equal to:
The sum of two numbers is 20 and their difference is 2.5. Ratio of these numbers will be:
If \(\rm \frac{\sqrt{19 - x \sqrt{12}}}{1} = \sqrt 4 - \sqrt 3\) then the value of x is equal to:
