If \(x\, = \,7\, + \,4\sqrt 3 \) , then what is the value of \(\sqrt x \, + \,\frac{1}{{\sqrt x }}\) ?
4
Let's find the value of the expression \( \sqrt x \, + \,\frac{1}{{\sqrt x }} \) when \( x = 7 + 4\sqrt 3 \). This problem involves simplifying radical expressions and rationalizing the denominator.
Simplifying the Expression for x
The given value of \( x \) is \( 7 + 4\sqrt 3 \). We need to find the square root of \( x \). Often, expressions like this can be written as the square of a binomial of the form \( (a+b)^2 \) or \( (a+\sqrt{b})^2 \).
Let's consider the form \( (a+b)^2 = a^2 + 2ab + b^2 \). We want to express \( 7 + 4\sqrt 3 \) in this form.
Comparing \( 7 + 4\sqrt 3 \) with \( a^2 + 2ab + b^2 \), we can see that the term \( 4\sqrt 3 \) must be related to \( 2ab \). We can write \( 4\sqrt 3 \) as \( 2 \times (2) \times (\sqrt 3) \). This suggests that \( a \) and \( b \) might be 2 and \( \sqrt 3 \).
Let's check if \( a=2 \) and \( b=\sqrt 3 \) satisfy the condition \( a^2 + b^2 = 7 \):
- \( a^2 = 2^2 = 4 \)
- \( b^2 = (\sqrt 3)^2 = 3 \)
- \( a^2 + b^2 = 4 + 3 = 7 \)
This matches the constant term in \( 7 + 4\sqrt 3 \). So, we can write \( 7 + 4\sqrt 3 \) as the square of \( (2 + \sqrt 3) \).
\( x = 7 + 4\sqrt 3 = (2)^2 + 2(2)(\sqrt 3) + (\sqrt 3)^2 = (2 + \sqrt 3)^2 \)
Calculating \(\sqrt x\)
Now we can find the square root of \( x \):
\( \sqrt x = \sqrt{(2 + \sqrt 3)^2} \)
The square root of a square is the absolute value, so \( \sqrt{(2 + \sqrt 3)^2} = |2 + \sqrt 3| \). Since \( 2 + \sqrt 3 \) is a positive number, \( |2 + \sqrt 3| = 2 + \sqrt 3 \).
So, \( \sqrt x = 2 + \sqrt 3 \).
Calculating \(\frac{1}{{\sqrt x }}\)
Next, we need to find the value of \( \frac{1}{{\sqrt x }} \). We substitute the value of \( \sqrt x \) we just found:
\( \frac{1}{{\sqrt x }} = \frac{1}{{2 + \sqrt 3 }} \)
To simplify this expression, we need to rationalize the denominator. We multiply the numerator and the denominator by the conjugate of \( 2 + \sqrt 3 \), which is \( 2 - \sqrt 3 \).
\( \frac{1}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = \frac{{1 \times (2 - \sqrt 3 )}}{{(2 + \sqrt 3 )(2 - \sqrt 3 )}} \)
In the denominator, we use the difference of squares formula, \( (a+b)(a-b) = a^2 - b^2 \). Here, \( a=2 \) and \( b=\sqrt 3 \).
\( (2 + \sqrt 3 )(2 - \sqrt 3 ) = 2^2 - (\sqrt 3)^2 = 4 - 3 = 1 \)
So the expression becomes:
\( \frac{1}{{\sqrt x }} = \frac{{2 - \sqrt 3 }}{1} = 2 - \sqrt 3 \)
Calculating \(\sqrt x \, + \,\frac{1}{{\sqrt x }}\)
Finally, we need to find the sum \( \sqrt x \, + \,\frac{1}{{\sqrt x }} \). We substitute the values we found for \( \sqrt x \) and \( \frac{1}{{\sqrt x }} \).
\( \sqrt x \, + \,\frac{1}{{\sqrt x }} = (2 + \sqrt 3) + (2 - \sqrt 3) \)
Combine like terms:
\( \sqrt x \, + \,\frac{1}{{\sqrt x }} = 2 + \sqrt 3 + 2 - \sqrt 3 \)
\( \sqrt x \, + \,\frac{1}{{\sqrt x }} = (2 + 2) + (\sqrt 3 - \sqrt 3) \)
\( \sqrt x \, + \,\frac{1}{{\sqrt x }} = 4 + 0 \)
\( \sqrt x \, + \,\frac{1}{{\sqrt x }} = 4 \)
The value of \( \sqrt x \, + \,\frac{1}{{\sqrt x }} \) is 4.
Comparing with Options
Let's compare our result with the given options:
- Option 1: 1
- Option 2: 2
- Option 3: 3
- Option 4: 4
Our calculated value is 4, which matches Option 4.
Revision Table: Key Steps for Solving Radical Expressions
This table summarizes the important techniques used in solving this problem.
| Step | Description | Technique Used |
|---|---|---|
| 1 | Simplify the expression under the square root. | Recognizing perfect squares (e.g., \((a+b)^2\)) |
| 2 | Calculate the square root. | \( \sqrt{y^2} = |y| \) |
| 3 | Calculate the reciprocal of the square root. | Finding \( \frac{1}{\sqrt{x}} \) |
| 4 | Rationalize the denominator if needed. | Multiplying by the conjugate (e.g., \( (a+b)(a-b)=a^2-b^2 \)) |
| 5 | Perform the required operation (addition/subtraction). | Combining like terms |
Additional Information: Working with Surds
Surds are expressions that include irrational roots. Here are some key concepts related to working with surds, as seen in this problem:
- Simplifying Surds: Try to express the number under the square root as a product of a perfect square and another number (e.g., \( \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \)). In this problem, we did the reverse by recognizing \( 7 + 4\sqrt 3 \) as a perfect square \( (2+\sqrt{3})^2 \).
- Rationalizing the Denominator: This is the process of removing a surd from the denominator of a fraction. If the denominator is of the form \( a + \sqrt b \) or \( a + c\sqrt d \), we multiply the numerator and denominator by its conjugate (e.g., \( a - \sqrt b \) or \( a - c\sqrt d \)). This uses the identity \( (x+y)(x-y) = x^2 - y^2 \), which eliminates the radical in the denominator.
- Adding/Subtracting Surds: You can only add or subtract 'like' surds, which are surds with the same number under the root sign (e.g., \( 2\sqrt 3 + 5\sqrt 3 = 7\sqrt 3 \), but \( 2\sqrt 3 + 5\sqrt 5 \) cannot be simplified further by combining the surd terms). In this problem, the \( \sqrt 3 \) terms cancelled out.
Understanding these techniques is crucial for solving problems involving surds and radicals effectively.
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Column II
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ii.
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c.
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d.
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