Delgado Powers
03/28/2023 · Middle School
Q. If \( x=\sqrt{3}+\sqrt{2} \), twen valu J \( \left(x+\frac{1}{x}\right)= \) ? \( \begin{array}{llll}\text { (a) } 2 & \text { (b) } 3 & \text { (c) } 2 \sqrt{2} & \text { (d) } 2 \sqrt{3}\end{array} \)
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Tutor-Verified Answer
Step-by-step Solution
To find the value of \( J = x + \frac{1}{x} \) where \( x = \sqrt{3} + \sqrt{2} \), we first need to calculate \( \frac{1}{x} \).
1. **Calculate \( \frac{1}{x} \)**:
\[
\frac{1}{x} = \frac{1}{\sqrt{3} + \sqrt{2}}
\]
To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{3} - \sqrt{2} \):
\[
\frac{1}{x} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}
\]
The denominator simplifies as follows:
\[
(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1
\]
Thus, we have:
\[
\frac{1}{x} = \sqrt{3} - \sqrt{2}
\]
2. **Calculate \( J = x + \frac{1}{x} \)**:
Now we can substitute \( x \) and \( \frac{1}{x} \) into the expression for \( J \):
\[
J = (\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2})
\]
Simplifying this gives:
\[
J = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3}
\]
Thus, the value of \( J \) is \( \boxed{2\sqrt{3}} \).
Quick Answer
The value of \( J \) is \( 2\sqrt{3} \).
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