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Question
question 13 of 25
how would the fraction \\(\frac{1}{3+\sqrt{2}}\\) be rewritten if its denominator is rationalized using difference of squares?
a. \\(\frac{3-\sqrt{2}}{5}\\)
b. \\(3 - \sqrt{2}\\)
c. \\(\frac{3-\sqrt{2}}{7}\\)
d. \\(-\frac{3+\sqrt{2}}{7}\\)
Step1: Identify conjugate of denominator
Conjugate of $3+\sqrt{2}$ is $3-\sqrt{2}$.
Step2: Multiply numerator and denominator by conjugate
$$\frac{1}{3+\sqrt{2}} \times \frac{3-\sqrt{2}}{3-\sqrt{2}} = \frac{3-\sqrt{2}}{(3)^2 - (\sqrt{2})^2}$$
Step3: Simplify denominator using difference of squares
$$(3)^2 - (\sqrt{2})^2 = 9 - 2 = 7$$
Step4: Write simplified fraction
$$\frac{3-\sqrt{2}}{7}$$
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C. $\frac{3-\sqrt{2}}{7}$