Question
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Rationalize the denominator using the conjugate: \( \frac{4}{4+\sqrt{2}} \)

Ask by Hamilton Osborne. in the United States
Mar 10,2025

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Tutor-Verified Answer

Answer

\[ \frac{8 - 2\sqrt{2}}{7} \]

Solution

We start with the fraction: \[ \frac{4}{4+\sqrt{2}} \] **Step 1: Multiply by the conjugate of the denominator.** The conjugate of \(4+\sqrt{2}\) is \(4-\sqrt{2}\). Multiply the numerator and denominator by \(4-\sqrt{2}\): \[ \frac{4}{4+\sqrt{2}} \times \frac{4-\sqrt{2}}{4-\sqrt{2}} = \frac{4(4-\sqrt{2})}{(4+\sqrt{2})(4-\sqrt{2})} \] **Step 2: Simplify the denominator using the difference of squares.** Recall that: \[ (a+b)(a-b) = a^2 - b^2 \] Let \(a = 4\) and \(b = \sqrt{2}\): \[ (4+\sqrt{2})(4-\sqrt{2}) = 4^2 - (\sqrt{2})^2 = 16 - 2 = 14 \] **Step 3: Simplify the numerator.** Multiply out the numerator: \[ 4(4-\sqrt{2}) = 16 - 4\sqrt{2} \] **Step 4: Write the rationalized form.** Thus, the expression becomes: \[ \frac{16-4\sqrt{2}}{14} \] **Step 5: Simplify by reducing common factors.** Both the numerator and denominator are divisible by 2: \[ \frac{16-4\sqrt{2}}{14} = \frac{2(8-2\sqrt{2})}{2 \cdot 7} = \frac{8-2\sqrt{2}}{7} \] The rationalized form of the given expression is: \[ \frac{8-2\sqrt{2}}{7} \]

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To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is \(4 - \sqrt{2}\). So we get: \[ \frac{4(4 - \sqrt{2})}{(4 + \sqrt{2})(4 - \sqrt{2})} \] Now, let's simplify the denominator: \[ (4 + \sqrt{2})(4 - \sqrt{2}) = 4^2 - (\sqrt{2})^2 = 16 - 2 = 14 \] And the numerator: \[ 4(4 - \sqrt{2}) = 16 - 4\sqrt{2} \] Putting it all together, we have: \[ \frac{16 - 4\sqrt{2}}{14} \] This can be simplified further: \[ \frac{16}{14} - \frac{4\sqrt{2}}{14} = \frac{8}{7} - \frac{2\sqrt{2}}{7} \] So the final result is: \[ \frac{8 - 2\sqrt{2}}{7} \]

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