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Question 2

Number 2

The simplified form of

\frac{3\sqrt{3} + \sqrt{7}}{\sqrt{7} - 2\sqrt{3}} = ....

$3 - \sqrt{21}$

$3 + \sqrt{21}$

$5 - \sqrt{21}$

$-5 + \sqrt{21}$

$-5 - \sqrt{21}$

Explanation

To simplify $\frac{3\sqrt{3} + \sqrt{7}}{\sqrt{7} - 2\sqrt{3}}$ , we rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is $\sqrt{7} + 2\sqrt{3}$

\frac{3\sqrt{3} + \sqrt{7}}{\sqrt{7} - 2\sqrt{3}} = \frac{3\sqrt{3} + \sqrt{7}}{\sqrt{7} - 2\sqrt{3}} \times \frac{\sqrt{7} + 2\sqrt{3}}{\sqrt{7} + 2\sqrt{3}}

= \frac{(3\sqrt{3} + \sqrt{7})(\sqrt{7} + 2\sqrt{3})}{(\sqrt{7} - 2\sqrt{3})(\sqrt{7} + 2\sqrt{3})}

= \frac{3\sqrt{3} \cdot \sqrt{7} + 3\sqrt{3} \cdot 2\sqrt{3} + \sqrt{7} \cdot \sqrt{7} + \sqrt{7} \cdot 2\sqrt{3}}{(\sqrt{7})^2 - (2\sqrt{3})^2}

= \frac{3\sqrt{21} + 6 \cdot 3 + 7 + 2\sqrt{21}}{7 - 4 \cdot 3}

= \frac{3\sqrt{21} + 18 + 7 + 2\sqrt{21}}{7 - 12}

= \frac{25 + 5\sqrt{21}}{-5}

= \frac{25}{-5} + \frac{5\sqrt{21}}{-5}

= -5 - \sqrt{21}

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Question 2Set 1

Exercises

Number 2

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