Question

rationalize the denominater​

Answer 1

Question :

$\mapsto \tt \dfrac{ \sqrt{3} + 1}{2 \sqrt{2} - 3}$

Solution :

$\quad\longmapsto \sf \dfrac{ \sqrt{3} + 1}{2 \sqrt{2} - 3} \\ \\ \quad\longmapsto \sf \frac{ \big( \sqrt{3} + 1 \big) \big(2 \sqrt{2} + 3 \big) }{ \big(2 \sqrt{2} -3 \big) \big(2 \sqrt{2} + 3 \big) } \\ \\ \quad\red{ \bigstar} \sf \: \: using \: identity \leadsto \\ \\ \quad \sf \overline{\underline{ \purple{(a - b)(a + b) = {a}^{2} - {b}^{2}}}}$

$\quad\longmapsto \sf \dfrac{ \big( \sqrt{3} + 1 \big) \big(2 \sqrt{2} + 3 \big) }{ \big(2 \sqrt{2} \big)^{2} - (3)^{2}} \\ \\ \quad \\ \longmapsto \sf \dfrac{ \big( \sqrt{3} + 1 \big) \big(2 \sqrt{2} + 3 \big) }{ \big(2 \sqrt{2} \big)(2 \sqrt{2}) - 9}$

$\quad\longmapsto \sf \dfrac{ \big( \sqrt{3} + 1 \big) \big(2 \sqrt{2} + 3 \big) }{ 8 - 9} \\ \\ \longmapsto \sf \dfrac{ \big( \sqrt{3} + 1 \big) \big(2 \sqrt{2} + 3 \big) }{ - 1} \\ \\ \longmapsto \sf \dfrac{\sqrt{3} \big(2 \sqrt{2} + 3 \big) + 1(2 \sqrt{2} + 3)}{ - 1}$

$\longmapsto \sf \dfrac{\sqrt{3} \big(2 \sqrt{2} + 3 \big) + 1(2 \sqrt{2} + 3)}{ - 1} \\ \\ \longmapsto \sf \dfrac{2 \sqrt{6} + 3 \sqrt{3} + 2 \sqrt{2} + 3}{ - 1} \\ \\ \quad \longmapsto \sf - \big(2 \sqrt{6} + 3 \sqrt{3} + 2 \sqrt{2} + 3 \big) \\ \\ \quad \longmapsto \sf - 2 \sqrt{6} - 3 \sqrt{3} - 2 \sqrt{2} - 3$

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Some useful identitites :-

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)

Signs are changed on the following bases :-

• (+) (+) = +
• (+) (-) = -
• (-) (-) = +
• (-) (+) = -