Question

Rationise the following denominators3- 2√2 / 3+2√2

Answer 1

$\sf\huge\blue{\underline{\underline{ Question : }}}$

Rationalise the following denominator 3- 2√2 / 3+2√2

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

$\tt\:\implies \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}$

• [Rationalise the numerator & denominator with 3 - 2√2 ]

$\tt\:\implies \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}\times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}}$

$\tt\:\implies \dfrac{(3 - 2\sqrt{2})^{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}$

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

$\tt\:\implies \dfrac{(3)^{2} + (2\sqrt{2})^{2} - 2(3)(2\sqrt{2})}{(3)^{2} - (2\sqrt{2})^{2}}$

$\tt\:\implies \dfrac{9 + 8 - 6\sqrt{2}}{9 - 8}$

$\tt\:\implies 17 - 6\sqrt{2}$

$\underline{\boxed{\rm{\purple{\therefore \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt {2}}=17 - 6\sqrt{2}.}}}}\:\orange{\bigstar}$

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Answer 2

$\sf \huge \bigstar \fbox \green{ \: \: answer \: \: : - }$

$\sf \huge{17 - 12 \sqrt{2} }$

$\sf \huge \bigstar \fbox \green{ \: \: solution \: \: : - }$

$\tt \huge \rightarrow \: \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} }$

Rationalising the denominator,

$\tt \huge \rightarrow \: \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

$\tt \huge \rightarrow \: \frac{ {(3 - 2 \sqrt{2} }^{2}) }{ {(3)}^{2} - {(2 \sqrt{2} })^{2} }$

$\tt \huge \rightarrow \: \frac{ {3}^{2} - 2(3)(2 \sqrt{2)} + {(2 \sqrt{2} )}^{2} }{9 - 8}$

$\tt \huge \rightarrow \: \frac{9 - 12 \sqrt{2} + 8 }{1}$

$\tt \huge \rightarrow \: 17 - 12 \sqrt{2}$

$\tt \huge \bigstar \fbox \green{ \: \: formula \: used \: \: : - }$

$\bf \huge \star \red{ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} }$

$\bf \huge \star \red{ \: {a}^{2} - {b}^{2} = (a + b)(a - b)}$