Question

3-2√2/3+2√2 solve this question​

Answer 1

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Answer 2

Given fraction is $\dfrac{3-2\sqrt2}{3+2\sqrt2}$. These types of questions can be easily solved, if we rationalise their denominator.

Here,

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

By Rationalisation ( multiply and dividing the fraction by the original denominator with opposite sign so that they satisfy the formula ( a - b )( a + b ) = a^2 - b^2.

$\implies \dfrac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } \times \dfrac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \implies \dfrac{(3 - 2 \sqrt{2})(3 - 2 \sqrt{2} )}{(3 + 2 \sqrt{2})(3 - 2 \sqrt{2} )}$

From the properties of expansion : -

• ( a + b )( a + b ) = ( a + b )^2 = a^2 + b^2 + 2ab
• ( a + b )( a - b ) = a^2 - b^2

Thus,

$\implies \dfrac{(3{}^{} - 2 \sqrt{2}) {}^{2} }{(3) {}^{2} - (2 \sqrt{2} ) {}^{2} } \\ \\ \implies \dfrac{9 + 8 - 12 \sqrt{2} }{9 - 8} \\ \\ \implies \dfrac{17 - 12 \sqrt{2} }{1}$

Hence the required value of ( 3 - 2√2 ) / ( 3 + 2√2 ) is 17 - 12√2.