Question

If a=1/7+4√3&b=1/7-4√3 find a²+b²​

Answer 1

Solution:-

Given

$\rm \: a = \dfrac{1}{7 + 4 \sqrt{3} } \: and \: \: b \: = \dfrac{1}{7 - 4 \sqrt{3} }$

To find

$\rm \: a {}^{2} + {b}^{2}$

Now to the value of a and b

$\bigg( \dfrac{1}{7 + 4 \sqrt{3} } \bigg) {}^{2} + \bigg( \dfrac{1}{7 - 4 \sqrt{3} } \bigg) {}^{2}$

$\dfrac{1}{( 7 + 4 \sqrt{3} ) {}^{2} } + \dfrac{1}{(7 - 4 \sqrt{3} ) {}^{2} }$

Using this identity

$\rm(a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\rm( {a}^{} - {b}^{} ) {}^{2} = {a}^{2} + {b}^{2} - 2ab$

we get

$\rm \dfrac{1}{ {7}^{2} + (4 \sqrt{3} ) {}^{2} + 2 \times 7 \times 4 \sqrt{3} } + \dfrac{1}{ {7}^{2} + (4 \sqrt{3}) {}^{2} - 2 \times 7 \times 4 \sqrt{3} }$

$\dfrac{1}{49 + 48 + 56 \sqrt{3} } + \dfrac{1}{49 + 48 - 56 \sqrt{3} }$

$\dfrac{1}{97 + 56 \sqrt{3} } + \dfrac{1}{97 - 56 \sqrt{3} }$

Taking lcm

$\dfrac{97 - 56 \sqrt{3} + 97 + 56 \sqrt{3} }{(97 + 56 \sqrt{3} )(97 - 56 \sqrt{3} )}$

$\dfrac{97 - \cancel{56 \sqrt{3} }+ 97 + \cancel{56 \sqrt{3} } }{(97 {}^{2} - ( 56 \sqrt{3} ){}^{2} )}$

$\dfrac{194}{9409 - 9408}$

$\dfrac{194}{1}$

Answer:- 194