Question

find the value a and b 5+2 root3/ 7+4 root3= a+b root3

Answer 1

Step-by-step explanation:

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }$

First we rationalize the denominator and then solve

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } = \frac{(5 + 2 \sqrt{3} ) \times (7 - 4 \sqrt{3}) }{(7 + 4 \sqrt{3} ) \times (7 - 4 \sqrt{3}) } = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24 }{ {7}^{2} - {(4 \sqrt{3} )}^{2} } = \frac{11 - 6 \sqrt{3} }{1}$

Identity used is

${x}^{2} - {y}^{2} = (x + y)(x - y)$

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Answer 2

Question :–

Find the value a and b $\tt \dfrac {5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} = a+b \sqrt{3}$

Solution :–

L.H.S. = $\tt \dfrac {5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} = a+b \sqrt{3}$

$\implies \tt \dfrac {(5 + 2 \sqrt{3})(7 - 4 \sqrt{3})}{(7 - 4 \sqrt{3})7 + 4 \sqrt{3}} = a+b \sqrt{3}$

$\implies \tt \dfrac {35-20 \sqrt{3} + 14 \sqrt{3} - 24}{{7}^{2} + {(4 \sqrt{3})}^{2}}$

$\implies \tt \dfrac {11 - 6 \sqrt{3}}{49 - 48}$

$\implies \tt \dfrac {11 - 6 \sqrt{3}}{1}$

$\implies \tt {11 - 6 \sqrt{3}}$

$\implies \tt {11 - 6 \sqrt{3} = a + b \sqrt{3} }$

Compare both sides,

$\boxed {\tt {a = 11 \: and \: b = -6}}$