Question

Find the rational numbers a and b such that 2+5√7/2-5√7=a+√7b

Answer 1

Given

$\frac{2 + 5 \sqrt{7} }{2 - 5 \sqrt{7} } = a + \sqrt{7} b$
$taking \: \: LHS \\ \\ = \frac{2 + 5 \sqrt{7} }{2 - 5 \sqrt{7} } \\ \\ = \frac{2 + 5 \sqrt{7} }{2 - 5 \sqrt{7} } \times \frac{2 + 5 \sqrt{7} }{2 + 5 \sqrt{7} } \: \\ \\ = \frac{(2 + 5 \sqrt{7})^{2} }{ {2}^{2} - (5 \sqrt{7})^{2} } \\ \\ = \frac{4+ 175 + 20 \sqrt{7} }{4 - 175} \\ \\ = \frac{179+ 20 \sqrt{7} }{ - 171}$
there, on comparing it with a + √7 b we have

a= 179 and b= 20

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Answer 2

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$\frac{2 + 5 \sqrt{7} }{2 - 5 \sqrt{7} } \\ \\ on \: rationalising : \\ \\ = > \frac{2 + 5 \sqrt{7} }{2 - 5 \sqrt{7} } \times \frac{2 + 5 \sqrt{7} }{2 + 5 \sqrt{7} } \\ \\ = > \frac{4 + 175 + 20 \sqrt{7} }{4 - 175} = > \frac{179 + 20 \sqrt{7} }{ - 171}$
by comparing

we have:


a=179

and

b=20