Question

p-√q=4+√2/4-√2 then p,q=?​

Answer 1

Given:

p-√q=$\frac{4+\sqrt{2} }{4-\sqrt{2}}$

To find:

the value of p and q

Solution:

RHS=$\frac{4+\sqrt{2} }{4-\sqrt{2}}$

=$\frac{4+\sqrt{2} }{4-\sqrt{2}}$×$\frac{4+\sqrt{2} }{4+\sqrt{2} }$

=$\frac{(4+\sqrt{2}) ^{2} }{(4+\sqrt{2} )(4-\sqrt{2} )}$

use (a+b)² and a²-b² formulas and solve

=$\frac{16+2+8\sqrt{2} }{4^{2} -(\sqrt{2} )^{2} }$

$=\frac{18+8\sqrt{2} }{16-2} \\=\frac{18+8\sqrt{2} }{14}$

take 2 common

=$\frac{9+4\sqrt{2} }{7}$

separate, and get

$\frac{9}{7} +\frac{4\sqrt{2} }{7}$

compare by LHS

Hence, p=$\frac{9}{7}$ and q=$\frac{4\sqrt{2} }{7}$