Question

If p - √q=4-√2/4+√2, then) (p, q) =​

Answer 1

Answer:

${\large{\underline{\sf{Solution-}}}}$

${\longrightarrow{\sf{p - \sqrt{q} = \frac{4 - \sqrt{2} }{4 + \sqrt{2} } }}}$

By Rationalizing,

${\longrightarrow{\sf{ \frac{4 - \sqrt{2} }{4 + \sqrt{2} } \times \frac{4 - \sqrt{2} }{4 - \sqrt{2} } }}}$

${\longrightarrow{\sf{ \frac{ {(4 - \sqrt{2}) }^{2} }{ {(4)}^{2} - {( \sqrt{2} })^{2} } }}}$

${\longrightarrow{\sf{ \frac{ {(4)}^{2} + {( \sqrt{2}) }^{2} - 2(4)( \sqrt{2} )}{16 - 2} }}}$

${\longrightarrow{\sf{ \frac{16 + 2 - 8 \sqrt{2} }{14} }}}$

${\longrightarrow{\sf{ \frac{18 - 8 \sqrt{2} }{14} }}}$

${\longrightarrow{\sf{ \frac{2(9 - 4 \sqrt{2}) }{2(7)} }}}$

${\longrightarrow{\sf{ \frac{9 - 4 \sqrt{2} }{7} }}}$

${\longrightarrow{\sf{ \frac{9}{7} - \frac{4 \sqrt{2} }{7} }}}$

By Comparing With p - √q,

${\longrightarrow{\sf{ \frac{9 }{7} - \frac{4 \sqrt{2} }{7} = p - \sqrt{q} }}}$

We Get,

${\longrightarrow{\sf{p = \frac{9}{7} }}}$

${\longrightarrow{\sf{ \sqrt{q} = \frac{4 \sqrt{2} }{7} }}}$

Now doing S.O.B.S for q,

${\longrightarrow{\sf{ {( \sqrt{q} )}^{2} = { \bigg( \frac{4 \sqrt{2} }{7} \bigg) }^{2} }}}$

${\longrightarrow{\sf{q = \frac{32}{49} }}}$

${ \therefore{ \underline{ \pmb{ \sf{ p = \frac{9}{7} \: \: and \: \: q = \frac{32}{49} }}}}}$

pls mark aa Brainliest :)