Question

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Answer 1

Answer:

This is not full but it may be helpful to you Sorry for inconvience

Answer 2

$\underline{\red{\sf{given \: that : - }}}$

• $\sf \: p = \dfrac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} } \:$
• $\sf \: q = \dfrac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} }$

$\underline{\blue{\sf{to \: find : - }}}$

• $\sf \: prove \: that \: the \: \purple{\sqrt{q} - \sqrt{p} - 2 \sqrt{pq} = 0}$

$\underline{\green{\sf{solution : - }}}$

In this question, we have been asked to prove that $\sf \: \sqrt{q} - \sqrt{p} - 2 \sqrt{pq} = 0$we have to mind that method of contrcidiction is used to prove a statement

it is given that the $\sf \: q \: = \dfrac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} - \sqrt{5} } \: \: \: and \: \: \: p = \dfrac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} }$

To show that ,

$\sf \orange{\longrightarrow} \: \: \sqrt{q} - \sqrt{p} - 2 \sqrt{pq} = 0 \\ \\ \\ \boxed{ \sf \orange{\longrightarrow} \sqrt{q} - \sqrt{p} = 2 \sqrt{pq} }$

now , we squaring both sides

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

$\\ \\ \tt \orange{\longrightarrow}q = \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} - \sqrt{5} } \\ \\ \\ \tt \orange{\longrightarrow} \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} - \sqrt{5} } \times \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} + \sqrt{5} } \\ \\ \\ \tt \orange{\longrightarrow} \frac{10 + 5 + 2 \sqrt{50} }{10 - 5} \\ \\ \\ \tt \orange{\longrightarrow} \frac{15 + 2 \times 5 \sqrt{50} }{10 - 5} \\ \\ \\ \tt \orange{\longrightarrow} \frac{ \cancel{15} + 2 \times \cancel5 \sqrt{2} }{ \cancel5}$

$\qquad{ \tt \orange{\longrightarrow}3 + 2 \sqrt{2} } \qquad \qquad - - - \bigg \lgroup \: eq(1) \bigg \rgroup$

now we find

$\sf \: \sqrt{p} = \dfrac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} }$

$\orange{\tt\longrightarrow}\tt \: \sqrt{p} = \dfrac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} } \\ \\ \\ \orange{\tt\longrightarrow}\tt \frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} } \times \frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} - \sqrt{5} } \\ \\ \\ \orange{\tt\longrightarrow}\tt \frac{10 + 5 - 2 \sqrt{50} }{10 - 5} \\ \\ \\ \orange{\tt\longrightarrow}\tt \frac{15 - 2 \times 5 \sqrt{2} }{10 - 5} \\ \\ \\ \orange{\tt\longrightarrow}\tt \frac{ \cancel{15} - 2 \times \cancel5 \sqrt{2} }{ \cancel5}$

$\qquad\orange{\tt\longrightarrow}\tt3 - 2 \sqrt{2} \qquad \qquad - - - \bigg \lgroup \: eq(2) \bigg \rgroup$

From eq (1) and eq (2)

$\sf \: pq\tt \: \: = \: \: (3 + 2 \sqrt{2})(3 - 2 \sqrt{2} ) \\ \\ \\ \orange{\tt\longrightarrow}\tt9 - 8 \\ \\ \\ \orange{\tt\longrightarrow}\tt1 - - - \bigg \lgroup \: eq(3) \bigg \rgroup$

putting value in q & p in

$\qquad \qquad \qquad \qquad \qquad \: q + p = 2 \sqrt{pq} = 4pq$

we get ,

$\orange{\tt\longrightarrow}\tt3 + 2 \sqrt{2} + 3 - 2\sqrt{1} = 4 \times 1 \qquad \: from \: eq(1) \: (2) \: and \: (3)$

now solving

$\orange{\tt\longrightarrow}\tt(3 + 2 \cancel{\sqrt{2}} )+( 3 - 2 \cancel{\sqrt{2} } ) - 2 \sqrt{1} = 4 \times\sqrt{ 1 }\\ \\ \\\orange{\tt\longrightarrow}\tt3 + 2 + 3 - 2 - 2 \cancel {\sqrt{1} } = 4 \cancel{ \sqrt{1} } \\ \\ \\ \orange{\tt\longrightarrow}\tt5 + 1 - 2 = 4\\ \\ \\ \orange{\tt\longrightarrow}\tt6 - 2 = 4 \\ \\ \\ \orange{\tt\longrightarrow}\tt4 = 4 \\ \\ \\ \orange{\tt\longrightarrow}\tt4 - 4 = 0 \\ \\ \\\orange{\tt\longrightarrow}\tt0 = 0 \\ \\ \\ \tt as \: \sqrt{} 0 = 0$

$\boxed{ \large \underline{ \frak{hence \: \sqrt{q} - \sqrt{p} - 2 \sqrt{pq} = 0}}}$

$\qquad \qquad \qquad \qquad \qquad \: \huge \tt \red{hence \: proved}$