[tex] \bf \underline{Given-} \\ [/tex]
[tex] \sf{x = \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } } \\ [/tex]
[tex] \bf \underline{To\: show-} \\ [/tex]
[tex] \sf{prove \: that : \: {qx}^{2} - px + q = 0 } \\ [/tex]
[tex] \bf \underline{Solution-} \\ [/tex]
[tex]\textsf{We have,}\\[/tex]
[tex] \sf{x = \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } } \\ [/tex]
[tex]\textsf{The denominator is : √(p+2q) - √(p-2q)}\\[/tex]
[tex]\textsf{We know that}\\[/tex]
[tex]\textsf{The rationalising factor of : √(a + b) - √(a-b) = √(a+b) + √(a-b).}\\[/tex]
[tex]\textsf{Therefore, the rationalising factor of: √(p+2q) - √(p-2q) = √(p+2q) + √(p-2q).}\\[/tex]
[tex]\textsf{On, rationalising the denominator,we get}\\[/tex]
[tex] \sf{x = \frac{ \sqrt{p + 2a} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } \times\frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} + \sqrt{p - 2q} } } \\ [/tex]
[tex] \sf{x = \frac{ (\sqrt{p + 2q} + \sqrt{p - 2q})( \sqrt{p + 2q} + \sqrt{p - 2q}) }{( \sqrt{p + 2q} - \sqrt{p - 2q})( \sqrt{p + 2q} + \sqrt{p - 2q} )} } \\ [/tex]
[tex] \sf{x = \frac{ (\sqrt{p + 2q} + \sqrt{p - 2q} {)}^{2} }{( \sqrt{p + 2q} - \sqrt{p - 2q})( \sqrt{p + 2q} + \sqrt{p - 2q} )} } \\ [/tex]
[tex]\textsf{★ Now, comparing the denominator with (a-b)(a+b), we get}\\[/tex]
[tex] \sf{ \: \: \: \: \: a = \sqrt{p + 2q} \: and \: b = \sqrt{p - 2q} } \\ [/tex]
[tex]\textsf{Using identity (a+b)(a-b) = a²-b², we get}\\[/tex]
[tex] \sf{x = \frac{ (\sqrt{p + 2q} + \sqrt{p - 2q} {)}^{2} }{( \sqrt{p + 2q} {)}^{2} - (\sqrt{p - 2q} {)}^{2} } } \\ [/tex]
[tex] \sf{x = \frac{ (\sqrt{p + 2q} + \sqrt{p - 2q} {)}^{2} }{p + 2q - (p + 2q) } } \\ [/tex]
[tex] \sf{x = \frac{ p + 2q + p - 2q + 2 \sqrt{ {p}^{2} - {4q}^{2} } }{p + 2q - (p + 2q) } } \\ [/tex]
[tex] \Rightarrow\sf{x = \frac{p + \sqrt{ {p}^{2} - {4q}^{2} } }{2q} } \\ [/tex]
[tex]\Rightarrow\sf{2qx =p + \sqrt{ {p}^{2} - {4q}^{2} } } \\ [/tex]
[tex]\Rightarrow\sf{2qx - p = \sqrt{ {p}^{2} - {4q}^{2} } } \\ [/tex]
[tex]\textsf{Squaring on both sides,we get}\\[/tex]
[tex] \sf{(2qx - p {)}^{2} = {p}^{2} - 4 {q}^{2} } \\ [/tex]
[tex]\Rightarrow\sf{ {4q}^{2} {x}^{2} + \cancel{{p}^{2}} - 4pqx - \cancel{ {p}^{2}} + {4q}^{2} =0} \\ [/tex]
[tex]\Rightarrow\sf{ {4q}^{2} {x}^{2} - 4pqx + {4q}^{2} =0 \: \: \Rightarrow\sf{4( {q}^{2} {x}^{2} - pqx + {4q}^{2} ) =0} } \\ [/tex]
[tex]\Rightarrow\sf{{q}^{2} {x}^{2} - pqx + {4q}^{2} =0 \: \: \: \: \: \Rightarrow\sf{q( {q}{x}^{2} - px + {q} ) =0} } \\ [/tex]
[tex]\Rightarrow\sf{{q}{x}^{2} - px + {q} =0} \\ [/tex]
[tex] \bf \underline{Hence, proved.} \\ [/tex]
