Question

If y = √a+2b +√a−2b/√a+2b− √a−2b , prove that by 2 – ay + b = 0 complete ex

Answer 1

Answer:

Given that ;

$x = \frac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} -\sqrt{a - 2b}}$

On rationalising it's denominator, we get ;

$\frac{\sqrt{a + 2b} + \sqrt{a - 2b}}{ \sqrt{a + 2b} -\sqrt{a - 2b}}\\ \\ \frac{\sqrt{a + 2b} + \sqrt{a - 2b}}{ \sqrt{a + 2b} -\sqrt{a - 2b}} * \frac{\sqrt{a + 2b} +\sqrt{a - 2b}} {\sqrt{a + 2b} + \sqrt{a - 2b} } \\\\\frac{(\sqrt{a + 2b} + \sqrt{a - 2b})^2 }{(\sqrt{a + 2b}) ^2-(\sqrt{a + 2b}) ^2}\\\\ \frac{(\sqrt{a + 2b}) ^2+ ( \sqrt{a -2b}) ^2+2( \sqrt{a + 2b} )( \sqrt{a -2b} )}{a +2b - a + 2b} \\\\ \frac{a + 2b + a - 2b + 2( \sqrt{a + 2b})( \sqrt{a -2b} )}{4b}$

$\frac{2a + 2 ( \sqrt{a + 2b})( \sqrt{a - 2b})}{4b}$

Now, taking 2 as common ;

$x = \frac{\cancel{2}( a + ( \sqrt{a + 2b}) ( \sqrt{a - 2b}) }{\cancel{4}b}\\\\ x = \frac{a+ ( \sqrt{a + 2b}) ( \sqrt{a - 2b})}{2b}$

On cross multiplying ;

$2bx = a+ ( \sqrt{a + 2b}) ( \sqrt{a - 2b})\\\\ 2bx - a = ( \sqrt{a + 2b}) ( \sqrt{a - 2b})$

On squaring both sides ;

$(2bx - a)^{2} =( \sqrt{a + 2b}) ( \sqrt{a -2b})^{2}$

⇒ 4b²x² + a² - 4bxa = ( a + 2b )( a - 2b )

⇒ 4b²x² + a² - 4bxa = a² - 4b²

⇒ 4b²x² + a² - 4bxa - a² + 4b² = 0

⇒ 4b²x² -  4xa + 4b² = 0

⇒ 4b ( bx² - xa + b ) = 0

⇒ bx² - ax + b = 0

                    Hence Proved.

Answer 2

$\bf \underline{Given-}$

$\sf{x = \frac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } }$

$\bf \underline{To\: find-}$

$\sf{prove \: that : \: {bx}^{2} - ax + b = 0 }$

$\bf \underline{Solution-}$

$\textsf{We have,}$

$\sf{x = \frac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } }$

$\textsf{The denominator is : √(a+2b) - √(a-2b)}$

$\textsf{We know that}$

$\textsf{The rationalising factor of : √(p + q) - √(p-q) = √(p+q) + √(p-q).}$

$\textsf{Therefore, the rationalising factor of: √(a+2b) - √(a-2b) = √(a+2b) + √(a-2b).}$

$\textsf{On, rationalising the denominator,we get}$

$\sf{x = \frac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } \times\frac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} + \sqrt{a - 2b} } }$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b})( \sqrt{a + 2b} + \sqrt{a - 2b}) }{( \sqrt{a + 2b} - \sqrt{a - 2b})( \sqrt{a + 2b} + \sqrt{a - 2b} )} }$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b} {)}^{2} }{( \sqrt{a + 2b} - \sqrt{a - 2b})( \sqrt{a + 2b} + \sqrt{a - 2b} )} }$

$\textsf{★ Now, comparing the denominator with (a-b)(a+b), we get}$

$\sf{ \: \: \: \: \: a = \sqrt{a + 2b} \: and \: b = \sqrt{a - 2b} }$

$\textsf{Using identity (a+b)(a-b) = a²-b², we get}$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b} {)}^{2} }{( \sqrt{a + 2b} {)}^{2} - (\sqrt{a - 2b} {)}^{2} } }$

$\sf{x = \frac{ (\sqrt{a + 2b} + \sqrt{a - 2b} {)}^{2} }{a + 2b - (a + 2b) } }$

$\sf{x = \frac{ a + 2b + a - 2b + 2 \sqrt{ {a}^{2} - {4b}^{2} } }{a + 2b - (a + 2b) } }$

$\Rightarrow\sf{x = \frac{a + \sqrt{ {a}^{2} - {4b}^{2} } }{2b} }$

$\Rightarrow\sf{2bx =a + \sqrt{ {a}^{2} - {4b}^{2} } }$

$\Rightarrow\sf{2bx - a = \sqrt{ {a}^{2} - {4b}^{2} } }$

$\textsf{Squaring on both sides,we get}$

$\sf{(2bx - a {)}^{2} = {a}^{2} - 4 {b}^{2} }$

$\Rightarrow\sf{ {4b}^{2} {x}^{2} + \cancel{{a}^{2}} - 4abx - \cancel{ {a}^{2}} + {4b}^{2} =0}$

$\Rightarrow\sf{ {4b}^{2} {x}^{2} - 4abx + {4b}^{2} =0 \: \: \Rightarrow\sf{4( {b}^{2} {x}^{2} - abx + {4b}^{2} ) =0} }$

$\Rightarrow\sf{{b}^{2} {x}^{2} - abx + {4b}^{2} =0 \: \: \: \: \: \Rightarrow\sf{b( {b}{x}^{2} - ax + {b} ) =0} }$

$\Rightarrow\sf{{b}{x}^{2} - ax + {b} =0}$

$\bf \underline{Hence, proved.}$