Question

If x=√a+2b+√a-2b / √a+2b-√a-2b , then show that bx^2-ax+b=0

Answer 1

Answer in attachment. Hope it helped.

Answer 2

Answer :

Step-by-step explanation :

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.