Question

$if \: \frac{ \sqrt{2a + 1} + \sqrt{2a - 1} }{ \sqrt{2a + 1} - \sqrt{2a - 1} } = x$


Prove that -:


${x}^{2} - 4ax + 1 = 0$



Answer it.......BOARDZ..!⚠​

Answer 1

$\frac{ \sqrt{2a + 1} + \sqrt{2a - 1} }{ \sqrt{2a + 1} - \sqrt{2a - 1}} \times \frac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} } \\ = \frac{{(\sqrt{2a + 1} + \sqrt{2a - 1})}^{2} }{2a + 1 - 2a + 1} \\ = \frac{2a + 1 + 2a - 1 + 2 \sqrt{4 {a}^{2} - 1} }{2} \\ = \frac{4a + 2 \sqrt{4 {a}^{2} - 1 } }{2} = 2a + \sqrt{4 {a}^{2} - 1 }$

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Answer 2

We have give that, $\sf\frac{ \sqrt{2a + 1} + \sqrt{2a - 1} }{ \sqrt{2a + 1} - \sqrt{2a - 1} }\: =\: x$

We have to use the above equations and prove that x² - 4ax + 1 = 0

Now,

Rationalize the denominator

$\Rightarrow\:\sf\frac{ \sqrt{2a + 1} + \sqrt{2a - 1} }{ \sqrt{2a + 1} - \sqrt{2a - 1} } \:\times\:\frac{ \sqrt{2a + 1} + \sqrt{2a +1} }{ \sqrt{2a + 1} + \sqrt{2a + 1} }\:=\:x$

$\Rightarrow\:\sf\dfrac{( \sqrt{2a + 1} + \sqrt{2a - 1} )^{2} }{ {2a + 1} -{2a + 1} }\:=\:x$

Used identities:

(a + b) (a - b) = a² - b²

(a + b) (a + b) = (a + b)²

$\Rightarrow\:\sf\dfrac{2a+1 + 2a-1+2\sqrt{(2a+1)(2a-1)}}{ 2 }\:=\:x$

+1 and -1 cancel out, we left with

$\Rightarrow\:\sf\dfrac{4a +2\sqrt{(2a+1)(2a-1)}}{ 2 }\:=\:x$

Take 2 as common

$\Rightarrow\:\sf\dfrac{2[2a+\sqrt{(2a+1)(2a-1)}]}{ 2 }\:=\:x$

$\Rightarrow\:\sf{2a+\sqrt{(2a+1)(2a-1)}\:=\:x}$

$\Rightarrow\:\sf{\sqrt{(2a+1)(2a-1)}\:=\:x-2a}$

Again, (2a+1)(2a-1) = 4a² - 1

Here, used identity is (a+b)(a-b) = a² - b²

$\Rightarrow\:\sf{\sqrt{(4a^2-1)}\:=\:x-2a}$

Squaring on both sides

$\Rightarrow\:\sf{\sqrt{(4a^2-1)}^2\:=\:(x-2a)^2}$

Used identity:

(a - b)² = a² + b² - 2ab

$\Rightarrow\:\sf{4a^2-1\:=\:x^2+4a^2-4ax}$

Both 4a² cancel out, as they are opposite in signs

$\Rightarrow\:\sf{-1\:=\:x^2-4ax}$

$\Rightarrow\:\sf{x^2-4ax+1\:=\:0}$

Hence, proved