Question

Rationalize :-
$\frac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} } + \frac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} }$
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Answer 1

ANSWER:

$\sf\dfrac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} } + \dfrac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} } = \dfrac{ 4x}{ \sqrt{4x^{2} - 25 } }$

GIVEN:

$\sf\dfrac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} } + \dfrac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} }$

TO RATIONALIZE:

$\sf\dfrac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} } + \dfrac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} }$

EXPLANATION:

$\sf Take \ A = \dfrac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} }$

$\sf Multiply\ and\ divide\ by\ \sqrt{2x - 5}$

$\sf \leadsto\dfrac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} } \\ \\ \\ \sf\leadsto\dfrac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} } \times \dfrac{ \sqrt{2x - 5} }{ \sqrt{2x - 5} } \\ \\ \\ \boxed{ \bold{\large{ \purple{(A - B)(A - B) = (A - B)^2}}}} \\ \\ \\\boxed{ \bold{\large{ \red{(A+B)(A-B) = A^2-B^2}}}}\\ \\ \\ \\ \sf\leadsto \dfrac{ (\sqrt{2x - 5})^{2} }{ \sqrt{(2x)^{2} - {5}^{2} } } \\ \\ \\ \sf\leadsto \dfrac{ 2x - 5 }{ \sqrt{4x^{2} - 25 } }$

$\sf Take\ B = \dfrac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} }$

$\sf Multiply\ and\ divide\ by\ \sqrt{2x + 5}$

$\sf \leadsto\dfrac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} } \\ \\ \\ \sf\leadsto\dfrac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} } \times \dfrac{ \sqrt{2x + 5} }{ \sqrt{2x + 5} } \\ \\ \\ \boxed{ \bold{\large{ \blue{(A + B)(A + B) = (A + B)^2}}}} \\ \\ \\\boxed{ \bold{\large{\orange{(A+B)(A-B) = A^2-B^2}}}} \\ \\ \\ \sf\leadsto \dfrac{ (\sqrt{2x + 5})^{2} }{ \sqrt{(2x)^{2} - {5}^{2} } } \\ \\ \\ \sf\leadsto \dfrac{ 2x + 5 }{ \sqrt{4x^{2} - 25 } }$

Add A and B

$\sf\leadsto \dfrac{ 2x - 5 }{ \sqrt{4x^{2} - 25 } } + \dfrac{ 2x + 5 }{ \sqrt{4x^{2} - 25 } }$

$\sf\leadsto \dfrac{ 2x - 5 + 2x + 5 }{ \sqrt{4x^{2} - 25 } }$

$\sf\leadsto \dfrac{ 4x}{ \sqrt{4x^{2} - 25 } }$

$\sf\dfrac{ \sqrt{2x - 5} }{ \sqrt{2x + 5} } + \dfrac{ \sqrt{2x + 5} }{ \sqrt{2x - 5} } = \dfrac{ 4x}{ \sqrt{4x^{2} - 25 } }$