Question

Plz try solving it by simple radical equation​

Answer 1

QUESTION

$\dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} } - \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} } = 2 \sqrt{3}$

ANSWER:

$\sqrt{3} (x - a) - 2 \sqrt{ax}= 0$

GIVEN:

$\dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} } - \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} } = 2 \sqrt{3}$

TO SIMPLIFY:

$\dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} } - \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} } = 2 \sqrt{3}$

EXPLANATION:

BY CROSS MULTIPLICATION:

$\dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} } - \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} } = 2 \sqrt{3}$

Cross multiply the numerator and denominator

$\dfrac{ (\sqrt{x} + \sqrt{a}) ^{2} - (\sqrt{x} - \sqrt{a})^{2} }{ (\sqrt{x} - \sqrt{a} )(\sqrt{x} + \sqrt{a})} = 2 \sqrt{3}$

$(A + B)^2 = A^2 + B^2 + 2 AB$

$(A - B)^2 = A^2 + B^2 - 2 AB$

$(A + B) (A - B) = A^2 - B^2$

$(\sqrt{x} + \sqrt{a}) ^{2} = ( { \sqrt{x} )}^{2} + ( { \sqrt{a} )}^{2} + 2 \sqrt{x} \sqrt{a}$

$(\sqrt{x} + \sqrt{a}) ^{2} =x + a+ 2 \sqrt{ax}$

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

$(\sqrt{x} - \sqrt{a}) ^{2} =x + a - 2 \sqrt{ax}$

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

$(\sqrt{x} - \sqrt{a} )(\sqrt{x} + \sqrt{a}) = x - a$

Substituting these values, we will get

$\dfrac{ x + a + 2 \sqrt{ax} - x - a + 2 \sqrt{ax} }{x - a} = 2 \sqrt{3}$

$\dfrac{ 4 \sqrt{ax} }{x - a} = 2 \sqrt{3}$

$2 \sqrt{ax}= \sqrt{3} (x - a)$

$\sqrt{3} (x - a) - 2 \sqrt{ax}= 0$

BY RATIONALIZING METHOD:

$\dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} } - \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} } = 2 \sqrt{3}$

$Take \: \: \dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} }$

Multiply by √x + √a on both numerator and denominator.

$\dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} - \sqrt{a} } \times \dfrac{ \sqrt{x} + \sqrt{a} }{ \sqrt{x} + \sqrt{a} }$

$\dfrac{ (\sqrt{x} + \sqrt{a}) ^{2} }{ (\sqrt{x})^{2} - (\sqrt{a}) ^{2} }$

$\dfrac{ ( { \sqrt{x} )}^{2} + ( { \sqrt{a} )}^{2} + 2 \sqrt{x} \sqrt{a} }{ x - a }$

$\dfrac{x + a + 2 \sqrt{xa} }{ x - a }$

$Take \: \: - \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} }$

Multiply by √x - √a on both numerator and denominator.

$- \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} + \sqrt{a} } \times \dfrac{ \sqrt{x} - \sqrt{a} }{ \sqrt{x} - \sqrt{a} }$

$- \dfrac{ (\sqrt{x} - \sqrt{a}) ^{2} }{ (\sqrt{x})^{2} - (\sqrt{a}) ^{2} }$

$- \dfrac{ ( { \sqrt{x} )}^{2} + ( { \sqrt{a} )}^{2} - 2 \sqrt{x} \sqrt{a} }{ x - a }$

$\dfrac{ - x - a + 2 \sqrt{xa} }{ x - a }$

Add the two terms and equate to 2√3.

$\dfrac{ x + a + 2 \sqrt{ax} - x - a + 2 \sqrt{ax} }{x - a} = 2 \sqrt{3}$

$\dfrac{ 4 \sqrt{ax} }{x - a} = 2 \sqrt{3}$

$2 \sqrt{ax}= \sqrt{3} (x - a)$

$\sqrt{3} (x - a) - 2 \sqrt{ax}= 0$

Answer 2

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