Question

please solve this for me.​

Answer 1

Solution

$8 \sqrt{ \dfrac{x}{x + 3} } - \sqrt{ \dfrac{x + 3}{x} } = 2$

Squaring on both sides

$\implies \bigg(8 \sqrt{ \dfrac{x}{x + 3} } - \sqrt{ \dfrac{x + 3}{x} } \bigg)^{2} = {2}^{2}$

$\implies \bigg(8 \sqrt{ \dfrac{x}{x + 3} } \bigg)^{2} - \bigg(\sqrt{ \dfrac{x + 3}{x} } \bigg)^{2} + 2 \bigg(8\sqrt{ \dfrac{x}{x + 3} } \bigg) \bigg(\sqrt{ \dfrac{x + 3}{x} } \bigg) = 4$

$\implies 64 \bigg( \dfrac{x}{x + 3} \bigg)- \bigg( \dfrac{x + 3}{x} \bigg) + 16 \bigg(\sqrt{ \dfrac{x(x + 3)}{(x + 3)x} } \bigg) = 4$

$\implies \dfrac{64x}{x + 3}- \dfrac{x + 3}{x} + 16 (\sqrt{1 }) = 4$

$\implies \dfrac{64x}{x + 3}- \dfrac{x + 3}{x} + 16= 4$

$\implies 16 - 4 = \dfrac{x + 3}{x} - \dfrac{64x}{x + 3}$

$\implies \dfrac{x + 3}{x} - \dfrac{64x}{x + 3} = 12$

Taking LCM

$\implies \dfrac{(x + 3)^{2} - 64x ^{2} }{x(x + 3)} = 12$

$\implies \dfrac{ {x}^{2} + {3}^{2} + 2(x)(3)- 64x ^{2} }{ {x}^{2} + 3x } = 12$

[ Because (a + b)² = a² + b² + 2ab ]

$\implies {x}^{2} + 9 + 6x- 64x ^{2} = 12( {x}^{2} + 3x)$

$\implies - 63{x}^{2} + 6x + 9 = 12{x}^{2} + 36x$

$\implies - 75 {x}^{2} + 6x + 9 - 36x = 0$

$\implies - 75{x}^{2} - 30x + 9 = 0$

$\implies - 3(25{x}^{2} + 10x - 3) = 0$

$\implies 25{x}^{2} + 10x - 3 = 0$

Factorising the equation, we get

Splitting the middle term

$\implies 25{x}^{2} + 15x - 5x - 3 = 0$

$\implies 5x(x + 3) - 5(x + 3) = 0$

$\implies (x + 3)(5x - 5) = 0$

$\implies x + 3 = 0 \quad or \quad 5x - 5= 0$

$\implies x = - 3 \quad or \quad 5x = 5$

$\implies x = - 3 \quad or \quad x = 1$

[ Neglecting x = - 3 as it is not possible ]

$\implies \boxed{x = 1}$

Hence the value of x is 1.