Question

6 root x upon X + 4 - 2 root x + 4 upon X equals to 11

Answer 1

$Given \: 6 \sqrt{\frac{x}{x+4}} - 2 \sqrt{\frac{x+4}{x}} = 11$

$\implies 6 \sqrt{\frac{x}{x+4}} - 2 \sqrt{\frac{1}{\frac{x}{x+4}}} = 11$

$Let \: a = \sqrt{\frac{x}{x+4} }\: --(1)$

$\implies 6a - \frac{2}{a} = 11$

$\implies 6a^{2} - 2 = 11a$

$\implies 6a^{2} -11a-2 = 0$

/* Splitting the middle term, we get */

$\implies 6a^{2} -12a+ 1a -2 = 0$

$\implies 6a(a-2)+1(a-2) = 0$

$\implies (a-2)(6a+1) = 0$

$\implies a-2 = 0 \: Or \: 6a+1= 0$

$\implies a = 2 \: Or \: a = \frac{-1}{6}$

Case 1:

$a = 2$

$\implies a^{2} = 4$

$\implies \frac{x}{x+4} = 4$

$\implies x = 4(x+4)$

$\implies x = 4x+16$

$\implies x - 4x = 16$

$\implies - 3x = 16$

$\implies x = \frac{-16}{3}$

Case 2:

$a = \frac{-1}{6}$

$\implies a^{2} = \frac{1}{36}$

$\implies \frac{x}{x+4} = \frac{1}{36}$

$\implies 36x = x+4$

$\implies 36x - x = 4$

$\implies 35x = 4$

$\implies x = \frac{4 }{35}$

Therefore.,

$\green { x = \frac{-16}{3} \: Or \: x = \frac{4 }{35}}$

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