(((√x) + (√(x − 4a)))/( (√x) − (√(x − 4a)))) = a ≠ 0 find “x” in terms of “a”.


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Question Number 183989 by HeferH last updated on 01/Jan/23

$\frac{\sqrt{x} + \sqrt{x - 4 a}}{\sqrt{x} - \sqrt{x - 4 a}} = a \neq 0$ $f i n d ‘ ‘ x^{″} i n t e r m s o f ‘ ‘ a^{″} .$
Commented by Frix last updated on 01/Jan/23

$I get x = {(a + 1)}^{2}$
	Answered by Rasheed.Sindhi last updated on 01/Jan/23

	$\frac{\sqrt{x} + \sqrt{x - 4 a}}{\sqrt{x} - \sqrt{x - 4 a}} = a \neq 0$ $f i n d ‘ ‘ x^{″} i n t e r m s o f ‘ ‘ a^{″} .$ $\frac{\sqrt{x} + \sqrt{x - 4 a}}{\sqrt{x} - \sqrt{x - 4 a}} + 1 = a + 1$ $\frac{2 \sqrt{x}}{\sqrt{x} - \sqrt{x - 4 a}} = a + 1$ $2 \sqrt{x} = a \sqrt{x} - a \sqrt{x - 4 a} + \sqrt{x} - \sqrt{x - 4 a}$ $a \sqrt{x} - \sqrt{x} = a \sqrt{x - 4 a} + \sqrt{x - 4 a}$ $\sqrt{x} (a - 1) = \sqrt{x - 4 a} (a + 1)$ $\frac{\sqrt{x - 4 a}}{\sqrt{x}} = \frac{a - 1}{a + 1}$ $\frac{x - 4 a}{x} = {(\frac{a - 1}{a + 1})}^{2}$ $1 - \frac{4 a}{x} = \frac{a^{2} - 2 a + 1}{a^{2} + 2 a + 1}$ $- \frac{4 a}{x} = \frac{a^{2} - 2 a + 1}{a^{2} + 2 a + 1} - 1 = \frac{- 4 a}{{(a + 1)}^{2}}$ $\frac{1}{x} = \frac{1}{{(a + 1)}^{2}} [∵ a \neq 0]$ $x = {(a + 1)}^{2}$
	Answered by a.lgnaoui last updated on 01/Jan/23

	$\frac{{(\sqrt{x} + \sqrt{x - 4 a})}^{2}}{4 a} = a$ $2 x - 4 a + 2 \sqrt{x (x - 4 a)} = 4 a^{2}$ $\sqrt{x (x - 4 a)} = 2 a^{2} + 2 a - x$ $x (x - 4 a) = 4 a^{4} + 4 a^{2} + x^{2} + 8 a^{3} - 4 a^{2} x - 4 a x$ $x^{2} - 4 a x = x^{2} - 4 a x (a + 1) + 4 a^{2} (a^{2} + 2 a + 1)$ $0 = 4 a^{2} {(a + 1)}^{2} - 4 a^{2} x$ $x = {(a + 1)}^{2}$
	Answered by manolex last updated on 01/Jan/23

	$m = \sqrt{x}$ $n = \sqrt{x - 4 a}$ $\frac{m + n}{m - n} = a$ $m^{2} = x$ $n^{2} = x - 4 a$ $m^{2} - n^{2} = 4 a$ $(m + n) (m - n) = 4 a$ $\frac{m + n}{m - n} = a m \neq n$ ${(m + n)}^{2} = 4 a^{2}$ $m + n = 2 a \lor m + n = - 2 a$ $m - n = 2 m - n = - 2$ $2 m = 2 a + 2 2 m = - 2 (a + 1)$ $m = a + 1 m = - (a + 1)$ $\sqrt{x} = a + 1 \sqrt{x} = - (a + 1)$ $x = {(a + 1)}^{2} x = {(a + 1)}^{2}$ $s o l u c i o n x = {(a + 1)}^{2}$ $c o m p r o b a c i o n$ $\frac{a + 1 + \sqrt{{(a + 1)}^{2} - 4 a}}{a + 1 - \sqrt{{(a + 1)}^{2} - 4 a}} = \frac{a + 1 + (a - 1)}{a + 1 - (a - 1)} = \frac{2 a}{2} = a c o r r e c t o$
	Answered by Rasheed.Sindhi last updated on 01/Jan/23

	$\frac{\sqrt{x} + \sqrt{x - 4 a}}{\sqrt{x} - \sqrt{x - 4 a}} = a \neq 0$ $f i n d ‘ ‘ x^{″} i n t e r m s o f ‘ ‘ a^{″} .$ $\sqrt{x} + \sqrt{x - 4 a} = a \sqrt{x} - a \sqrt{x - 4 a}$ $a \sqrt{x} - \sqrt{x} = \sqrt{x - 4 a} + a \sqrt{x - 4 a}$ $\sqrt{x} (a - 1) = \sqrt{x - 4 a} (a + 1)$ $\frac{\sqrt{x - 4 a}}{\sqrt{x}} = \frac{a - 1}{a + 1}$ $\frac{x - 4 a}{x} = \frac{a^{2} - 2 a + 1}{a^{2} + 2 a + 1}$ $\frac{x - 4 a}{x} - 1 = \frac{a^{2} - 2 a + 1}{a^{2} + 2 a + 1} - 1$ $\frac{- 4 a}{x} = \frac{- 4 a}{{(a + 1)}^{2}}$ $\frac{1}{x} = \frac{1}{{(a + 1)}^{2}} [∵ a \neq 0]$ $x = {(a + 1)}^{2}$
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