Question

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Answer 1

let √(x/ x-3) = t

according to the question,

t + 1/t = 5/2

=> t^2 + 1 = (5/2)t

=> 2t^2 - 5t + 2 = 0

=> 2t^2 - 4t - t + 2 = 0

=> 2t (t - 2) - 1 (t - 2) = 0

=> (2t - 1) ( t - 2 ) = 0

so, (2t - 1) = 0. or, ( t - 2 ) = 0

=> t = 1/2. or, t = 2

=> √(x/ x-3) = 1/2 or, √(x/ x-3) = 2

=> (x/ x-3) = 1/4 or, (x/ x-3) = 4

=> 4x = x-3. or, 4x - 12 = x

=> 3x = -3. or, 3x = 12

=> x = (-1). or, x = 4

value of x is (-1) or, 4

Answer 2

Answer:

The value of x = 4, -1

Step-by-step explanation:

Given question :

$\sqrt{\dfrac{x}{x-3}} + \sqrt{\dfrac{x-3}{x}} = \dfrac{5}{2} \ ; \ \text{when x \neq 0, 3 }$

To Find :

The solutions of x (or) the values of x.

Solution :

Let us assume

$\dfrac{x}{x-3} = t^2$

So, the given equation, turns,

$t + \dfrac{1}{t} = \dfrac{5}{2}$

$: \implies 2 (t^2 + 1) = 5t$

$: \implies 2t^2 + 2 = 5t$

$: \implies 2t^2 - 5t + 2 = 0$

$: \implies 2t^2 - 4t -t + 2 = 0$

$: \implies 2t (t-2) -1 (t-2) = 0$

$: \implies (t-2)(2t-1) = 0$

$\mathbf {\therefore \ t = 2 \ (or) \ \dfrac{1}{2} }$

But we assumed,

$\sqrt{\dfrac{x}{x-3} } = t$

Thus,

$\sqrt{\dfrac{x}{x-3} } = \mathbf {2 \ (or) \ \dfrac{1}{2} }$

• Case (i) --->

$\sqrt{\dfrac{x}{x-3} } = 2$    

Squaring on both sides,

$: \implies x = 4(x-3)$

$: \implies x = 4x - 12$

$: \implies 3x = 12$

$: \implies \underline {\boxed {\mathbf {x = 4}}}$

• Case (ii) --->

$\sqrt{\dfrac{x}{x-3} } = \dfrac{1}{2}$

Squaring on both sides,

$: \implies 4x = x-3$

$: \implies 3x = -3$

$: \implies \underline {\boxed {\mathbf {x = -1}}}$

Hope it helps!!