Question

Solve using quadratic formula
$\sqrt{ \frac{x}{x - 3} } + \sqrt{ \frac{x - 3}{x} } = \frac{5}{2}$
solve the quadratic equation ​

Answer 1

Answer ⤴️

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Answer 2

ANSWER:

To Solve:

$\:\:\bullet\:\:\sqrt{\dfrac{x}{x-3}\,}+\sqrt{\dfrac{x-3}{x}\,}=\dfrac{5}{2}$

Solution:

METHOD 1:

We are given that,

$\implies\sqrt{\dfrac{x}{x-3}\,}+\sqrt{\dfrac{x-3}{x}\,}=\dfrac{5}{2}$

Squaring both sides,

$\implies\left(\sqrt{\dfrac{x}{x-3}\,}+\sqrt{\dfrac{x-3}{x}\,}\right)^2=\left(\dfrac{5}{2}\right)^2$

$\implies\left(\sqrt{\dfrac{x}{x-3}\,}\right)^2+\left(\sqrt{\dfrac{x-3}{x}\,}\right)^2+2\left(\sqrt{\dfrac{x}{x-3}\,}\right)\left(\sqrt{\dfrac{x-3}{x}\,}\right)=\dfrac{5^2}{2^2}$

$\implies\dfrac{x}{x-3}+\dfrac{x-3}{x}+2\left(\sqrt{\dfrac{x}{x-3}\times\dfrac{x-3}{x}\,}\right)=\dfrac{25}{4}$

$\implies\dfrac{x}{x-3}+\dfrac{x-3}{x}+2=\dfrac{25}{4}$

$\implies\dfrac{x^2+(x-3)^2}{x(x-3)}=\dfrac{25}{4}-2$

$\implies\dfrac{x^2+x^2+9-6x}{x^2-3x}=\dfrac{25-8}{4}$

$\implies\dfrac{2x^2-6x+9}{x^2-3x}=\dfrac{17}{4}$

$\implies4(2x^2-6x+9)=17(x^2-3x)$

$\implies8x^2-24x+36=17x^2-51x$

$\implies8x^2-24x+36-17x^2+51x=0$

$\implies-9x^2+27x+36=0$

$\implies-9(x^2-3x-4)=0$

$\implies x^2-3x-4=0$

$\implies x^2-4x+x-4=0$

$\implies x(x-4)+1(x-4)=0$

$\implies (x-4)(x+1)=0$

Hence,

$\implies x=4,\:-1$

METHOD 2:

We are given that,

$\implies\sqrt{\dfrac{x}{x-3}\,}+\sqrt{\dfrac{x-3}{x}\,}=\dfrac{5}{2}$

Let us assume that,

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

So,

$\implies\sqrt{\dfrac{x}{x-3}\,}+\sqrt{\dfrac{x-3}{x}\,}=\dfrac{5}{2}$

$\implies\sqrt{\dfrac{x-3}{x}\,}=\dfrac{1}{t}$

Hence,

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

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

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

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

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

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

$\implies 2t(t+2)+1(t+2)=0$

$\implies (2t+1)(t+2)=0$

So,

$\implies t=\dfrac{-1}{2},\:-2$

We had,

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

Case 1: Taking t = -1/2

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

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

Squaring both sides,

$\implies\dfrac{x}{x-3}=\dfrac{1}{4}$

$\implies4x=x-3$

$\implies4x-x=-3$

$\implies3x=-3$

$\implies x=-1$

Case 2: Taking t = -2

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

$\implies\sqrt{\dfrac{x}{x-3}\,}=-2$

Squaring both sides,

$\implies\dfrac{x}{x-3}=4$

$\implies x=4x-12$

$\implies12=4x-x$

$\implies3x=12$

$\implies x=4$

Hence,

$\implies x=4,\:-1$

Therefore from both methods,

$\implies\bf\large x=4,\:-1$