What is one of the values of x in the equation √(x/1-x)+√ (1-x/x)=13/6

Question

What is one of the values of x in the equation \(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}\) = \(\frac{13}{6}\)

(a) \(\frac{5}{13}\) 

(b) \(\frac{7}{13}\)

(c) \(\frac{9}{13}\)

(d) \(\frac{11}{3}\)

Answer 1

(c) \(\frac{9}{13}.\)

Let \(\sqrt{\frac{x}{1-x}}\) = y. Then, the given equation reduces to

y + \(\frac{1}{y}=\frac{13}{6}\) ⇒ 6 (y² + 1) = 13 y 

⇒ 6y² – 13y + 6 = 0 ⇒ 6y² – 9y – 4y + 6 = 0 

⇒ 3y (2y – 3) – 2(2y – 3) = 0 

⇒ (3y – 2) (2y – 3) = 0 ⇒ y = \(\frac{2}{3}\) and \(\frac{3}{2}\)

when y = \(\frac{2}{3}\), \(\sqrt{\frac{x}{1-x}}\) = \(\frac{2}{3}\) ⇒ \(\frac{x}{1-x}\) = \(\frac{4}{9}\)

⇒ 9x = 4 – 4x ⇒ 13x = 4 ⇒ x = \(\frac{4}{13}\)

when y = \(\frac{3}{2}\), \(\sqrt{\frac{x}{1-x}}\) = \(\frac{3}{2}\) ⇒ \(\frac{x}{1-x}\) = \(\frac{9}{4}\)

⇒ 4x = 9 – 9x ⇒ 13x = 9 ⇒ x = \(\frac{9}{13}.\)