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

1.	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» (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 = \(\frac{9}{13}.\)

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»

(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 = \(\frac{9}{13}.\)