1-x-x-x-1-x-13-6-

Question Number 117026 by bemath last updated on 08/Oct/20

\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1 - x}} = \frac{13}{6}

Answered by 1549442205PVT last updated on 09/Oct/20

We need the condition for the equation

is defined as 0 < x < 1 (*)

\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1 - x}} = \frac{13}{6} (1)

(1) \Leftrightarrow \frac{1 - x + x}{\sqrt{x (1 - x)}} = \frac{13}{6}

\Leftrightarrow 6 = 13 \sqrt{x (1 - x)} \Leftrightarrow 36 = 169 (x - x^{2})

\Leftrightarrow {169 x}^{2} - 169 x + 36 = 0

Δ = 169^{2} - 4.169.36 = 169.25 = 65^{2}

\Rightarrow x = \frac{169 \pm 65}{2.169} = \frac{13 \pm 5}{2.13} \in {\frac{4}{13}, \frac{9}{13}}

both satisfy (*)

Thus, the given equation has two roots

x \in {\frac{4}{13}, \frac{9}{13}}

Commented by bemath last updated on 09/Oct/20

gave kudos

Answered by TANMAY PANACEA last updated on 09/Oct/20

a + \frac{1}{a} = \frac{13}{6} \to 6 a^{2} - 13 a + 6 = 0

6 a^{2} - 9 a - 4 a + 6 = 0

3 a (2 a - 3) - 2 (2 a - 3) = 0

(2 a - 3) (3 a - 2) = 0

a = \frac{3}{2}, \frac{2}{3}

a^{2} = \frac{1 - x}{x} = \frac{9}{4} a n d \frac{1 - x}{x} = \frac{4}{9}

9 x + 4 x = 4 \to x = \frac{4}{13} a n d 13 x = 9 x = \frac{9}{13}