((√(1−x))/( (√x))) + ((√x)/( (√(1−x)))) = ((13)/6)


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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
	$We need the condition for the equation is defined as 0<x<1(∗) ((√(1−x))/( (√x))) + ((√x)/( (√(1−x)))) = ((13)/6) (1) (1)⇔((1−x+x)/( (√(x(1−x)))))=((13)/6) ⇔6=13(√(x(1−x)))⇔36=169(x−x^2 ) ⇔169x^2 −169x+36=0 Δ=169^2 −4.169.36=169.25=65^2 ⇒x=((169±65)/(2.169))=((13±5)/(2.13))∈{(4/(13)),(9/(13))} both satisfy (∗) Thus,the given equation has two roots x∈{(4/(13)),(9/(13))}$
	$We need the condition for the equation$ $is defined as 0 < x < 1 ()$ $\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1 - x}} = \frac{13}{6} (1)$ $(1) \Leftrightarrow \frac{1 - x + x}{\sqrt{x (1 - x)}} = \frac{13}{6}$ $\Leftrightarrow 6 = 13 \sqrt{x (1 - x)} \Leftrightarrow 36 = 169 (x - x^{2})$ $\Leftrightarrow {169 x}^{2} - 169 x + 36 = 0$ $Δ = 169^{2} - 4.169.36 = 169.25 = 65^{2}$ $\Rightarrow x = \frac{169 \pm 65}{2.169} = \frac{13 \pm 5}{2.13} \in {\frac{4}{13}, \frac{9}{13}}$ $both satisfy ()$ $Thus, the given equation has two roots$ $x \in {\frac{4}{13}, \frac{9}{13}}$
	Commented by bemath last updated on 09/Oct/20

	$gave kudos$
	Answered by TANMAY PANACEA last updated on 09/Oct/20

	$a + \frac{1}{a} = \frac{13}{6} \to 6 a^{2} - 13 a + 6 = 0$ $6 a^{2} - 9 a - 4 a + 6 = 0$ $3 a (2 a - 3) - 2 (2 a - 3) = 0$ $(2 a - 3) (3 a - 2) = 0$ $a = \frac{3}{2}, \frac{2}{3}$ $a^{2} = \frac{1 - x}{x} = \frac{9}{4} a n d \frac{1 - x}{x} = \frac{4}{9}$ $9 x + 4 x = 4 \to x = \frac{4}{13} a n d 13 x = 9 x = \frac{9}{13}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum