1-x-x-2-2a-1-x-x-1-0-solve-for-x-

Question Number 53491 by ajfour last updated on 22/Jan/19

{(\frac{1 + x}{\sqrt{x}})}^{2} + 2 a (\frac{1 + x}{\sqrt{x}}) + 1 = 0

s o l v e f o r x .

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

k^{2} + 2 a k + a^{2} + 1 = a^{2}

{(k + a)}^{2} = a^{2} - 1

k = - a \pm \sqrt{a^{2} - 1}

\frac{1 + x}{\sqrt{x}} = - a \pm \sqrt{a^{2} - 1}

1 + x = \frac{\sqrt{x}}{b} [\frac{1}{b} = - a \pm \sqrt{a^{2} - 1}]

x = b^{2} x^{2} + 2 b^{2} x + b^{2}

x^{2} (b^{2}) + x (2 b^{2} - 1) + b^{2} = 0

x = \frac{- (2 b^{2} - 1) \pm \sqrt{{(2 b^{2} - 1)}^{2} - 4 b^{2} \times b^{2}}}{2 b^{2}}

x = \frac{- (2 b^{2} - 1) \pm \sqrt{4 b^{4} - 4 b^{2} + 1 - 4 b^{4}}}{2 b^{2}}

x = \frac{- (2 b^{2} - 1) \pm \sqrt{1 - 4 b^{2}}}{2 b^{2}}

p l s s u b d t i t u d e t h e v a l u e o f b \dots

Answered by malwaan last updated on 22/Jan/19

\frac{1 + x}{\sqrt{x}} = \frac{- 2 a \pm \sqrt{4 a^{2} - 4}}{2}

= - a \pm \sqrt{a^{2} - 1}

∴ \frac{1 + 2 x + x^{2}}{x} = a^{2} \pm 2 a \sqrt{a^{2} - 1} + a^{2} - 1

\frac{1 + x^{2}}{x} + 2 = 2 a^{2} \pm 2 a \sqrt{a^{2} - 1} - 1

1 + x^{2} = (2 a^{2} \pm 2 a \sqrt{a^{2} - 1} - 3) x

x^{2} - (2 a^{2} \pm 2 a \sqrt{a^{2} - 1} - 3) x + 1 = 0

x = \frac{2 a^{2} \pm 2 a \sqrt{a^{2} - 1} - 3 \pm \sqrt{{(2 a^{2} \pm 2 a \sqrt{a^{2} - 1} - 3)}^{2} - 4}}{2}

A m I r i g h t ?

Answered by mr W last updated on 22/Jan/19

l e t u = \frac{1 + x}{\sqrt{x}} = \frac{1}{\sqrt{x}} + \sqrt{x} ⩾ 2

u^{2} + 2 a u + 1 = 0

{(u + a)}^{2} = a^{2} - 1

u = - a \pm \sqrt{a^{2} - 1} ⩾ 2

\sqrt{a^{2} - 1} ⩾ a + 2

\Rightarrow a ⩽ - \frac{5}{4}

x - u \sqrt{x} + 1 = 0

\sqrt{x} = \frac{u \pm \sqrt{u^{2} - 4}}{2}

x = \frac{u^{2} - 2 \pm u \sqrt{u^{2} - 4}}{2}

\Rightarrow x = \frac{2 a^{2} - 3 \pm 2 a \sqrt{a^{2} - 1} \pm (- a + \sqrt{a^{2} - 1}) \sqrt{2 a^{2} - 5 \pm 2 a \sqrt{a^{2} - 1}}}{2}

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