2x-4x-2-1-3y-9y-2-1-1-4x-6y-3-Show-full-solution-

Question Number 180565 by depressiveshrek last updated on 13/Nov/22

(2 x + \sqrt{4 x^{2} + 1}) (3 y + \sqrt{9 y^{2} + 1}) = 1

{(4 x + 6 y)}^{3} = ?

S h o w f u l l s o l u t i o n

Commented by Frix last updated on 13/Nov/22

\frac{1}{2 x + \sqrt{4 x^{2} + 1}} = - 2 x + \sqrt{4 x^{2} + 1} = 3 y + \sqrt{9 y^{2} + 1}

\Rightarrow y = - \frac{2 x}{3}

\Rightarrow answer is 0

Answered by Rasheed.Sindhi last updated on 14/Nov/22

\underset{{(4 x + 6 y)}^{3} = ?}{(2 x + \sqrt{4 x^{2} + 1}) (3 y + \sqrt{9 y^{2} + 1}) = 1}

2 x + \sqrt{4 x^{2} + 1} - \frac{1}{3 y + \sqrt{9 y^{2} + 1}} = 0

2 x + \sqrt{4 x^{2} + 1} - \frac{1}{3 y + \sqrt{9 y^{2} + 1}} \cdot \frac{3 y - \sqrt{9 y^{2} + 1}}{3 y - \sqrt{9 y^{2} + 1}} = 0

2 x + \sqrt{4 x^{2} + 1} - \frac{3 y - \sqrt{9 y^{2} + 1}}{9 y^{2} - 9 y^{2} - 1} = 0

2 x + \sqrt{4 x^{2} + 1} + 3 y - \sqrt{9 y^{2} + 1} = 0

2 x + 3 y + \sqrt{4 x^{2} + 1} - \sqrt{9 y^{2} + 1} = 0

2 x + 3 y + \frac{4 x^{2} + 1 - 9 y^{2} - 1}{(\sqrt{4 x^{2} + 1} + \sqrt{9 y^{2} + 1})}

2 x + 3 y - \frac{(2 x + 3 y) (2 x - 3 y)}{\sqrt{9 y^{2} + 1} + \sqrt{4 x^{2} + 1}} = 0

(2 x + 3 y) (1 - \frac{2 x - 3 y}{\sqrt{9 y^{2} + 1} + \sqrt{4 x^{2} + 1}}) = 0

2 x + 3 y = 0 \Rightarrow 4 x + 6 y = 0 \Rightarrow {(4 x + 6 y)}^{3} = 0

Answered by cortano2 last updated on 14/Nov/22

(2 x + \sqrt{4 x^{2} + 1}) (3 y + \sqrt{9 y^{2} + 1}) = 1

\Rightarrow 2 x + \sqrt{4 x^{2} + 1} = 1 \Rightarrow x = 0

\Rightarrow 3 y + \sqrt{9 y^{2} + 1} = 1 \Rightarrow y = 0

\Rightarrow {(4 x + 6 y)}^{3} = 0

Commented by MJS_new last updated on 14/Nov/22

{that}^{'} s not enough, {it}^{'} s just one case . what if

2 x + \sqrt{4 x^{2} + 1} \neq 1 ?

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

3 y + \sqrt{9 y^{2} + 1} = \frac{1}{a} \Rightarrow y = \frac{1 - a^{2}}{6 a}

\Rightarrow 4 x + 6 y = 0

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