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

$${\left(\frac{1+x}{\sqrt{x}}\right)}^2+2a\left(\frac{1+x}{\sqrt{x}}\right)+1=0$$$$solveforx.$$

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

$$k^2+2ak+a^2+1=a^2$$$${(k+a)}^2=a^2-1$$$$k=-a\pm \sqrt{a^2-1}$$$$\frac{1+x}{\sqrt{x}}=-a\pm \sqrt{a^2-1}$$$$1+x=\frac{\sqrt{x}}{b}[\frac{1}{b}=-a\pm \sqrt{a^2-1}]$$$$x=b^2{x}^2+2b^{2}x+b^2$$$$x^2\left(b^2\right)+x(2b^2-1)+b^2=0$$$$x=\frac{-(2b^2-1)\pm \sqrt{{(2b^2-1)}^2-4b^2\times b^2}}{2b^2}$$$$x=\frac{-(2b^2-1)\pm \sqrt{4b^4-4b^2+1-4b^4}}{2b^2}$$$$x=\frac{-(2b^2-1)\pm \sqrt{1-4b^2}}{2b^2}$$$$plssubdtitudethevalueofb...$$

Answered by malwaan last updated on 22/Jan/19

$$\frac{1+x}{\sqrt{x}}=\frac{-2a\pm \sqrt{4a^2-4}}{2}$$$$=-a\pm \sqrt{a^2-1}$$$$\therefore \frac{1+2x+x^2}{x}=a^2\pm 2a\sqrt{a^2-1}+a^2-1$$$$\frac{1+x^2}{x}+2=2a^2\pm 2a\sqrt{a^2-1}-1$$$$1+x^2=(2a^2\pm 2a\sqrt{a^2-1}-3)x$$$$x^2-(2a^2\pm 2a\sqrt{a^2-1}-3)x+1=0$$$$x=\frac{2a^2\pm 2a\sqrt{a^2-1}-3\pm \sqrt{{(2a^2\pm 2a\sqrt{a^2-1}-3)}^2-4}}{2}$$$$\mathit{A}\mathit{m}\mathit{I}\mathit{r}\mathit{i}\mathit{g}\mathit{h}\mathit{t}?$$

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

$$letu=\frac{1+x}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}⩾2$$$$u^2+2au+1=0$$$${(u+a)}^2=a^2-1$$$$u=-a\pm \sqrt{a^2-1}⩾2$$$$\sqrt{a^2-1}⩾a+2$$$$\Rightarrow a⩽-\frac{5}{4}$$$$x-u\sqrt{x}+1=0$$$$\sqrt{x}=\frac{u\pm \sqrt{u^2-4}}{2}$$$$x=\frac{u^2-2\pm u\sqrt{u^2-4}}{2}$$$$\Rightarrow x=\frac{2a^2-3\pm 2a\sqrt{a^2-1}\pm (-a+\sqrt{a^2-1})\sqrt{2a^2-5\pm 2a\sqrt{a^2-1}}}{2}$$

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