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Question Number 82644 by Power last updated on 23/Feb/20

Commented by Power last updated on 23/Feb/20

$$\mathrm{sorry}\mathrm{it}\mathrm{worked}$$

Commented by john santu last updated on 23/Feb/20

$$\sqrt{x}=t\Rightarrow t^4+12t-5=0$$$$letx+2\sqrt{x}=t^2+2t=k$$$$withHornermethodweget$$$$t^4+12t-5=(t^2+2t-k)(t^2-2t+k+4)+(4-4k)t+k^2+4k-5$$

Answered by MJS last updated on 23/Feb/20

$$x^2+12\sqrt{x}=5$$$$12\sqrt{x}=5-x^2$$$$\mathrm{squaring}\Rightarrow \mathrm{we}\mathrm{might}\mathrm{get}\mathrm{false}\mathrm{solutions}$$$$...$$$$x^4-10x^2-144x+25=0$$$$(x^2-6x+1)(x^2+6x+25)=0$$$$\Rightarrow x=3\pm 2\sqrt{2}\mathrm{but}x=3+2\sqrt{2}\mathrm{is}\mathrm{false}$$$$\Rightarrow x=3-2\sqrt{2}$$$$\Rightarrow \mathrm{answer}\mathrm{is}1$$

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