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Question Number 68835 by MJS last updated on 16/Sep/19

$$\mathrm{reposting}\mathrm{this}\mathrm{because}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{yet}\mathrm{solved}$$$$\mathrm{correctly}.\mathrm{I}\mathrm{criticize}\mathrm{that}\mathrm{most}\mathrm{of}\mathrm{you}\mathrm{people}$$$$\mathrm{do}\mathrm{not}\mathrm{test}\mathrm{if}\mathrm{your}\mathrm{solutions}\mathrm{fit}\mathrm{the}\mathrm{given}$$$$\mathrm{equations}\mathrm{in}\mathrm{many}\mathrm{cases}$$$$$$$$x^2+1-\sqrt{x^3+x}=6x$$$$$$$$\mathrm{please}\mathrm{determine}$$$$\left(1\right)\mathrm{how}\mathrm{many}\mathrm{real}\mathrm{solutions}$$$$\left(2\right)\mathrm{how}\mathrm{many}\mathrm{complex}\mathrm{solutions}$$$$\mathrm{we}\mathrm{can}\mathrm{expect}$$$$\left(3\right)\mathrm{solve}\mathrm{it}$$$$\left(4\right)\mathrm{test}\mathrm{your}\mathrm{solutions}$$

Commented by Prithwish sen last updated on 16/Sep/19

$${\mathrm{x}}^2-6\mathrm{x}+1=\sqrt{{\mathrm{x}}^3+\mathrm{x}}$$$$\mathrm{Squaring}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}$$$${\mathbf{x}}^4-13{\mathbf{x}}^3+38{\mathbf{x}}^2-13\mathbf{x}+1=0$$$$\mathbf{and}\mathbf{this}\mathbf{is}\mathbf{the}\mathbf{reciprocal}\mathbf{equation}\mathbf{of}{1}^{\mathbf{st}}\mathbf{type}$$$$\therefore ({\mathrm{x}}^4+1)-13\mathrm{x}({\mathrm{x}}^2+1)+{38\mathrm{x}}^2=0$$$$\mathrm{Dividing}\mathrm{the}\mathrm{equation}\mathrm{by}{\mathrm{x}}^2\mathrm{we}\mathrm{get}$$$$({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2})-13(\mathrm{x}+\frac{1}{\mathrm{x}})+38=0$$$${(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-13(\mathrm{x}+\frac{1}{\mathrm{x}})+36=0$$$$\mathrm{putting}\mathrm{x}+\frac{1}{\mathrm{x}}=\mathbf{y}\mathrm{we}\mathrm{get}$$$${\mathrm{y}}^2-13\mathrm{y}+36=0$$$$(\mathrm{y}-4)(\mathrm{y}-9)=0\mathrm{and}\mathrm{i}.\mathrm{e}$$$${\mathrm{x}}^2-4\mathrm{x}+1=0\mathrm{and}{\mathrm{x}}^2-9\mathrm{x}+1=0$$$$\mathrm{Solving}\mathrm{we}\mathrm{get}$$$$\mathbf{x}=2\pm \sqrt{3},\frac{9\pm \sqrt{77}}{2}$$$$\mathbf{now}2-\sqrt{3}\mathbf{and}2+\sqrt{3}\mathbf{are}\mathbf{reciprocal}$$$$\mathbf{on}\mathbf{the}\mathbf{other}\mathbf{hand}\frac{9+\sqrt{77}}{2}\mathbf{and}\frac{9-\sqrt{77}}{2}\mathbf{are}\mathbf{reciprocal}.$$$$\mathbf{now}2+\sqrt{3}\backsim 3.73\mathbf{and}2-\sqrt{3}\backsim 0.268$$$$\frac{9-\sqrt{77}}{2}\backsim 0.11\mathbf{and}\frac{9+\sqrt{77}}{2}\backsim 8.89$$$$\mathbf{the}\mathbf{equation}\mathbf{satisfies}\mathbf{only}\mathbf{on}0.11\mathbf{and}8.89$$$$\mathbf{MJS}\mathbf{Sir},\mathbf{please}\mathbf{share}\mathbf{your}\mathbf{valuable}\mathbf{comments}.$$

Commented by MJS last updated on 16/Sep/19

$$\mathrm{good},\mathrm{thank}\mathrm{you}!\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{test},\mathrm{because}$$$$\mathrm{with}\mathrm{squaring}\mathrm{we}\mathrm{get}\mathrm{false}\mathrm{solutions}$$

Commented by Prithwish sen last updated on 16/Sep/19

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

Commented by Rasheed.Sindhi last updated on 16/Sep/19

$$\mathrm{Slightly}\mathrm{different}(?)\mathrm{way}$$$${\mathrm{x}}^2+1-\sqrt{{\mathrm{x}}^3+\mathrm{x}}=6\mathrm{x}$$$$\mathrm{x}(\mathrm{x}+\frac{1}{\mathrm{x}})-\mathrm{x}\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}}-6\mathrm{x}=0$$$$\mathrm{x}\{(\mathrm{x}+\frac{1}{\mathrm{x}})-\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}}-6\}=0$$$$\mathrm{x}=0\left(\mathrm{Not}\mathrm{valid}\right)\mid (\mathrm{x}+\frac{1}{\mathrm{x}})-\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}}-6=0$$$$\mathrm{Let}\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}}=\mathrm{y}$$$${\mathrm{y}}^2-\mathrm{y}-6=0$$$$(\mathrm{y}-3)(\mathrm{y}+2)=0$$$$\mathrm{y}=3\mid \mathrm{y}=-2$$$$\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}}=3\mid \sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}}=-2\left(\mathrm{false}\right)$$$$\mathrm{x}+\frac{1}{\mathrm{x}}=9$$$${\mathrm{x}}^2-9\mathrm{x}+1=0$$$$\mathrm{x}=\frac{9\pm \sqrt{81-4}}{2}=\frac{9\pm \sqrt{77}}{2}$$$$\mathrm{Substituting}\mathrm{in}\mathrm{original}\mathrm{equation}\mathrm{we}$$$$\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{these}\mathrm{both}\mathrm{are}\mathrm{valid}.$$

Answered by ajfour last updated on 16/Sep/19

$${(x^2+1-6x)}^2=x^3+x$$$$dividingby{x}^2$$$${(x+\frac{1}{x}-6)}^2=x+\frac{1}{x}$$$$ifweletx+\frac{1}{x}=t$$$$t^2-13t+36=0$$$$t=\frac{13}{2}\pm \sqrt{\frac{169}{4}-36}$$$$t=\frac{13}{2}\pm \frac{5}{2}$$$$\Rightarrow x+\frac{1}{x}=4,9$$$$x^2-4x+1=0\Rightarrow$$$$x=2\pm \sqrt{4-1}=2\pm \sqrt{3}>0$$$$andfrom{x}^2-9x+1=0$$$$x=\frac{9}{2}\pm \sqrt{\frac{81}{4}-1}$$$$=\frac{9}{2}\pm \frac{\sqrt{77}}{2}>0$$$$x^3-4x^2+x=0\Rightarrow x^3+x=4x^2$$$$or{x}^3-9x^2+x=0\Rightarrow x^3+x=9x^2$$$$testing$$$$x^2+1-6x=4\mid x\mid andasall$$$$foundrootsare>0$$$$\Rightarrow x^2-10x+1=0$$$$\Rightarrow x=5\pm \sqrt{25-1}=5\pm 2\sqrt{6}$$$$\Rightarrow x=2\pm \sqrt{3}cannotbetrue$$$$solutions.$$$$x^2+1-6x=3\mid x\mid givesforx>0$$$$x^2-9x+1=0(sameeq.)$$$$\Rightarrow sox=\frac{9\pm \sqrt{77}}{2}aretruesolutions.$$

Commented by MJS last updated on 16/Sep/19

$$\mathrm{good},\mathrm{thank}\mathrm{you}!$$

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