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Question Number 203941 by Panav last updated on 02/Feb/24

$$\mathit{I}\mathit{f}{\mathit{a}}^2-\mathit{a}+2=0\mathit{a}\mathit{n}\mathit{d}{\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{x}?$$$$\mathit{I}\mathit{I}\mathit{T}-\mathit{J}\mathit{E}\mathit{E}\mathit{b}\mathit{a}\mathit{s}\mathit{e}\mathit{d}\mathit{q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}.\mathit{F}\mathit{i}\mathit{n}\mathit{d}{\mathit{s}\mathit{o}\mathit{l}}^{\mathit{n}}.$$

Answered by Rasheed.Sindhi last updated on 02/Feb/24

$$\mathit{I}\mathit{f}{\mathit{a}}^2-\mathit{a}+2=0\mathit{a}\mathit{n}\mathit{d}{\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{x}?$$$${\mathit{a}}^2-\mathit{a}+2=0\Rightarrow a^2=a-2$$$$\Rightarrow a^3=a^2-2a=(a-2)-2a=-a-2$$$$\Rightarrow a^4=-a^2-2a=-(a-2)-2a=-3a+2$$$$\Rightarrow a^5=-3a^2+2a=-3(a-2)+2a=-a+6$$$$\Rightarrow a^6=-a^2+6a=-(a-2)+6a=5a+2$$$$$$$${\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2$$$$=(5a+2)+2(-3a+2)+(a-2)$$$$=4$$$$x=\pm 2$$

Commented by siyathokoza last updated on 06/Feb/24

$$\mathit{I}\mathit{f}{\mathit{a}}^2-\mathit{a}+2=0\mathit{a}\mathit{n}\mathit{d}{\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{x}?$$$${\mathit{a}}^2-\mathit{a}+2=0\Rightarrow a^2=a-2$$$$\Rightarrow a^3=a^2-2a=(a-2)-2a=-a-2$$$$\Rightarrow a^4=-a^2-2a=-(a-2)-2a=-3a+2$$$$\Rightarrow a^5=-3a^2+2a=-3(a-2)+2a=-a+6$$$$\Rightarrow a^6=-a^2+6a=-(a-2)+6a=5a+2$$$$$$$${\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2$$$$=(5a+2)+2(-3a+2)+(a-2)$$$$=4$$$$x=\pm 2$$$$\mathbf{BY}:\mathit{s}\mathit{i}\mathit{y}\mathit{t}\mathit{h}\mathit{o}\mathit{k}\mathit{o}\mathit{z}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{m}\mathit{a}...\left(\mathit{N}\mathit{S}\mathit{Q}\right)$$$$$$$$$$$$$$

Answered by Rasheed.Sindhi last updated on 02/Feb/24

$$\mathit{I}\mathit{f}{\mathit{a}}^2-\mathit{a}+2=0\mathit{a}\mathit{n}\mathit{d}{\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{x}?$$$${\mathit{a}}^2-\mathit{a}+2=0\Rightarrow a^2=a-2$$$$$$$$x^2=a^2(a^4+2a^2+1)$$$$=a^2{(a^2+1)}^2$$$$x=\pm a(a^2+1)$$$$=\pm a(a-2+1)$$$$=\pm a(a-1)$$$$=\pm (a^2-a)$$$$=\pm (\cancel{a}-2-\cancel{a})$$$$=\mp 2$$$$x=\pm 2$$

Commented by Panav last updated on 10/Feb/24

$$Correct$$

Answered by Rasheed.Sindhi last updated on 02/Feb/24

$$\mathit{I}\mathit{f}{\mathit{a}}^2-\mathit{a}+2=0\mathit{a}\mathit{n}\mathit{d}{\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{x}?$$$${\mathit{a}}^2-\mathit{a}+2=0\Rightarrow \{\begin{array}{l}2=a-a^2\\ a^2=a-2\end{array}$$$$$$$${\mathit{x}}^2={\mathit{a}}^6+2{\mathit{a}}^4+{\mathit{a}}^2=a^6+(a-a^2)a^4+a^2$$$$=\cancel{a^6}+a^5-\cancel{a^6}+a^2=a^2(a^3+1)$$$$=a^2(a(a-2)+1)$$$$=a^2(a^2-2a+1)$$$$x=\pm a(a-1)$$$$=\pm (a^2-a)$$$$=\pm (a-2-a)$$$$=\mp 2$$$$x=\pm 2$$

Answered by deleteduser1 last updated on 02/Feb/24

$$a^2=a-2\Rightarrow a^4=a^2-4a+4=a^2-a+2-3a+2=2-3a$$$$\Rightarrow x^2=a^6+4-6a+a-2=a^6-5a+2$$$$a^3=a^2-2a=a^2-a+2-a-2=-2-a$$$$\Rightarrow a^6=a^2+4a+4=a^2-a+2+5a+2=5a+2$$$$\Rightarrow x^2=5a+2-5a+2=4\Rightarrow x=\underset{-}{+}2$$

Answered by esmaeil last updated on 02/Feb/24

$$x^2={(a^3+a)}^2\to x=a(a^2+1)\stackrel{a^2-a+2}{=}a(a-1)=$$$$a=\frac{1\pm i\sqrt{7}}{2}\to a(a-1)=x$$$$\left(\frac{1+i\sqrt{7}}{2}\right)\left(\frac{i\sqrt{7}-1}{2}\right)=\frac{-7-1}{4}=-2=x$$$$\left(\frac{1-i\sqrt{7}}{2}\right)\left(\frac{-i\sqrt{7}-1}{2}\right)=\frac{8}{4}=2=x$$

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