Question Number 11661 by @ANTARES_VY last updated on 29/Mar/17
![8x^3 −6x+1=0. Solves... equation x_1 =? x_2 =?](https://www.tinkutara.com/question/Q11661.png)
$$\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{6}\boldsymbol{\mathrm{x}}+\mathrm{1}=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{Solves}}…\:\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{1}} =?\:\:\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} =? \\ $$
Answered by Joel576 last updated on 29/Mar/17
![8x^3 − 6x + 2 = 0 ⇒ (x + 1)(8x^2 − 8x + 2) = 0 ⇒ x + 1 = 0 or 8x^2 − 8x + 2 = 0 • x + 1 = 0 x = −1 • 2(4x^2 − 4x + 1) = 0 (2x − 1)^2 = 0 x = (1/2)](https://www.tinkutara.com/question/Q11665.png)
$$\mathrm{8}{x}^{\mathrm{3}} \:−\:\mathrm{6}{x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\left({x}\:+\:\mathrm{1}\right)\left(\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\:\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\:\:\:\:\mathrm{or}\:\:\:\:\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\:\mathrm{2}\:=\:\mathrm{0}\:\:\: \\ $$$$\bullet\:{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:\:{x}\:=\:−\mathrm{1} \\ $$$$\bullet\:\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{2}} \:−\:\mathrm{4}{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\left(\mathrm{2}{x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\:\:\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by @ANTARES_VY last updated on 29/Mar/17
![?????](https://www.tinkutara.com/question/Q11673.png)
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Commented by @ANTARES_VY last updated on 29/Mar/17
![?????](https://www.tinkutara.com/question/Q11674.png)
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Commented by @ANTARES_VY last updated on 29/Mar/17
![?????](https://www.tinkutara.com/question/Q11675.png)
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Commented by Joel576 last updated on 30/Mar/17
![Hmm, before that you wrote 8x^3 − 6x + 2 = 0](https://www.tinkutara.com/question/Q11697.png)
$$\mathrm{Hmm},\:\mathrm{before}\:\mathrm{that}\:\mathrm{you}\:\mathrm{wrote} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} \:−\:\mathrm{6}{x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$