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Question Number 114653 by bemath last updated on 20/Sep/20

solve 6x^4 −25x^3 +12x^2 −25x+6=0

$${solve}\:\mathrm{6}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{6}=\mathrm{0} \\ $$

Answered by bobhans last updated on 20/Sep/20

⇔6(x^4 +1)−25(x^3 +x)+12x^2 =0 assume x ≠ 0 6(x^2 +(1/x^2 ))−25(x+(1/x))+12=0 let x+(1/x) = b ⇔ 6(b^2 −2)−25b+12=0 6b^2 −25b = 0 ⇒ { ((b=0⇒x+(1/x)=0; x^2 +1=0; x=±i)),((b=((25)/6)⇒x+(1/x)=((25)/6);6x^2 −25x+6=0)) :} x = ((25 ± (√(625−144)))/(12)) = ((25 ± (√(481)))/(12))

$$\Leftrightarrow\mathrm{6}\left({x}^{\mathrm{4}} +\mathrm{1}\right)−\mathrm{25}\left({x}^{\mathrm{3}} +{x}\right)+\mathrm{12}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${assume}\:{x}\:\neq\:\mathrm{0} \\ $$$$\mathrm{6}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\mathrm{25}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{12}=\mathrm{0} \\ $$$${let}\:{x}+\frac{\mathrm{1}}{{x}}\:=\:{b} \\ $$$$\Leftrightarrow\:\mathrm{6}\left({b}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{25}{b}+\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{6}{b}^{\mathrm{2}} −\mathrm{25}{b}\:=\:\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{b}=\mathrm{0}\Rightarrow{x}+\frac{\mathrm{1}}{{x}}=\mathrm{0};\:{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0};\:{x}=\pm{i}}\\{{b}=\frac{\mathrm{25}}{\mathrm{6}}\Rightarrow{x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{25}}{\mathrm{6}};\mathrm{6}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{6}=\mathrm{0}}\end{cases} \\ $$$$\:{x}\:=\:\frac{\mathrm{25}\:\pm\:\sqrt{\mathrm{625}−\mathrm{144}}}{\mathrm{12}}\:=\:\frac{\mathrm{25}\:\pm\:\sqrt{\mathrm{481}}}{\mathrm{12}} \\ $$

Answered by MJS_new last updated on 20/Sep/20

x^4 −((25)/6)x^3 +2x^2 −((25)/6)x+1=0 (x^2 +1)(x^2 −((25)/6)x+1)=0 x_(1, 2) =((25±(√(481)))/(12)) x_(3, 4) =±i

$${x}^{\mathrm{4}} −\frac{\mathrm{25}}{\mathrm{6}}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{6}}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{6}}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1},\:\mathrm{2}} =\frac{\mathrm{25}\pm\sqrt{\mathrm{481}}}{\mathrm{12}} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =\pm\mathrm{i} \\ $$

Commented by bobhans last updated on 21/Sep/20

gave kudos sir MJS −new

$${gave}\:{kudos}\:{sir}\:{MJS}\:−{new} \\ $$

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