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24x-3-26x-2-9x-1-0-solve-

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Question Number 31677 by gyugfeet last updated on 12/Mar/18
24x^3 −26x^2 +9x−1=0(solve)
$$ \\ $$$$\mathrm{24}{x}^{\mathrm{3}} −\mathrm{26}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{1}=\mathrm{0}\left({solve}\right) \\ $$
Answered by Joel578 last updated on 12/Mar/18
Trial and error x^3 − ((13)/(12))x^2 + (3/8)x − (1/(24)) = 0 x = (1/2) → (1/8) − ((13)/(12 . 4)) + (3/(8 . 2)) − (1/(24)) = 0 (True) x = −(1/2) → −(1/8) − ((13)/(12 . 4)) − (3/(8 . 2)) − (1/(24)) ≠ 0 x = (1/3) → (1/(27)) − ((13)/(12 . 9)) + (3/(8 . 3)) − (1/(24)) = 0 (True) x = −(1/3) → −(1/(27)) − ((13)/(12 . 9)) − (3/(8 . 3)) − (1/(24)) ≠ 0 x = (1/4) → (1/(64)) − ((13)/(12 . 16)) + (3/(8 . 4)) − (1/(24)) = 0 (True) 24x^3 −26x^2 +9x−1 = (2x − 1)(3x − 1)(4x −1) = 0 x = (1/2) ∨ x = (1/3) ∨ x = (1/4)
$$\mathrm{Trial}\:\mathrm{and}\:\mathrm{error} \\ $$$${x}^{\mathrm{3}} \:−\:\frac{\mathrm{13}}{\mathrm{12}}{x}^{\mathrm{2}} \:+\:\frac{\mathrm{3}}{\mathrm{8}}{x}\:−\:\frac{\mathrm{1}}{\mathrm{24}}\:=\:\mathrm{0} \\ $$$$ \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\rightarrow\:\:\frac{\mathrm{1}}{\mathrm{8}}\:−\:\frac{\mathrm{13}}{\mathrm{12}\:.\:\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{8}\:.\:\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{24}}\:=\:\mathrm{0}\:\:\left({True}\right) \\ $$$${x}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\rightarrow\:−\frac{\mathrm{1}}{\mathrm{8}}\:−\:\frac{\mathrm{13}}{\mathrm{12}\:.\:\mathrm{4}}\:−\:\frac{\mathrm{3}}{\mathrm{8}\:.\:\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{24}}\:\neq\:\mathrm{0}\:\:\: \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\rightarrow\:\frac{\mathrm{1}}{\mathrm{27}}\:−\:\frac{\mathrm{13}}{\mathrm{12}\:.\:\mathrm{9}}\:+\:\frac{\mathrm{3}}{\mathrm{8}\:.\:\mathrm{3}}\:−\:\frac{\mathrm{1}}{\mathrm{24}}\:=\:\mathrm{0}\:\:\left({True}\right) \\ $$$${x}\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\rightarrow\:−\frac{\mathrm{1}}{\mathrm{27}}\:−\:\frac{\mathrm{13}}{\mathrm{12}\:.\:\mathrm{9}}\:−\:\frac{\mathrm{3}}{\mathrm{8}\:.\:\mathrm{3}}\:−\:\frac{\mathrm{1}}{\mathrm{24}}\:\neq\:\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\rightarrow\:\frac{\mathrm{1}}{\mathrm{64}}\:−\:\frac{\mathrm{13}}{\mathrm{12}\:.\:\mathrm{16}}\:+\:\frac{\mathrm{3}}{\mathrm{8}\:.\:\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{24}}\:=\:\mathrm{0}\:\:\left({True}\right) \\ $$$$ \\ $$$$\mathrm{24}{x}^{\mathrm{3}} −\mathrm{26}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{1}\:=\:\left(\mathrm{2}{x}\:−\:\mathrm{1}\right)\left(\mathrm{3}{x}\:−\:\mathrm{1}\right)\left(\mathrm{4}{x}\:−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\vee\:\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\vee\:\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by MJS last updated on 12/Mar/18
you could solve the quadratic equation after you found one root (x^3 −((13)/(12))x^2 +(3/8)x−(1/(24)))/(x−(1/2))= =x^2 −(7/(12))x+(1/(12)) x=(7/(24))±(√(((49)/(576))−(1/(12))))=(7/(24))±(1/(24)) x_2 =(1/4); x_3 =(1/3)
$$\mathrm{you}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{quadratic} \\ $$$$\mathrm{equation}\:\mathrm{after}\:\mathrm{you}\:\mathrm{found}\:\mathrm{one} \\ $$$$\mathrm{root} \\ $$$$\left({x}^{\mathrm{3}} −\frac{\mathrm{13}}{\mathrm{12}}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{8}}{x}−\frac{\mathrm{1}}{\mathrm{24}}\right)/\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$={x}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{12}}{x}+\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${x}=\frac{\mathrm{7}}{\mathrm{24}}\pm\sqrt{\frac{\mathrm{49}}{\mathrm{576}}−\frac{\mathrm{1}}{\mathrm{12}}}=\frac{\mathrm{7}}{\mathrm{24}}\pm\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}};\:{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by Joel578 last updated on 12/Mar/18
thanks for suggestion
$${thanks}\:{for}\:{suggestion} \\ $$

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