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2x-4-4x-3-22x-2-24x-36-0-




Question Number 211597 by MrGaster last updated on 14/Sep/24
                         2x^4 −4x^3 −22x^2 +24x+36=0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{{x}}^{\mathrm{4}} −\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{22}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{24}\boldsymbol{{x}}+\mathrm{36}=\mathrm{0} \\ $$$$ \\ $$
Answered by Ghisom last updated on 14/Sep/24
x^4 −2x^3 −11x^2 +12x+18=0  (x−(1/2))^4 −((25)/2)(x−(1/2))^2 +((337)/(16))=0  (x−(1/2))^2 =((25)/4)±3(√2)  x=(1/2)±((√(25±12(√2)))/2)
$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{11}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{18}=\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} −\frac{\mathrm{25}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{337}}{\mathrm{16}}=\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{4}}\pm\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{25}\pm\mathrm{12}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$

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