Réitigh 4(x^4+1)=5x^2

$$4 ( x ^ { 4 } + 1 ) = 5 x ^ { 2 }$$

Réitigh do x. (complex solution)

$$x=e^{\frac{-\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}\approx -0.901387819+0.433012702i$$

$$x=e^{-\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}}\approx 0.901387819-0.433012702i$$

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}\approx -0.901387819-0.433012702i$$

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}}\approx 0.901387819+0.433012702i$$

Céimeanna ag Baint Úsáid as an bhFoirmle Chearnach

Úsáid an t-airí dáileach chun 4 a mhéadú faoi x^{4}+1.

$$4x^{4}+4=5x^{2}$$

Bain 5x^{2} ón dá thaobh.

$$4x^{4}+4-5x^{2}=0$$

Cuir t in ionad x^{2}.

$$4t^{2}-5t+4=0$$

Is féidir gach cothromóid i bhfoirm ax^{2}+bx+c=0 a réiteach ach an fhoirmle chearnach seo a úsáid: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Cuir 4 in ionad a, -5 in ionad b agus 4 in ionad c san fhoirmle chearnach.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Déan áirimh.

$$t=\frac{5±\sqrt{-39}}{8}$$

Réitigh an chothromóid t=\frac{5±\sqrt{-39}}{8} nuair is ionann ± agus luach deimhneach agus ± agus luach diúltach.

$$t=\frac{5+\sqrt{39}i}{8} t=\frac{-\sqrt{39}i+5}{8}$$

Más x=t^{2}, is féidir teacht ar na réitigh ach x=±\sqrt{t} a mheas i gcomhair gach t.

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}} x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}} x=e^{-\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}} x=e^{\frac{-\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}$$