preciso de uma explicação á respeito de equações biquadradas
Problema -> x elevado á 4 - 8x elevado ao quadrado + 15 = 0
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Bom dia Emanuelle!
Solução!
Para resolver uma equação biquadrática temos que fazer uma substituição de x por y.
Veja como é simples!
![x^{4}-8x+15=0\\\\\\
x^{2} . x^{2} -8x+15=0\\\\\
x^{2}=y\\\\\\
y . y -8x+15=0\\\\\
y^{2}-8y+15=0](https://tex.z-dn.net/?f=+x%5E%7B4%7D-8x%2B15%3D0%5C%5C%5C%5C%5C%5C%0A+x%5E%7B2%7D+.+x%5E%7B2%7D+-8x%2B15%3D0%5C%5C%5C%5C%5C%0A+x%5E%7B2%7D%3Dy%5C%5C%5C%5C%5C%5C%0Ay+.+y+-8x%2B15%3D0%5C%5C%5C%5C%5C%0A+y%5E%7B2%7D-8y%2B15%3D0+ "x^{4}-8x+15=0\\\\\\
x^{2} . x^{2} -8x+15=0\\\\\
x^{2}=y\\\\\\
y . y -8x+15=0\\\\\
y^{2}-8y+15=0")
Usando a formula de Baskara!
![y= \dfrac{-(-8)\pm \sqrt{(-8)^{2}-4.1.15 } }{2.1}\\\\\\\
y= \dfrac{8\pm \sqrt{64-60} }{2}\\\\\\\\
y= \dfrac{8\pm \sqrt{4} }{2}\\\\\\\\
y= \dfrac{8\pm2 }{2}\\\\\\\\
Raizes\\\\\\
y_{1}= \dfrac{8+2}{2}= \dfrac{10}{2}=5\\\\\\\\\
y_{2}= \dfrac{8-2}{2}= \dfrac{6}{2}=3\\\\\\\\
Veja~~que encontramos~~os~~ valores~~ de~~ y ~~mas\\\\\\\
o ~~que ~~nos~~ interessa~~s\~ao~~os valores de x.~\\\\\\\
Ent\~ao ~~vamos~~ fazer ~~a~~ substituic\~ao.](https://tex.z-dn.net/?f=y%3D+%5Cdfrac%7B-%28-8%29%5Cpm+%5Csqrt%7B%28-8%29%5E%7B2%7D-4.1.15+%7D+%7D%7B2.1%7D%5C%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B8%5Cpm+%5Csqrt%7B64-60%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C+%0Ay%3D+%5Cdfrac%7B8%5Cpm+%5Csqrt%7B4%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C+%0Ay%3D+%5Cdfrac%7B8%5Cpm2+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C%0ARaizes%5C%5C%5C%5C%5C%5C%0A%0A++y_%7B1%7D%3D+%5Cdfrac%7B8%2B2%7D%7B2%7D%3D+%5Cdfrac%7B10%7D%7B2%7D%3D5%5C%5C%5C%5C%5C%5C%5C%5C%5C+++%0Ay_%7B2%7D%3D+%5Cdfrac%7B8-2%7D%7B2%7D%3D+%5Cdfrac%7B6%7D%7B2%7D%3D3%5C%5C%5C%5C%5C%5C%5C%5C%0A%0AVeja%7E%7Eque+encontramos%7E%7Eos%7E%7E+valores%7E%7E+de%7E%7E+y+%7E%7Emas%5C%5C%5C%5C%5C%5C%5C%0Ao+%7E%7Eque+%7E%7Enos%7E%7E+interessa%7E%7Es%5C%7Eao%7E%7Eos+valores+de+x.%7E%5C%5C%5C%5C%5C%5C%5C%0AEnt%5C%7Eao+%7E%7Evamos%7E%7E+fazer+%7E%7Ea%7E%7E+substituic%5C%7Eao.++ "y= \dfrac{-(-8)\pm \sqrt{(-8)^{2}-4.1.15 } }{2.1}\\\\\\\
y= \dfrac{8\pm \sqrt{64-60} }{2}\\\\\\\\
y= \dfrac{8\pm \sqrt{4} }{2}\\\\\\\\
y= \dfrac{8\pm2 }{2}\\\\\\\\
Raizes\\\\\\
y_{1}= \dfrac{8+2}{2}= \dfrac{10}{2}=5\\\\\\\\\
y_{2}= \dfrac{8-2}{2}= \dfrac{6}{2}=3\\\\\\\\
Veja~~que encontramos~~os~~ valores~~ de~~ y ~~mas\\\\\\\
o ~~que ~~nos~~ interessa~~s\~ao~~os valores de x.~\\\\\\\
Ent\~ao ~~vamos~~ fazer ~~a~~ substituic\~ao.")
![x^{2} =y\\\\\
x^{2} =5\\\\\\
\boxed{x=\pm \sqrt{5}}\\\\\\\
x^{2} =y\\\\\
x^{2} =3\\\\\
x =\pm \sqrt{3} \\\\\\\\\\\
\boxed {Resposta:S=\{\pm \sqrt{3},\pm \sqrt{5}\}}](https://tex.z-dn.net/?f=+x%5E%7B2%7D+%3Dy%5C%5C%5C%5C%5C%0A+x%5E%7B2%7D+%3D5%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D%5Cpm+%5Csqrt%7B5%7D%7D%5C%5C%5C%5C%5C%5C%5C%0A++x%5E%7B2%7D+%3Dy%5C%5C%5C%5C%5C%0A+x%5E%7B2%7D+%3D3%5C%5C%5C%5C%5C%0A+x+%3D%5Cpm+%5Csqrt%7B3%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed+%7BResposta%3AS%3D%5C%7B%5Cpm+%5Csqrt%7B3%7D%2C%5Cpm+%5Csqrt%7B5%7D%5C%7D%7D%0A "x^{2} =y\\\\\
x^{2} =5\\\\\\
\boxed{x=\pm \sqrt{5}}\\\\\\\
x^{2} =y\\\\\
x^{2} =3\\\\\
x =\pm \sqrt{3} \\\\\\\\\\\
\boxed {Resposta:S=\{\pm \sqrt{3},\pm \sqrt{5}\}}")
Bom dia!
Boms estudos!
Solução!
Para resolver uma equação biquadrática temos que fazer uma substituição de x por y.
Veja como é simples!
Usando a formula de Baskara!
Bom dia!
Boms estudos!
EmmanuelleGaudencio:
Muito obrigada
Dê nada! Qualquer dúvida comente aqui.Ok!
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