quantas raízes tem essa funçao x²-4x+10,-6 < x < 10 , -10 < y < 25
Usuário anônimo:
Bom dia Mary! Essa função tem duas raizes,porem as raizes são complexas.Você esta estudando sobre o assunto?
oi,eu estou sim,entrei na faculdade ja em semana de trabalçhos,estou estudando,mas eh muita coisa atrasa que nao lembro mais,eu sabia que era duas raizes,mas estava cm muitas duvidas,eu fiz os graficos,mas tava complicado entender!
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Bom dia Mary!
Solução!
Vamos substituir esses coeficientes a,b e c na formula de Bhaskara.
![x= \dfrac{-b\pm \sqrt{b^{2} -4.a.c} }{2.a}\\\\\\
x= \dfrac{-(-4)\pm \sqrt{(-4)^{2} -4.1.10} }{2.1}\\\\\\
x= \dfrac{4\pm \sqrt{16-40} }{2}\\\\\\
x= \dfrac{4\pm \sqrt{-24} }{2}\\\\\\
x= \dfrac{4\pm i\sqrt{24} }{2}\\\\\\
x= \dfrac{4\pm 2i\sqrt{6} }{2}\\\\\\
x=2( \dfrac{2\pm i\sqrt{6} }{2})\\\\\\
x=(2\pm i\sqrt{6})\\\\\\\\\\
Resposta~~~\\\\\\
Duas~~Raizes~~complexas\\\\\
x_{1}=2+i \sqrt{6}\\\\\
x_{2}=2-i \sqrt{6}](https://tex.z-dn.net/?f=x%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7Bb%5E%7B2%7D+-4.a.c%7D+%7D%7B2.a%7D%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B-%28-4%29%5Cpm+%5Csqrt%7B%28-4%29%5E%7B2%7D+-4.1.10%7D+%7D%7B2.1%7D%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B4%5Cpm+%5Csqrt%7B16-40%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B4%5Cpm+%5Csqrt%7B-24%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B4%5Cpm+i%5Csqrt%7B24%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B4%5Cpm+2i%5Csqrt%7B6%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0Ax%3D2%28+%5Cdfrac%7B2%5Cpm+i%5Csqrt%7B6%7D+%7D%7B2%7D%29%5C%5C%5C%5C%5C%5C%0Ax%3D%282%5Cpm+i%5Csqrt%7B6%7D%29%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AResposta%7E%7E%7E%5C%5C%5C%5C%5C%5C%0ADuas%7E%7ERaizes%7E%7Ecomplexas%5C%5C%5C%5C%5C%0A+x_%7B1%7D%3D2%2Bi+%5Csqrt%7B6%7D%5C%5C%5C%5C%5C%0A+x_%7B2%7D%3D2-i+%5Csqrt%7B6%7D++++%0A%0A "x= \dfrac{-b\pm \sqrt{b^{2} -4.a.c} }{2.a}\\\\\\
x= \dfrac{-(-4)\pm \sqrt{(-4)^{2} -4.1.10} }{2.1}\\\\\\
x= \dfrac{4\pm \sqrt{16-40} }{2}\\\\\\
x= \dfrac{4\pm \sqrt{-24} }{2}\\\\\\
x= \dfrac{4\pm i\sqrt{24} }{2}\\\\\\
x= \dfrac{4\pm 2i\sqrt{6} }{2}\\\\\\
x=2( \dfrac{2\pm i\sqrt{6} }{2})\\\\\\
x=(2\pm i\sqrt{6})\\\\\\\\\\
Resposta~~~\\\\\\
Duas~~Raizes~~complexas\\\\\
x_{1}=2+i \sqrt{6}\\\\\
x_{2}=2-i \sqrt{6}")
Bom dia!
Bons estudos!
Solução!
Vamos substituir esses coeficientes a,b e c na formula de Bhaskara.
Bom dia!
Bons estudos!
Anexos:
obrigadooooo,muito obrigadooo..
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