Question

Determine o conjunto solução da seguinte equações biquadradas:

b) X^4 - 5x^2 + 4 = 0

c) Y^4 - 7y^2 + 12 = 0

d) x^4 + 4x^2 + 3 = 0

e) 16y^4 -1 = 0

f) 2x^4 - 5x^2 + 2 = 0

Answer 1

Resposta:

$b)\\ \\x^{4}-5x^{2}+4=0\\ \\(x^{2})^{2}-5x^{2}+4=0\\ \\atribua~o~valor~de~(x^{2})~como~sendo~(y)~e~substitua\\ \\y^{2}-5y+4=0\\ \\delta=b^{2}-4.a.c\\ \\delta=25-4.1.4\\ \\delta=25-16=9\\ \\y=\frac{-b+-\sqrt{delta}}{2.a}\\ \\y=\frac{5+-\sqrt{9}}{2}\\ \\y=\frac{5+-3}{2}\\ \\y'=\frac{5+3}{2}=4\\ \\y''=\frac{5-3}{2}=1\\ \\para~y=4\\ \\x^{2}=y\\ \\x^{2}=4\\ \\x=\sqrt{4}\\ \\x=2\\ \\para~y=1\\ \\x^{2}=y\\ \\x^{2}=1\\ \\x=\sqrt{1}\\ \\x=1$

$S~~\left \{ {{1;~~2} \}} \right.$

$y^{4}-7y^{2}+12=0\\ \\(y^{2})^{2}-7y^{2} +12=0\\ \\substitua~~~y^{2}~~pela~incognita~x\\ \\x^{2}-7x+12=0\\ \\delta=b^{2}-4.a.c\\ \\delta=49-4.1.12\\ \\delta=49-48=1\\ \\x=\frac{-b+-\sqrt{delta}}{2.a}\\ \\x=\frac{7+-\sqrt{1}}{2}\\ \\x'=\frac{7+1}{2}=\frac{8}{2}=4\\ \\x''=\frac{7-1}{2}=\frac{6}{2}=3\\ \\para~x=4\\ \\y^{2}=x\\ \\y^{2}=4\\ \\y=\sqrt{4}\\ \\y=2\\ \\para~x=3\\ \\y^{2}=x\\ \\y^{2}=3\\ \\y=\sqrt{3}\\ \\S~~\left \{ {{\sqrt{3};~~2} \}} \right.$

$d)\\ \\x^{4}+4x^{2}+3=0\\ \\(x^{2})^{2}+4x^{2}+3=0\\ \\substitua~~x^{2} ~~por~y\\ \\y^{2}+4y+3=0\\ \\delta=b^{2}-4.a.c\\ \\delta=16-4.1.3\\ \\delta=4\\ \\y=\frac{-b+-\sqrt{delta}}{2.a}\\ \\y=\frac{-4+-2}{2}\\ \\y'=\frac{-4+2}{2}=\frac{-2}{2}=-1\\ \\y''=\frac{-4-2}{2}=\frac{-6}{2}=-3\\ \\para~y=-1\\ \\x^{2}=y\\ \\x^{2}=-1\\ \\x=\sqrt{-1}\\ \\para~y=-3\\ \\x^{2}=y\\ \\x^{2}=-3\\ \\x=\sqrt{-3}\\ \\ S~~nao~existem~raizes~reais~para~a~equacao$

$e)\\ \\16y^{4}-1=0\\ \\(4y^{2})^{2}-1=0\\ \\substitua~y^{2}~por~x\\ \\4x^{2}-1=0\\ \\equacao~do~2~grau~incompleta~b=0\\ \\x^{2}=\frac{-c}{a}\\ \\x^{2}=\frac{1}{4}\\ \\x=\sqrt{\frac{1}{4}}\\ \\x=\frac{1}{2}\\ \\para~~x=\frac{1}{2}\\ \\y^{2}=x\\ \\y^{2}=\frac{1}{2}\\ \\ y=\sqrt{\frac{1}{2} }\\ \\ S~~~~\left \{ {{\sqrt{\frac{1}{2}} } \}} \right.$

Desculpe-me, não deu tempo para a f), mas espero ter ajudado