Question

encontre o conjunto de solução de cada equação abaixo
1) x-3=√(2x+2)
2) y4-10y2+9=0


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Answer 1

Resposta:

1)

$x-3=\sqrt{2x+2} \\ \\ (x-3)^{2} =2x+2\\ \\ x^{2} -6x+9=2x+2\\ \\ x^{2} -8x+7=0$

$delta=b^{2} -4ac\\ \\ delta=(-8)^{2} -4.1.7\\ \\ delta=64-28\\ \\ delta=36$

$x=\frac{-b+-\sqrt{delta} }{2a} \\ \\ x=\frac{-(-8)+-\sqrt{36} }{2} \\ \\ x=\frac{8+-6}{2} \\ \\x'=\frac{8-6}{2} =\frac{2}{2} =1\\ \\ x''=\frac{8+6}{2} =\frac{14}{2} =7\\ \\ S~~\left \{ {{1;~~7} {}} \}$

2)

$y^{4}-10y^{2} +9=0~~~~~~equacao~biquadratica\\ \\ y^{4}= x^{2} ~~e~~y^{2}=x\\ \\x^{2} -10x+9=0\\ \\ delta=b^{2} -4.a.c\\ \\ delta=(-10)^{2} -4.1.9\\ \\ delta=100-36\\ \\delta=64$

$x=\frac{-b+-\sqrt{delta} }{2a} \\ \\ x=\frac{-(-10)+-\sqrt{64} }{2} \\ \\ x=\frac{10+-8}{2} \\ \\ x'=\frac{10-8}{2} =\frac{2}{2} =1\\ \\ x''=\frac{10+8}{2} =\frac{18}{2} =9$

$para~x=1\\ \\ y^{2} =x\\ \\ y^{2} =1\\ \\ y=\sqrt{1} \\ \\ y'=1\\ \\ \\ para~x=9\\ \\ y^{2} =x\\ \\ y^{2} =9\\ \\ y=\sqrt{9} \\ \\ y''=3$