Question

8x a quarta menos 10x ao quadrado + 3 = 0

Answer 1

Resposta:

$\displaystyle S=\{-\frac{\sqrt{2}}{2},~\frac{\sqrt{2}}{2}, ~-\frac{\sqrt{3}}{2}},~ \frac{\sqrt{3}}{2}}\}$

Explicação passo-a-passo:

8x⁴-10x²+3=0

8(x²)²-10x²+3=0

Fazendo: y=x² e substituindo

8y²-10y+3=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~8y^{2}-10y+3=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=8{;}~b=-10~e~c=3\\\\\Delta=(b)^{2}-4(a)(c)=(-10)^{2}-4(8)(3)=100-(96)=4\\\\y^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-10)-\sqrt{4}}{2(8)}=\frac{10-2}{16}=\frac{8}{16}=\frac{1}{2}\\\\y^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-10)+\sqrt{4}}{2(8)}=\frac{10+2}{16}=\frac{12\div4}{16\div4}=\frac{3}{4}\\\\S=\{\frac{1}{2},~\frac{3}{4}\}$

Para y=1/2

$\displaystyle y=x^2\\\\\frac{1}{2} =x^2\\\\x=\pm\sqrt{\frac{1}{2} } =\pm\frac{\sqrt{1}}{\sqrt{2}} =\pm\frac{1}{\sqrt{2}}.\frac{\sqrt{2}}{\sqrt{2}} =\pm\frac{\sqrt{2}}{2}$

Para y=3/4

$\displaystyle y=x^2\\\\\frac{3}{4} =x^2\\\\x=\pm\sqrt{\frac{3}{4} } =\pm\frac{\sqrt{3}}{\sqrt{4}} =\pm\frac{\sqrt{3}}{2}}$