Question

3x2-7x+4=0 me ajudem
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Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~3x^{2}-7x+4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=3{;}~b=-7~e~c=4\\\\\Delta=(b)^{2}-4(a)(c)=(-7)^{2}-4(3)(4)=49-(48)=1\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-7)-\sqrt{1}}{2(3)}=\frac{7-1}{6}=\frac{6}{6}=1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-7)+\sqrt{1}}{2(3)}=\frac{7+1}{6}=\frac{8\div2}{6\div2}=\frac{4}{3}\\\\S=\{1,~\frac{4}{3}\}$

Answer 2

Para solucionar o problema vamos utilizar a fórmula de bhaskara:

∆=b^2 - 4.a.c
∆=(-7)^2 - 4.1.4
∆=33

x-1=0
x=4/3

X1=1
x2=4/3