Question

Determine a solução da equação do 2º grau 2x² + x – 3 = 0 ​

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~2x^{2}+x-3=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=2{;}~b=1~e~c=-3\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(2)(-3)=1-(-24)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{25}}{2(2)}=\frac{-1-5}{4}=\frac{-6\div2}{4\div2}=-\frac{3}{2}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{25}}{2(2)}=\frac{-1+5}{4}=\frac{4}{4}=1\\\\S=\{-\frac{3}{2},~1\}$

Answer 2

Explicação passo-a-passo:

2x² + x - 3 = 0

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∆ = b² - 4ac

∆ = 1² - 4.2.(-3)

∆ = 1 + 24

∆ = 25

x = - b ± √∆/2a

x = - (1) ± √25/2.2

x = - 1 ± 5/4

x' = - 1+5/4 = 4/4 = 1

x" = - 1-5/4 = - 6/4 = -3/2

S = {- 3/2 ; 1}