Question

determine s raizes de (x)=×2+16x+39

Answer 1

Resposta:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+16x+39=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=16~e~c=39\\\\\Delta=(b)^{2}-4(a)(c)=(16)^{2}-4(1)(39)=256-(156)=100\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(16)-\sqrt{100}}{2(1)}=\frac{-16-10}{2}=\frac{-26}{2}=-13\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(16)+\sqrt{100}}{2(1)}=\frac{-16+10}{2}=\frac{-6}{2}=-3\\\\S=\{-13,~-3\}$

Answer 2

Resposta:

S = {- 3; - 13}

Explicação passo-a-passo:

x² + 16x + 39 = 0

Coeficientes:

a = 1; b = 16; c = 39

Calculando delta

Δ = b² - 4ac

Δ = 16² - 4.1.39

Δ = 256 - 156

Δ = 100

Há 2 raízes reais

Aplicando Bhaskara

x = - b ± √Δ/2a

x = - (16 ± √100)/2.1

x = - 16 ± 10/2

x = - 16 + 10/2

x = - 6/2

x' = - 3

x = - 16 - 10/2

x = - 26/2

x" = - 13

S = {- 3; - 13}

Bons estudos.