Question

X2-16x-36=0 pergunta para meu trabalho de matemática

Answer 1

Resposta:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-16x-36=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-16~e~c=-36\\\\\Delta=(b)^{2}-4(a)(c)=(-16)^{2}-4(1)(-36)=256-(-144)=400\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-16)-\sqrt{400}}{2(1)}=\frac{16-20}{2}=\frac{-4}{2}=-2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-16)+\sqrt{400}}{2(1)}=\frac{16+20}{2}=\frac{36}{2}=18\\\\S=\{-2,~18\}$

Answer 2

Δ = (-16)² - 4.1.(-36)

Δ = 256 + 144 = 400

x¹ = 16 + √400 /2

x¹ = 36 /2 = 18

x² = 16 - 20 /2

x² = 4/2 = 2