Resposta :
Resposta
x²-15x+50=0
coeficientes
a = 1
b = -15
c =50
Δ = b² - 4ac
Δ = ( -15)² -4(1)(50)
Δ = 225 - 200
Δ = 25
[tex]x=\dfrac{-b\pm\sqrt{\Delta}}{2a}} \\ \\ \\ x=\dfrac{-(-15)\pm\sqrt{25} }{2(1)}=\dfrac{15\pm5}{2}\\ \\ \\ x'=\dfrac{15-5}{2}=\dfrac{10}{2}=5\\ \\ \\ x"= \dfrac{15+5}{2}=\dfrac{20}{2}=10\\ \\ \\\boxed{ S=\{5,10\}}[/tex]
Resolva a equação do 2 grau abaixo:
x²-15x+50=0
a = 1; b = - 15; c = 50
/\= b^2 - 4ac
/\ = (-15)^2 - 4.1.50
/\= 225 - 200
/\= 25
X = [ -(-15) +/- \/25]/2.1
X = (15 +/- 5)/2
X ' = (15+5)/2= 20/2= 10
X " = (15-5)/2= 10/2= 5
R.:
S = {5 , 10}
x²-15x+50=0
a = 1; b = - 15; c = 50
/\= b^2 - 4ac
/\ = (-15)^2 - 4.1.50
/\= 225 - 200
/\= 25
X = [ -(-15) +/- \/25]/2.1
X = (15 +/- 5)/2
X ' = (15+5)/2= 20/2= 10
X " = (15-5)/2= 10/2= 5
R.:
S = {5 , 10}