Question

alguem poderia me ajudar a calcular as delta pfvr:

a) x2+3x+4=0

b) x2+5x+3=0

c) x2+10x+4=0

d) 2x+4x+ 1=0

e) x2-6x+5=0​

Answer 1

Resposta:

a) x²+3x+4=0

∆=3²-4.1.4 = 9 - 16 = -7

b) x²+5x+3=0

∆=5²-4.1.3 = 25 - 12 = 13

c) x²+10x+4=0

∆=10²-4.1.4 = 100 - 16 = 84

d) 2x²+4x+ 1=0

∆=4²-4.2.1 = 16 - 8 = 8

e) x²-6x+5=0

∆=(-6)²-4.1.5 = 36 - 20 = 16