Question

Utilizando o processo algébrico de Bhaskara, determine as raízesUtilizando o processo algébrico de Bhaskara, determine as raízes das equações do 2º grau no conjunto dos números reais:
a) x²-10x+9=0
b) x²+x-6=0
c) x²+4x-5=0
d) x²-10x+24=0

Answer 1

$x^2-10x+9=0 \\ \\ D=10^2-4.1.9 \\ \\ D=100-36 \\ \\ D=64 \\ \\ \frac{-b+/- \sqrt{D} }{2.a} \\ \\ x'=\frac{10-8}{2}\>\>\>\>\>\>\>\>\>x'= 1 \\ \\ x"= \frac{10+8}{2} \>\>\>\>\>\>\>\>\>x"=9$
__________________________________________________________

$x^2+x-6=0 \\ \\ D=1^2-4.1.(-6) \\ \\ D=1+24 \\ \\ D=25 \\ \\ \frac{-b+/- \sqrt{D} }{2.a} \\ \\ x'=\frac{-1- {5 }}{2} \>\>\>\>\>\>\>\>\>\ x'=-3 \\ \\ x"= \frac{-1+5}{2} \>\>\>\>\>\>\>\>\>\ x"=2$
___________________________________________________________

$x^2+4x-5=0 \\ \\ D=4^2-4.1.(-5) \\ \\ D=16+20 \\ \\ D=36 \\ \\ \frac{-b+/- \sqrt{D} }{2} \\ \\ x'=\frac{-4-6}{2}\>\>\>\>\>\>\>\>\>\ x'=-5 \\ \\ x"=\frac{-4+6 }{2}\>\>\>\>\>\>\>\>\>\ x"=1$
___________________________________________________________

$x^2-10x+24=0 \\ \\ D=(-10)^2-4.1.24 \\ \\ D=100-96 \\ \\ D=4 \\ \\ \frac{-b+/- \sqrt{D} }{2.a} \\ \\x'= \frac{10-2}{2} \>\>\>\>\>\>\>\>\>\ x'=4 \\ \\ x"=\frac{10+2}{2}\>\>\>\>\>\>\>\>\>\ x"=6$