Question

Resolva as equações do 2° grau usando formula resolutiva (formula de bhaskara)
2
A) X - 5x + 6 = 0

2
B) x - 8x + 12 = 0

Answer 1

A) $x^{2} -5x+6=0$
a= 1
b= -5
c= 6
$\frac{-b+/- \sqrt{ b^{2}-4*a*b } }{2}$
$\frac{-5+/- \sqrt{ (-5)^{2}-4*1*6 } }{2}$
$\frac{5+/- \sqrt{ 25-24 } }{2}$
$\frac{5+/- \sqrt{ 1 } }{2}$
$\frac{5+/- 1 }{2}$
$x^{1} = \frac{5+1}{2} =\ \textgreater \ x^{1} = \frac{6}{2} =\ \textgreater \ x^{1} =3$
$x^{2} = \frac{5-1}{2} =\ \textgreater \ x^{2} = \frac{4}{2} =\ \textgreater \ x^{2} =2$

B)$x^{2} -8x+12=0$
a= 1
b= -8
c= 12
$\frac{-b+/- \sqrt{ b^{2}-4*a*b } }{2}$
$\frac{-(-8)+/- \sqrt{ (-8)^{2}-4*1*12 } }{2}$
$\frac{ 8+/- \sqrt{ 64-48 } }{2}$
$\frac{ 8+/- \sqrt{ 16 } }{2}$
$\frac{ 8+/- 4 }{2}$
$\frac{8+4}{2} =\ \textgreater \ x^{1} = \frac{12}{2} =\ \textgreater \ x^{1} =6$
$\frac{8-4}{2} =\ \textgreater \ x^{2} = \frac{4}{2} =\ \textgreater \ x^{2} =2$