Question

fórmula de bhaskara

x elevado ao quadrado -5x+6=0

Answer 1

a)
$x^{2} -5x+6=0$

Δ=b²-4ac
Δ=(-5)²-4(6)
Δ=25-24
Δ=1
√Δ=√1

$x= \frac{-b\pm \sqrt{\triangle} }{2a} = \frac{-(-5)\pm \sqrt{1} }{2} = \frac{5\pm1}{2} =$

$x'= \frac{5+1}{2} = \frac{6}{2} =3$

$x"= \frac{5-1}{2} = \frac{4}{2} =2$

S={2,3}