Question

use a fórmula de Bhaskara para resolver as equações abaixo:

A) 2x²-6x+5
B) 2x²-5x+2=0
C)-x²+2x-1=0

Answer 1

$a) x = \frac{-b \pm \sqrt{b^2 - 4.a.c}}{2a} \\ \\ x = \frac{6 \pm \sqrt{6^2 - 4.2.5}}{2.2} \\ \\ x = \frac{6 \pm \sqrt{36 - 40}}{4} \\ \\ x = \frac{6 \pm \sqrt{-4}}{4} \\ \\ x = \frac{6 \pm 2i}{4} \\ \\ x' = \frac{3}{2} + \frac{i}{2} \\ \\ x'' = \frac{3}{2} - \frac{i}{2}$


$b) x = \frac{-b \pm \sqrt{b^2 - 4.a.c}}{2a} \\ \\ x = \frac{5 \pm \sqrt{5^2 - 4.2.2}}{2.2} \\ \\ x = \frac{5 \pm \sqrt{25 - 16}}{4} \\ \\ x = \frac{5 \pm \sqrt{9}}{4} \\ \\ x = \frac{5 \pm 3}{4} \\ \\ x' = \frac{8}{4} = 2 \\ \\ x'' = \frac{5-3}{4} = \frac{1}{2}$


$c) x = \frac{-b \pm \sqrt{b^2 - 4.a.c}}{2a} \\ \\ x = \frac{-2 \pm \sqrt{2^2 - 4.(-1).(-1)}}{2.(-1)} \\ \\ x = \frac{-2 \pm \sqrt{4 - 4}}{-2} \\ x = \frac{-2 \pm \sqrt{0}}{-2} \\ \\ x = \frac{-2 \pm 0}{-2} \\ \\ x = 1 \: \text{(Unica Solucao)}$

Answer 2

a) 2x²-6x+5=0
∆=b²-4ac
∆=(-6)²-4.2.5
∆=36-40
∆=-4
delta negativo, não há raízes reais.
b) 2x²-5x+2=0
∆=b²-4ac
∆=(-5)²-4.2.2
∆=25-16
∆=9
x=-b±√∆/2a
x=-(-5)-3/2.2
x=5-3/4
x=2/4 (÷2)
x=1/2
x=-(-5)+3/2.2
x=5+3/4
x=8/4
x=2
c) -x²+2x-1=0
∆=b²-4ac
∆=2²-4.(-1).(-1)
∆=4-4
∆=0
x=-b±√∆/2a
x=-2-0/2.(-1)
x=-2/-2
x=1