Question

Quais são as raízes das equações abaixo:

a)X2-5x +4 =0
b)X2 +4x + 4=0
c)X--X-6=0
d)X2 +8X +12=0

Answer 1

Resposta:

$a)x^{2}-5x+4=0 \\ x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4}}{2} \\ x=\frac{-\left(-5\right)±\sqrt{9}}{2} \\ x=\frac{5±3}{2} \\ x1=\frac{8}{2} = 4 \\ x2 =\frac{2}{2} = 1$

$b)x^{2}+4x+4=0 \\ x=\frac{-4±\sqrt{4^{2}-4\times 4}}{2} \\ x=\frac{-4±\sqrt{0}}{2} \\ x=-\frac{4}{2} \\ x=-2$

$a \: alternativa \: c \: eu \: não \: entendi \: então \: \\ fiz \: dessas \: duas \: formas \\ primeira \: \\ x - ( - x) - 6 = 0 \\ x + x = 6 \\ 2x = 6 \\ x = \frac{6}{2} = 3 \\ segunda \\ x^{2}-x-6=0 \\ x=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{2} \\ x=\frac{-\left(-1\right)±\sqrt{25}}{2} \\ x=\frac{1±5}{2} \\ x1=\frac{6}{2} = 3 \\ x2=\frac{-4}{2} = - 2$

$d)x^{2}+8x+12=0 \\ x=\frac{-8±\sqrt{8^{2}-4\times 12}}{2} \\ x=\frac{-8±\sqrt{16}}{2} \\ x=\frac{-8±4}{2} \\ x1=\frac{-4}{2} = - 2 \\ x2=\frac{-12}{2} = - 6$