Question

Quais as raízes de x 2  – (k + 4)x + 4k = 0 , sendo x a variável?

Answer 1

vamos lá....

x² -( k+ 4)x +4k=0

a=1
b=-(k+4)
c=4k

Δ=b²-4ac
Δ=[-(k+4)]²-4(1)(4k)
Δ=(k+4)²-16k
Δ=k²+8k+16-16k
Δ=k²-8k+16
Δ=(k-4)²

$x= \frac{-[-(k+4)\pm \sqrt{(k-4)^2} }{2} = \frac{k+4\pm(k-4)}{2} \\ \\ x'= \frac{k+4+(k-4)}{2} = \frac{k+\not4+k-\not4}{2} = \frac{\not2k}{\not2} =k \\ \\ x"= \frac{k+4-(k-4)}{2} = \frac{\not k+4-\not k+4}{2} = \frac{8}{2} =4 \\ \\ S=\{4,k\}$