Question

Determine os valores de k para os quais a equa¸c˜ao (2k − 3)x2− (5k + 6)x + k + 4 = 0 (a) Tenha duas ra´ızes reais distintas. (b) Tenha apenas uma solu¸c˜ao real.

Answer 1

Olá.

$\ (2k-3)*2-(5k+6)x+k+4=0\\\\\ 4k - 6 - (5k+6)x+k+4=0\\\\4k - 6 - (5kx + 6x) + k + 4 = 0\\\\4k - 6 - 5kx - 6x + k + 4 = 0\\\\5k - 6 - 5kx - 6x + 4 = 0\\\\5k - 2 - 5kx - 6x = 0\\\\5k - 5kx - 6x = 2\\\\5k - 5kx = 2 + 6x\\\\(5 - 5x)k = 2 + 6x\\\\(5 - 5x)k = 6x+ 2\\\\\frac{(5-5x)k}{5-5x} = \frac{6x+2}{5-5x} \\\\k = \frac{2(3x+1)}{5(1-x)}$

$x = -\frac{2-5k}{5k+6}$