Question

Determine x real de modo que se tenha:
a) 3x² + 3x -6 = 0
b) 2x2 + 4x – 16=0
c) 4x + 11x = 30

Answer 1

Resposta:

a)

${3x}^{2} + 3x - 6 = 0$

a = 3 b = -3 c = -6

$d = {b}^{2} - 4ac$

$d = {( - 3)}^{2} - 4 \times 3 \times ( - 6)$

$d = 9 + 72$

$d = 81$

$x = \frac{ - b + - \sqrt{d} }{2a}$

$x = \frac{ - ( - 3) + - \sqrt{81} }{2 \times 3}$

$x = \frac{3 + - 9}{6}$

$x1 = \frac{3 + 9}{6} = \frac{12}{6} = 2$

$x2 = \frac{3 - 9}{6} = \frac{ - 6}{6} = - 1$

b)

${2x}^{2} + 4x - 16 = 0$

a = 2 b = 4 c = -16

$d = {b}^{2} - 4ac$

$d = {4}^{2} - 4 \times 2 \times ( - 16)$

$d = 16 + 128$

$d = 144$

$x = \frac{ - b + - \sqrt{d} }{2a}$

$x = \frac{ - 4 + - \sqrt{144} }{2 \times 2}$

$x = \frac{ - 4 + - 12}{4}$

$x1 = \frac{ - 4 + 12}{4} = \frac{8}{4} = 2$

$x2 = \frac{ - 4 - 12}{4} = \frac{ - 16}{4} = - 4$

c)

$4x + 11 \times = 30$

$15x = 30$

$x = \frac{30}{15}$

$x = 2$