Question

calcule os zeros da função

a)
$f(x) = x {}^{2} - 6 \times x + 9$
B)
$f(x) = x {}^{2} + 2 \times x + 1$
c)
$y = 4 \times x {}^{2} + 2 \times x - 6$
​

Answer 1

Explicação passo-a-passo:

a) f(x) = x² - 6x + 9

$x = \frac{ - ( - 6) + - \sqrt{ {( - 6)}^{2} - 4 \times 1 \times 9} }{2 \times 1} \\ \\ x = \frac{6 + - \sqrt{36 - 36} }{2} \\ \\ x = \frac{6 + - \sqrt{0} }{2} \\ \\ x1 = x2 = \frac{6}{2} = 3$

b) f(x) = x² + 2x + 1

$x = \frac{ - 2 + - \sqrt{ {2}^{2} - 4 \times 1 \times 1 } }{2 \times 1} \\ \\ x = \frac{ - 2 + - \sqrt{4 - 4} }{2} \\ \\ x = \frac{ - 2 + - \sqrt{0} }{2} \\ \\ x1 = x2 = - \frac{ 2}{2} = - 1$

c) f(x) = 4x² + 2x - 6

$x = \frac{ - 2 + - \sqrt{ {2}^{2} - 4 \times 4 \times ( - 6) } }{2 \times 4} \\ \\ x = \frac{ - 2 + - \sqrt{4 + 96} }{8} \\ \\ x = \frac{ - 2 + - \sqrt{100} }{8} \\ \\ x1 = \frac{ - 2 + 10}{8} = \frac{8}{8} = 1 \\ \\ x2 = \frac{ - 2 - 10}{8} = - \frac{12}{8} = - \frac{3}{2}$