Question

alguém prfv poderia me ajudar????​

Answer 1

Resposta:

b. $x^2 + 16x = 0 \Rightarrow x(x + 16) = 0$. Assim, $x = 0$ ou $x + 16 = 0 \Rightarrow x = -16$. Portanto, $S = \{-16, 0\}$

c. $(x-3)(x+6)= 0$. Assim, $x - 3 = 0 \Rightarrow x = 3$ ou $x + 6 = 0 \Rightarrow x = -6$. Portanto, $S = \{-6, 3\}$.

d.

$x^2 + 2x - 35 = 0\\x = \frac{-2 \pm \sqrt{2^2 -4(1)(-35)}}{2(1)}\\x = \frac{-2 \pm \sqrt{4 + 140}}{2}\\x = \frac{-2 \pm \sqrt{149}}{2}\\x' = \frac{-2 + \sqrt{149}}{2}\\x'' = \frac{-2 - \sqrt{149}}{2}\\S = \{\frac{-2 - \sqrt{149}}{2}, \frac{-2 + \sqrt{149}}{2}\}$

e.

$x^2 - 6x + 5 = 0\\x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)}\\x = \frac{6 \pm \sqrt{36 - 20}}{2}\\x= \frac{6 \pm \sqrt{16}}{2}\\x = \frac{6 \pm 4}{2}\\x = 3 \pm 2\\x' = 3+2=5\\x''=3-2=1\\S = \{1,5\}$