Question

Determine as raízes dos seguintes polinômios:
a) P(x) = 2x + 6
b) P(x) = x² - 4x – 5
c) P(x) = x² - 16
d) P(x) = x² - 6x
e) P(x) = 3x - 9
​

Answer 1

Caso tenha problemas para visualizar a resposta experimente abrir pelo navegador.https://brainly.com.br/tarefa/33343893

===============================================================

$\tt{a)}~\sf{p(x)=2x+6}\\\sf{2x+6=0}\\\sf{2x=-6}\\\sf{x=-\dfrac{6}{2}}\\\large\boxed{\boxed{\boxed{\boxed{\sf{x=-3}}}}}\\\tt{b)}~\sf{p(x)=x^2-4x-5}\\\sf{\Delta=b^2-4ac}\\\sf{\Delta=(-4)^2-4\cdot1\cdot(-5)}\\\sf{\Delta=16+20}\\\sf{\Delta=36}\\\sf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\sf{x=\dfrac{-(-4)\pm\sqrt{36}}{2\cdot1}}\\\sf{x=\dfrac{4\pm6}{2}}\begin{cases}\sf{x_1=\dfrac{4+6}{2}=\dfrac{10}{2}=5}\\\sf{x_2=\dfrac{4-6}{2}=-\dfrac{2}{2}=-1}\end{cases}$

$\tt{c)}~\sf{p(x)=x^2-16}\\\sf{x^2-16=0}\\\sf{x^2=16}\\\sf{x=\pm\sqrt{16}}\\\sf{x=\pm2}\\\tt{d)}~\sf{p(x)=x^2-6x}\\\sf{x^2-6x=0}\\\sf{x\cdot(x-6)=0}\\\sf{x=0}\\\sf{x-6=0}\\\sf{x=6}\\\tt{e)}~\sf{p(x)=3x-9}\\\sf{3x-9=0}\\\sf{3x=9}\\\sf{x=\dfrac{9}{3}}\\\sf{x=3}$

Answer 2

Oie, Td Bom?!

a)

$P(x) = 2x + 6$

$0 = 2x + 6$

$- 2x = 6$

$x = \frac{6}{ - 2}$

$x = - \frac{6}{2}$

$x = - 3$

$S = \left \{ - 3\right \}$

b)

$P(x) = x {}^{2} - 4x - 5$

$0 = x {}^{2} - 4x - 5$

$x {}^{2} - 4x - 5 = 0$

$x {}^{2} + x - 5x - 5 = 0$

$x \: . \: (x + 1) - 5(x + 1) = 0$

$(x + 1) \: . \: (x - 5) = 0$

• $x + 1 = 0⇒x = - 1$

• $x - 5 = 0⇒x = 5$

$S = \left \{ - 1 \: ,\: 5 \right \}$

c)

$P(x) = x {}^{2} - 16$

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

$- x {}^{2} = - 16 \: . \: ( - 1)$

$x {}^{2} = 16$

$x = ± \sqrt{16}$

$x = ±4$

$S = \left \{ - 4 \:, \: 4 \right \}$

d)

$P(x) = x {}^{2} - 6x$

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

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

$x \: . \: (x - 6) = 0$

• $x = 0$

• $x - 6 = 0⇒x = 6$

$S = \left \{ 0 \: ,\: 6\right \}$

e)

$P(x) = 3x - 9$

$0 = 3x - 9$

$- 3x = - 9$

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

$x = \frac{9}{3}$

$x = 3$

$S = \left \{ 3\right \}$

Att. Makaveli1996