Question

resolva a equação do 2 grau.

a)
$x {}^{2} - 49 = 0$
$b)3x {}^{2} - 243 = 0$
$c = x {}^{2} + 3 - 28 = 0$
​

Answer 1

Resposta:

a) x' = 7; x'' = -7

b) x' = 9; x'' = -9

c) x' = +5; x'' = -5

Explicação passo-a-passo:

a)

x² - 49 = 0

x² = 49

x = ± $\sqrt{49}$

x = ± 7

x' = 7; x'' = -7

b)

3x² - 243 = 0

3x² = 243

x² = 243/3

x² = 81

x = ± $\sqrt{81}$

x = ± 9

x' = 9; x'' = -9

c)

x² + 3 - 28 = 0

x² = -3 + 28

x² = 25

x = ± $\sqrt{25}$

x = ± 5

x' = +5; x'' = -5

Bons estudos.

Answer 2

• A ↔

$\sf x^2-49=0\\\\\\x^2-49+49=0+49\\\\\\x^2=49\\\\\\x=\sqrt{49},\:x=-\sqrt{49}\\\\\\\to\boxed{\sf x=7}\\\to\boxed{\sf x=-7}$

• B ↔

$\sf 3x^2-243=0\\\\\\3x^2-243+243=0+243\\\\\\3x^2=243\\\\\\\dfrac{3x^2}{3}=\dfrac{243}{3}\\\\\\x^2=81\\\\\\x=\sqrt{81},\:x=-\sqrt{81}\\\\\\\to\boxed{\sf x=9}\\\to\boxed{\sf x=-9}$

• C ↔

$\sf x^2+3-28=0\\\\\\x^2+3-28-\left(3-28\right)=0-\left(3-28\right)\\\\\\x^2=25\\\\\\x=\sqrt{25},\:x=-\sqrt{25}\\\\\\\to \boxed{\sf x=5 }\\\to\boxed{\sf x=-5 }$