Question

Determine o valor de x que se revolvem nas seguintes equações. Equação do segundo grau

X2-225=0
X2+19=100
3X2-13=35
(X+3)2=64
(X-5)2=121
(X+11)2=324

Answer 1

Resposta:

resposta abaixo na explicação

Explicação passo-a-passo:

a) X²-225=0  => X²= 225 => $x = \sqrt{225} => x = 15$  => x = ±15

b) X²+19=100  => X²=100  - 19 => X²= 81 =>  $x = \sqrt{81} => x = 19$ => x = ±9

c) 3X²-13=35  =>3X²=35  + 13 => 3X²=48 =>   $x^{2} = \frac{48}{3} => x^{2} = 16 => x =\sqrt{16} => x = 4$ => x = ±4

d) (X+3)²=64  => x² +6x + 9 - 64 = 0  =>  x² + 6x - 55 = 0  

Δ=b² - 4ac=> Δ= (6)² - 4(1)(-55)=> Δ= 36 + 220 => Δ= 256

x = -b ± √Δ/2a => x = -6 ± √256/2(1) => x = -6 ± 16/2 => $x_{1} = \frac{-6+16}{2} =>x_{1} = \frac{10}{2} => x_{1} = 5\\x_{2} = \frac{-6-16}{2} =>x_{2} = \frac{-22}{2} => x_{2} = -11$

e) (X-5)²=121 => x² - 10x + 25 =121  => x² - 10x + 25 -121= 0 => x² - 10x - 96= 0 =>

Δ=b² - 4ac=> Δ= (-10)² - 4(1)(-96)=> Δ= 100 + 384 => Δ= 484

x = -b ± √Δ/2a => x = 10 ± √484/2(1) => x = 10 ± 22/2 => $x_{1} = \frac{10+22}{2} =>x_{1} = \frac{32}{2} => x_{1} = 16\\x_{2} = \frac{10-22}{2} =>x_{2} = \frac{-12}{2} => x_{2} = -6$  

f) (X+11)²=324 => x² + 22x + 121 - 324 = 0 => x² + 22x - 203 = 0 =>

Δ=b² - 4ac=> Δ= (22)² - 4(1)(-203)=> Δ= 484 + 812 => Δ= 1296

x = -b ± √Δ/2a => x = -22 ± √1296/2(1) => x = -22 ± 36/2 =>

$x_{1} = \frac{-22+36}{2} =>x_{1} = \frac{12}{2} => x_{1} = 6\\x_{2} = \frac{-22-36}{2} =>x_{2} = \frac{-58}{2} => x_{2} = -29$