Question

x² + 2x - 8 = 0.

Resolva a equação 


(R: 2, -4)

(R: 2, 3)

(R: 2, 8)

(R: 7, 3)

Outro: ....

ajuda aí ​

Answer 1

x^2 + 2x - 8 = 0

a= 1. b= 2. c= -8.

DELTA= b^2 - 4 × a × c

DELTA= 2^2 - 4 × 1 × (-8)

DELTA= 4 - 4 × (-8)

DELTA= 4 + 32

DELTA= 36

x1.2= -b +- Raíz quadrada DO DELTA / 2×a

x1.2= -2 +- Raíz quadrada de 36 / 2×1

x1.2= -2 +- 6/ 2

x1= -2+6/2

x1= 4/2= 2

x2= -2-6/2

x2= -8/2= -4

RESPOSTA= (R. 2, -4)

Answer 2

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+2x-8=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=2~e~c=-8\\\\\Delta=(b)^{2}-4(a)(c)=(2)^{2}-4(1)(-8)=4-(-32)=36\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(2)-\sqrt{36}}{2(1)}=\frac{-2-6}{2}=\frac{-8}{2}=-4\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(2)+\sqrt{36}}{2(1)}=\frac{-2+6}{2}=\frac{4}{2}=2\\\\S=\{-4,~2\}$