Question

Resolva as equações do 2° grau em R (Reais) pela fórmula de Báskara:

1) x²+2x-3=0
2) 4x²-7x+3=0
3) x²+6x-7=0
4) x²+4x-12=0
​

Answer 1

Explicação passo-a-passo:

1)

x² +2x - 3 =0

a=1

b=2

c=-3

Δ=b² -4ac

Δ=2²-4(1)(-3)

Δ=4+12

Δ = 16

$x={-b\pm\sqrt{\Delta} \over2a}\\ \\ x={-2\pm\sqrt{16} \over2(1)}={-2\pm4\over2}\\ \\ x'={-2-4\over2}=-{6\over2}=-3\\ \\ x"={-2+4\over2}={2\over2}=1$

-----------------------------------------------

2)

4x² - 7x + 3 =0

a=4

b=-7

c=3

Δ=b²-4ac

Δ=(-7)²-4(4)(3)

Δ=49 -48

Δ = 1

$x={-(-7)\pm\sqrt{1} \over2(4)}={7\pm1\over8}\\ \\ x'={7-1\over8}={6\over8}={3\over4}\\ \\ x"={7+1\over8}={8\over8}=1$

--------------------------------------

3)

x² + 6x - 7 =0

a=1

b=6

c=-7

Δ= 6² -4(1)(-7)

Δ=36+28

Δ=64

$x={-6\pm\sqrt{64} \over2(1)}={-6\pm8\over2}\\ \\ x'={-6-8\over2}-{14\over2}=-7\\ \\ x"={-6+8\over2}={2\over2}=1$

------------------------------------

4)

x² + 4x - 12 =0

a=1

b=4

c=-12

Δ=4²-4(1)(-12)

Δ=16+48

Δ=64

$x={-4\pm\sqrt{64} \over2(1)}={-4\pm8\over2}\\ \\ x'={-4-8\over2}=-{12\over2}=-6\\ \\ x"={-4+8\over2}={4\over2}=2$