(x^2+2x)^2-4(x+1)^2+7=0

Question

(x^2+2x)^2-4(x+1)^2+7=0

Answer 1

(x^2+2x)^2-4(x+1)^2+7=0

(x^2+2x)^2-4(x^2+2x+1)+7=0

x^2+2x=t

t^2 - 4(t + 1) + 7 = 0

t^2 - 4t + 3 = 0

D=16-12=4

t12=(4+-2)/2 = 1  3

1/ t=1

x^2+2x=1

x^2+2x-1=0

D=4+4=8

x12=(-2+-√8)/2 = -1 +- √2

2/ t=3

x^2+2x=3

x^2+2x-3=0

D=4+12=16

x34=(-2+-4)/2 = -3  1

ответ -3  1  -1+-√2

Comments

Большое спасибо

Answer 2

I hope this helps you

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