(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
I hope this helps you
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Answers & Comments
(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
Verified answer
I hope this helps you