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михаил678547
@михаил678547
July 2022
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(a-1)x^2+2x(2a+1)+(4a+3)=0
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mikael2
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(a-1)x²+2(2a+1)x+(4a+3) =0 D=4(2a+1)²-4(4a+3)(a-1)=
= 4[4a²+4a+1-4a²-3a+4a+3]=4[5a+4] при 5а+4≥0 а≥-4/5 корни есть.
√D=2√(5a+4)
a≠1 x1=1/2(a-1)[-2(2a+1)-√D] x2=1/2(a-1)[-2(2a+1)+√D]
при а=1 имеем простое уравнение 2(2a+1)x+4a+3=0
x=-(4a+3)/2(2a+1)
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Answers & Comments
Verified answer
(a-1)x²+2(2a+1)x+(4a+3) =0 D=4(2a+1)²-4(4a+3)(a-1)== 4[4a²+4a+1-4a²-3a+4a+3]=4[5a+4] при 5а+4≥0 а≥-4/5 корни есть.
√D=2√(5a+4)
a≠1 x1=1/2(a-1)[-2(2a+1)-√D] x2=1/2(a-1)[-2(2a+1)+√D]
при а=1 имеем простое уравнение 2(2a+1)x+4a+3=0
x=-(4a+3)/2(2a+1)