Решить неравенство (x2+4x-5)(x2+4x+3)

July 2022 1 8 Report

Решить неравенство (x2+4x-5)(x2+4x+3)<105

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Verified answer

$(x^2+4x-5)(x^2+4x+3)\ \textless \ 105$

Замена:
$x^2+4x=t \\ \\ (t-5)(t+3)\ \textless \ 105 \\ t^2-2t-120\ \textless \ 0 \\ \\ t^2-2t-120=0 \\ t_1+t_2=2 \cup t_1t_2=-120 \\ t_1=-10 \cup t_2=12 \\ \\a\ \textgreater \ 0 \Rightarrow t \in (-10;12)$

Обратная замена:
$1) \\ x^2+4x\ \textgreater \ -10 \\ x^2+4x+10\ \textgreater \ 0 \\ \\ x^2+4x+10=0 \\ D=16-40\ \textless \ 0$
нет решений

$2) \\ x^2+4x\ \textless \ 12 \\ x^2+4x-12\ \textless \ 0 \\ \\ x^2+4x-12=0 \\ x_1+x_2=-4 \cup x_1x_2=-12 \\ x_1=-6 \cup x_2=2 \\ \\ a\ \textgreater \ 0 \Rightarrow x \in (-6;2)$

Ответ: $x \in (-6;2)$

1 votes Thanks 1

ujujuj Спасибо, за помощь!

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