Cho pt x^2+ax+1=0. Xác định a để pt có hai nghiệm x1;x2 thoả mãn (x1/x2)^2+(x2/x1)^2>7. Giúp mở mình với ạ!!! 25/09/2021 Bởi Samantha Cho pt x^2+ax+1=0. Xác định a để pt có hai nghiệm x1;x2 thoả mãn (x1/x2)^2+(x2/x1)^2>7. Giúp mở mình với ạ!!!
Đáp án: \(\left[ \begin{array}{l}a > \sqrt 5 \\a < – \sqrt 5 \end{array} \right.\) Giải thích các bước giải: Để phương trình có 2 nghiệm ⇒Δ≥0 \(\begin{array}{l} \to {a^2} – 4.1 > 0\\ \to \left( {a – 2} \right)\left( {a + 2} \right) > 0\\ \to \left[ \begin{array}{l}a > 2\\a < – 2\end{array} \right.\\Có:\\{\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right)^2} + {\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right)^2} > 7\\ \to \dfrac{{{x_1}^4 + {x_2}^4}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\ \to \dfrac{{{x_1}^4 + {x_2}^4 + 2{x_1}^2{x_2}^2 – 2{x_1}^2{x_2}^2}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\ \to \dfrac{{{{\left( {{x_1}^2 + {x_2}^2} \right)}^2} – 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\ \to \dfrac{{{{\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2}} \right]}^2} – 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\ \to \dfrac{{{{\left( {{a^2} – 2} \right)}^2} – 2}}{1} > 7\\ \to {a^4} – 4{a^2} + 4 – 2 > 7\\ \to {a^4} – 4{a^2} – 5 > 0\\ \to \left( {{a^2} – 5} \right)\left( {{a^2} + 1} \right) > 0\\ \to {a^2} – 5 > 0\left( {do:{a^2} + 1 > 0\forall a \in R} \right)\\ \to \left( {a – \sqrt 5 } \right)\left( {a + \sqrt 5 } \right) > 0\\ \to \left[ \begin{array}{l}a > \sqrt 5 \\a < – \sqrt 5 \end{array} \right.\\KL:\left[ \begin{array}{l}a > \sqrt 5 \\a < – \sqrt 5 \end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
a > \sqrt 5 \\
a < – \sqrt 5
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm
⇒Δ≥0
\(\begin{array}{l}
\to {a^2} – 4.1 > 0\\
\to \left( {a – 2} \right)\left( {a + 2} \right) > 0\\
\to \left[ \begin{array}{l}
a > 2\\
a < – 2
\end{array} \right.\\
Có:\\
{\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right)^2} + {\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right)^2} > 7\\
\to \dfrac{{{x_1}^4 + {x_2}^4}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{{x_1}^4 + {x_2}^4 + 2{x_1}^2{x_2}^2 – 2{x_1}^2{x_2}^2}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{{{\left( {{x_1}^2 + {x_2}^2} \right)}^2} – 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{{{\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2}} \right]}^2} – 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{{{\left( {{a^2} – 2} \right)}^2} – 2}}{1} > 7\\
\to {a^4} – 4{a^2} + 4 – 2 > 7\\
\to {a^4} – 4{a^2} – 5 > 0\\
\to \left( {{a^2} – 5} \right)\left( {{a^2} + 1} \right) > 0\\
\to {a^2} – 5 > 0\left( {do:{a^2} + 1 > 0\forall a \in R} \right)\\
\to \left( {a – \sqrt 5 } \right)\left( {a + \sqrt 5 } \right) > 0\\
\to \left[ \begin{array}{l}
a > \sqrt 5 \\
a < – \sqrt 5
\end{array} \right.\\
KL:\left[ \begin{array}{l}
a > \sqrt 5 \\
a < – \sqrt 5
\end{array} \right.
\end{array}\)