Ta có: \(x^4-\left(a^2+1\right)x^2+4a^2=0\)(1)
Đặt \(b=x^2\)
Phương trình sẽ trở thành là: \(b^2-\left(a^2+1\right)b+4a^2=0\) (2)
\(\Delta=\left(a^2+1\right)^2-4\cdot1\cdot4a^2=a^4+2a^2+1-16a^2\)
\(=a^4-14a^2+1\)
Theo Vi-et, ta có: \(b_1+b_2=-\frac{b}{a}=a^2+1;b_1b_2=4a^2\)
Để (1) có nghiệm thì (2) phải có hai nghiệm không âm
=>\(b_1b_2\ge0;b_1+b_2\ge0;\Delta\ge0\)
=>\(\begin{cases}a^2+1\ge0\\ 4a^2\ge0\\ a^4-14a^2+1\ge0\end{cases}\Rightarrow a^4-14a^2+1\ge0\)
=>\(a^4-14a^2+49\ge48\)
=>\(\left(a^2-7\right)^2\ge48\)
=>\(\left[\begin{array}{l}a^2-7\ge4\sqrt3\\ a^2-7\le-4\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2\ge4\sqrt3+7\\ a^2\le7-4\sqrt3\end{array}\right.\)
=>\(\left[\begin{array}{l}a^2\ge\left(2+\sqrt3\right)^2\\ a^2\le\left(2-\sqrt3\right)^2\end{array}\right.\Rightarrow\left[\begin{array}{l}a\ge2+\sqrt3\\ a\le-2-\sqrt3\\ -2+\sqrt3\le a\le2-\sqrt3\end{array}\right.\)
