Question

Giải phương trình sau:

4x^4 - 7x^2 - 5x - 1 = 0

 

Answer 1

\(4x^4-7x^2-5x-1=0\) 

<=> \((4x^4+4x^3+x^2)-\left(4x^3+4x^2+x\right)-\left(4x^2+4x+1\right)=0\) 

<=> \(x^2\left(4x^2+4x+1\right)-x\left(4x^2+4x+1\right)-\left(4x^2+4x+1\right)=0\) 

<=> \(\left(4x^2+4x+1\right)\left(x^2-x-1\right)=0\) 

<=> \(\left(x+1\right)^2\left(x^2-x-1\right)=0\) => \(\orbr{\begin{cases}x+1=0\\x^2-x-1=0\end{cases}}\) 

(+) \(x+1=0=>x=-1\) 

(+) \(x^2-x-1=0\) 

=> \(x_1=\frac{1+\sqrt{5}}{2};x_2=\frac{1-\sqrt{5}}{2}\)