Question
If x^2-3x+1=0 , find the value of x^4+\dfrac{1}{x^4}
Answer
\because x^2-3x+1=0
x+\dfrac{1}{x}=3
x^2+\dfrac{1}{x^2} =7
\therefore x^4+\dfrac{1}{x^4} = 47
If x^2-3x+1=0 , find the value of x^4+\dfrac{1}{x^4}
\because x^2-3x+1=0
x+\dfrac{1}{x}=3
x^2+\dfrac{1}{x^2} =7
\therefore x^4+\dfrac{1}{x^4} = 47