Question

Miscellaneous Problem.
Calculate $x_1^4,+x_2^4$ where $x_{1}and{x}_2$ are the roots of the equation $3x^2$ - 5x + 1 = 0.

Answer 1

${3x}^2-5x+1=0({ax}^2+bx+c=0)$

Let roots be $x_1$ & $x_2$

Sum of roots i.e. $x_1+x_2=\frac{-b}{a}=-\frac{(-5)}{3}=\frac{5}{3}$

Products of roots i.e. $x_1{x}_2=\frac{c}{a}=\frac{1}{3}$

Now, $(x_1^2+x_2^2)={(x_1+x_2)}^2-2x_1{x}_2$

$={\left(5/3\right)}^2-2\left(1/3\right)=19/9$

$x_1^4+x_2^4={(x_1^2+x_2^2)}^2-2x_1^2{x}_2^2$

$={\left(19/9\right)}^2-2{\left(1/3\right)}^2=343/81$