Question

(a) x${}^4$+$\frac{8}{9}$x${}^2$+1 = 3x${}^3$+3x.
(b) (x${}^4$ + 1) - 4x (x${}^2$ +1) - 3x${}^2$ = 0.

Answer 1

We rewrite the equation as
(x${}^4$+1) -3x(x${}^2$+1)+$\frac{8}{9}{x}^2$ =0.
Dividing by x${}^2$ , we get
$(x^2+\frac{1}{x^2})-3(x+\frac{1}{x})+\frac{8}{9}$=0
Now put $x+\frac{1}{x}$ = y so that $(x^2+\frac{1}{x^2})$ +2 =y${}^2$
Hence y${}^2$ -2-3y +$\frac{8}{9}$= 0
or 9y${}^2$-27y-10 = 0 or (3y-10) (3y+ 1) = 0.
y = $\frac{10}{3}$ or -$\frac{1}{3}$
Taking y =$\frac{10}{3}$, we get x + $\frac{1}{x}$ =-$\frac{10}{3}$
or 3x${}^2$ -10x + 3 = 0, this gives x = $\frac{1}{3}$, 3.
Again taking y =-$\frac{1}{3}$, we get
x + $\frac{1}{x}$= -$\frac{1}{3}$ or 3x${}^2$ +x+3 = 0
x= $\frac{-1\pm \sqrt{1-36}}{6}$= $\frac{-1\pm \sqrt{-35}}{6}$x =$\frac{1}{3}$, 3, $\frac{-1\pm \sqrt{-35}}{6}$
Ans. x =$\frac{1}{3}$, 3, $\frac{-1\pm \sqrt{-35}}{6}$
(b) Proceed as in part (a).
Ans. x =$\frac{5\pm \sqrt{21}}{6}$, $\frac{-1\pm \sqrt{-3}}{2}$