Question

Find all the real values of x $ϵ$ (-2 , 3) which sastify the equation
$(x^2+x+1{)}^2+1=(x^2+x+1\left)\right(x^2+x+5)$

Answer 1

On dividing by $x^2$ $+x+1$ we have
$y+\frac{1}{y}=x^2$ $-x-5$
where $y=x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}=+ive$
Also x $ϵ$(-2,3) $\Rightarrow$ (x + 2)(x - 3) < 0
or $x^2$ $-x-6<0$
or $x^2$ $-x-5<1$ or $y+\frac{1}{y}<1$
Now, $y+\frac{1}{y}\ge 2$
A.M. $\ge$ G.M.
$\therefore$ L.H.S. $\ge$ But L.H.S. < 1 by (1) and (2)
Thus the equality can never arise. Hence no value of x $ϵ$(-2,3) exists which satisfies the given equation.