Solve x4 + x3 + x2 + x = 4.553.

Step-by-step solution

1. 1

   Factor to count the real roots first. Since

   $g(x)=x^4+x^3+x^2+x=x(x+1)(x^2+1)$

   and $x^2+1>0$, the function is negative only on $(-1,0)$ and rises like $x^4$ on both sides. A horizontal line at height $4.553>0$ therefore meets the graph exactly twice: once on the right branch and once on the left. Knowing this in advance prevents stopping after one root.

2. 2

   Bracket the positive root. $g(1)=1+1+1+1=4$, below the target, and $g(1.1)=1.4641+1.331+1.21+1.1=5.1051$, above it. So the root is in $(1,1.1)$.

3. 3

   Refine it. Bisecting (or one Newton step from $1.05$) gives

   $x_1=1.0524735$

   Check: $g(1.0524735)=4.55300$. The often-quoted $1.054381$ gives $g=4.5742$, too big by $0.021$ — the derivative here is about $11$, so a $0.002$ error in $x$ shows up clearly.

4. 4

   Bracket the negative root. $g(-1.69)=4.4966$ and $g(-1.70)=4.6291$, straddling $4.553$. So the second root lies in $(-1.70,-1.69)$.

5. 5

   Refine it.

   $x_2=-1.6942810,\qquad g(x_2)=4.55300$

   A reported value of $-1.853257$ is far outside this bracket — indeed $g(-1.853257)=7.0124$, which is not close to $4.553$ at all.

6. 6

   State the answer. The real solutions are

   $x\approx 1.0524735\quad\text{and}\quad x\approx -1.6942810$

   and the remaining two roots form a complex-conjugate pair, since a quartic has four roots in total.