Not many solutions? Part 1

Algebra Level 2

Let the functions $f(x)$ and $g(x)$ be $f(x)= x^{3}-2x^{2}+x-2$ and $g(x)= x^{2}+1$

Find the sum of all real solutions to the equation

$\frac{f(x)}{g(x)}=1$ .

1 solution

Tarmo Taipale
Jun 18, 2016

The fraction $\frac{f(x)}{g(x)}$ has a value with all real $x$ , because $x^2+1>0$ with all real values of x. We get a new equation:

$f(x)=g(x)$

$x^{3}-2x^{2}+x-2=x^{2}+1$

$x^{3}-3x^{2}+x-3=0$

$(x^{2}+1)(x-3)=0$

$x=3$ or $x^2+1=0$ . The second equation of these two has no real solutions, so we get

$x=3$

The only solution is 3, which means the sum of all real solutions is also $\boxed{3}$ .

Slight error.

The second factor $x^2+1=0$ has no real solutions. It is different from having no solutions.

The last line should also be changed. Write "real solutions" instead of just "solutions"

Hung Woei Neoh - 4 years, 12 months ago