Given the following biquadratic equation:
$$x^4-3x^3-2x^2-3x+1=0$$
Find the sum of its real roots.
Let $$f(x)=x^4-3x^3-2x^2-3x+1$$
By observing the behaviour of $f^{'}(x)$, I was able to deduce that the equation has only two real roots, both of then being positive.
Also, using a graphing calculator, the sum is found to be approximately $4$. The answer ia 4.
By Vieta's theorem the sum of roots comes out to be 3. So the sum of the non real roots must be -1.
One potentially useful representation of the equation(I have no idea how it is actually useful) was
$$(x^2-1)^2=3(x^2+1)$$, which clearly shows x cannot be negative, if it is to be a solution.
How do I proceed, without referring to the results found by the graphing calculator.
Best Answer
You can have $$x^4-3x^3-2x^2-3x+1=(x^2-4x+1)(x^2+x+1)$$ with $$x^2+x+1=\left(x+\frac 12\right)^2+\frac 34\gt 0.$$