How do we find the number of distinct real roots of the equation x^4-4x^3+12x^2+x-1=0 without using graph?
1 Answer
This quartic has exactly two real roots.
Explanation:
Given:
x^4-4x^3+12x^2+x-1 = 0
Let:
f(x) = x^4-4x^3+12x^2+x-1
Note that the pattern of signs of the coefficients is
The pattern of signs of the coefficients of
Also note that:
x^4-4x^3+12x^2+x-1
= (x-1)^4 + 6(x-1)^2 + 17(x-1) + 9
Note that the coefficients of all the expressions in
Hence we can deduce that any real zeros of
f'(x) = 4x^3-12x^2+24x+1
color(white)(f'(x)) = 4(x^3-3x^2+3x-1)+12(x-1)+17
color(white)(f'(x)) = 4(x-1)^3+12(x-1)+17
which is positive for all
Hence
Now let's see the graph...
graph{x^4-4x^3+12x^2+x-1 [-1.2, 1.2, -1.25, 1.25]}