Question

explain why the function f(x) = x^3 +2x-4 only has a root

Answer 1

because it is a depressed cubic

Answer 2

LOL what does that mean?

Answer 3

Not sure if I understand the question.

There are 3 roots, 1 real, 2 complex

Answer 4

A depressed cubic was extensively analyzed by Italian mathematicians in the 16th century and their work was picked up by Descartes and euler in the following century.
a standard cubic is in the form

ax^3+bx^2+cx+d
a depressed cubic has no x^2 term in it

Answer 5

LOL thank you mr historian!

Answer 6

suppose we have three real solutions.
\[x1+x2+x3=0\]
\[x1.x2+x2.x3+x3.x2=2 \]
so
\[(x1+x2+x3)^{2}=x1^{2}+x2^{2}+x3^{2}+2(x1x2+x2x3+x3x1)=0\]
so
\[x1^{2}+x2^{2}+x3^{2}+2(2)=0\]
that's impossible so our soulutions are not real,then just one is real

Answer 7

suppose that this function has 2 root, we call them that are \[x_1\] and \[x_2\].
Because they are root, we have \[x^3_1+2x_1-4=x^3_2+2x_2-4\]
Then we can deduce that \[x_1=x_2\]