Show that the equation x^5+2x^3+3x=7 has at most one solution?

Question

If anyone could help me out with this question, that would be super!

Could you also show your work so that I may learn how to do it
Thanks!

Answer 1

Hi,
x^5+2x^3+3x=7
x^5+2x^3+3x-7=0
Now, if we use Descartes' rules of signs, there is only one sign change in the function. So, it has only one possible positive REAL root.
Then if we try f(-x) and see how many sign changes that has.
f(-x) = -x^5 -2x^3-x -7 (Substitute "-x" for each x.)
So, there are no possible negative REAL roots.
So, there is only one real root, but, incidentally, it has also has four complex roots.

Just in case you don't remember Descartes' rule of signs, I'll paraphrase it:
In a polynomial with real coefficients, the number of possible positive real roots is equal to the number of sign changes or less than that by an even number.
For f(-x) (substituting -x for x in the polynomial) the same thing applies for the number of negative roots.

FE

Answer 2

Call y= x^5+2x^3+3x-7
The derivative is y´= 5x^4+6x^2+3 is always positive(all summands are positive)
so the function is increasing and as lim y x==> -infinity is - infinity and lim y x==>+infinity is positive(+infinity) there is only one root.
By the way.The expression is NOT odd
you call a function odd if f(-x) = -f(x)
In this case all terms but - 7changes sign
f(1)= 1+2+3-7=-4
f(-1)= -1-2-3-7=-10

Answer 3

P(x) =x^5+2x^3+3x-7 =0

This is a polynomial of order 5 (odd degree property of real polynomials) with real coefficients.

It has at least 1 real root

Answer 4

its' an odd function- the powers (exponents)
add up to an odd number so it cuts the x axis - so it has 1 solution.
you might want to see www.purplemath.com for other ideas