1. ## newtons method(finding solutions)

we first used newtons method to find the approximate value of 200^(1/7) and got it equal to the awnser in the caclulator which is = 2.131663117

i need some help in the right direction on how to solve the question after solving the above equation using newtons method!

Question:
the fundamental theorem of algebra guarantees that the equation x^7-200 = 0 must have 7 solutions. You have just found one of these. Are there any more real solutions? Use your tools of calculus to sustain your awnser.

2. Do you know Descartes rule of sign?

3. ## k

ok so in other words using decartes rule states that
x^(7)-200=0

has only 1 possible positive real zero? the other 6 are imaginary/complex values?

4. Yes because we obtain 1 (+), 0 (-), so the last solutions are 3 pairs of complex congujates

5. ## y

thx

6. I don't think you need DesCarte's rule of signs. Just observe that the derivative of $\displaystyle x^7- 200$, $\displaystyle 7x^6$ is always non-negative so it is a "one to one" function on the real numbers.

7. Originally Posted by dwsmith
Yes because we obtain 1 (+), 0 (-), so the last solutions are 3 pairs of complex congujates
Descartes rule of signs tells you there is exactly one positive solution. Substitute $\displaystyle u=-x$ and you get $\displaystyle -u^7-1$ and then Descartes rule of signs tells you there are no positive solutions for $\displaystyle u$ and so no negative solutions for $\displaystyle x$ to the original equation. Hence there is exactly 1 real solution to $\displaystyle x^7-1=0$.

But the question requires that you use HallsofIvy's method not the rule of signs.

CB