Hello sorry for the lengthy question. I was preparing for my exams and I came across a question where it required me to count the number of real roots for the expression: x^3 - 300x - 3000 = 0. I used the method described in the video and I came up with 1 positive real root, and 2 negative real roots. The answer was 1 real root only. I do realize that it is possible for there to be either 2 negative real roots or 0 negative real roots, but i'd like to ask how do we determine when to subtract? in this case it is not as obvious to me because 1 positive and 2 negative real roots add up to exactly 3 number of real roots
For f(x), you get 1 sign change so 1 positive root For f(-x), you get 2 sign changes so 2 negative roots or 0 negative roots When you say when do we determine when to subtract? You set up a table with the possibilities as done in the video. It tells you what is possible. Here it is possible to have 2 negative roots or 0 roots and you see you get 0 negative roots.
If you did not get it, it is because you can not have a negative number of negative solutions. Therefore by Descartes rule of signs we must have 1 negative solution since 1-2= -1 and again we cannot have a negative number or solutions.
THANK YOU SO MUCH, I AM SO CONFUSED ON WHAT TO DO AND THIS HELPED ME
Glad it helped!
Thankyou So much you helped me a lot.It was easy after u explained it.
You are welcome! 😎
What do you do when you dont have a number with no x at the end?
What's the example you are working with?
It was a very helpful video to learn decarts rule of sign....
glad you found it helpful.
Hello sorry for the lengthy question. I was preparing for my exams and I came across a question where it required me to count the number of real roots for the expression: x^3 - 300x - 3000 = 0. I used the method described in the video and I came up with 1 positive real root, and 2 negative real roots. The answer was 1 real root only. I do realize that it is possible for there to be either 2 negative real roots or 0 negative real roots, but i'd like to ask how do we determine when to subtract? in this case it is not as obvious to me because 1 positive and 2 negative real roots add up to exactly 3 number of real roots
For f(x), you get 1 sign change so 1 positive root
For f(-x), you get 2 sign changes so 2 negative roots or 0 negative roots
When you say when do we determine when to subtract? You set up a table with the possibilities as done in the video. It tells you what is possible. Here it is possible to have 2 negative roots or 0 roots and you see you get 0 negative roots.
Sir in 2nd example why there is only 1 possiblity why other possiblity 0,0 ,4 doesn't exist plzz explain🥺
Did you get this yet?
If you did not get it, it is because you can not have a negative number of negative solutions. Therefore by Descartes rule of signs we must have 1 negative solution since 1-2= -1 and again we cannot have a negative number or solutions.
And 0,0,4 isn't a possibility either since you have a sign change in f(x)
So in short, we just find the amount of possibilities?
Yes, that is the idea. Usually, you are using this as part of a full series of steps designed to allow you to find the zeros of a polynomial function.
Thanks for the help. I understand it now
You're welcome!
Great video! Thank you
Glad you liked it!
Thank you sir 👍
Welcome!
Well done
Thank you for the nice comment! 😎