# Number of roots

1. May 4, 2008

### DeanBH

how do i find the number of roots for a curve that has dy/dx

2x^4 -20x^2 + 50.

if i substitute y=x^2 and use the discriminant formula i get

b^2 - 4ac = 400 - 4 x 2 x 50
= 400 - 400
= 0

This way says there 1 root, answers say it has 2. Which method am i meant to use for this?

if i factorise i get 2(y-5)^2

which is also 1?

Last edited: May 4, 2008
2. May 4, 2008

### malty

Do you know of a way to find how many times the curve crosses the x-axis?

Think of info needed to sketch this curve. .

Last edited: May 4, 2008
3. May 4, 2008

### DeanBH

yes i know 2 ways... factorizing and discriminant. and they both say 1 root..

4. May 4, 2008

### malty

Are you familar with critical points (Maxima, Minima, points of inflection . .) of a curve?

It looks to me like you are only finding the roots of the tangent to the curve?

5. May 4, 2008

### HallsofIvy

Are you look for the zeroes of y= 2x^4 -20x^2 + 50 or y such that dy/dx= 2x^4 -20x^2 + 50?

Obviously, 2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)(x^2+ 5) has two real roots- they are $\pm\sqrt{5}$.

But if you mean y such that dy/dx= 2x^4 -20x^2 + 50, there is no way of telling. You lose an additive constant when you differentiate y and how many times y is 0 depends on that constant.

6. May 4, 2008

### DeanBH

2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)(x^2+ 5) Is wrong.?

2(x^2- 5)(x^2+ 5) = 2(X^4 -25) not 2(x^4 - 10x + 25)

7. May 4, 2008

### epenguin

It should be easy to sketch what a curve of dy/dx gainst x and y against x when dy/dx is what you said (in the form you factorised it). The height of the second of these is not determined, as mentioned, unless that was given too in your original problem. But basically you should then see there is only one general possibility, plus one special case.

8. May 4, 2008

### HallsofIvy

Ooops! That's embarrassing! 2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)^2 which has 2 distinct roots, each a double root.

Thanks, DeanBH.

I'm still wondering what the original problem really was!

9. May 4, 2008

### DeanBH

it still makes no sense.

it says find stationary points on the curve blahblahblah.
the point is the curve has dy/dx 2x^4 -20x^2 + 50.

how the hell do i go about finding it has 2.

it looks like it has 1.

10. May 4, 2008

### Hootenanny

Staff Emeritus
Ahh, the question make sense now!

Why do you think that there is only one root? Both you and Halls have shown that it has two roots and therefore two stationary points.

11. May 4, 2008

### rocomath

$$2(x^2-5)^2=0$$

Can you find those 2 roots?

12. May 4, 2008

### DeanBH

that looks like one root to me, why is that 2.

13. May 4, 2008

### Hootenanny

Staff Emeritus
Cancel the two and take the square root of both sides. Does that make it any easier?

14. May 4, 2008

### DeanBH

no.

i dont know what you are talking about

15. May 4, 2008

### rocomath

Solve ...

$$x^2-5=0$$

What do you get?

16. May 4, 2008

### DeanBH

oooooooooooooooooooohhhhhhhhhhhhh craaaaaaaaaaaaaaaappppppppppp>

i was ignoring the fact it was X^2. taking it as X
me being retarded

17. May 4, 2008

### rocomath

Yeah I was getting worried for a sec :p hehe, don't worry about it! We all have brain farts, just pray it isn't during an exam!!!