Real and distinct roots of a cubic equation

Question

The real values of $a$ for which the equation $x^3-3x+a=0$ has three real and distinct roots is

I found f'(x) put it equal to 0 got x=1 and -1. I put the values of -1 and 1 in f(x) got x=2 and -2 — kdfghdkljfgh, Jan 30 '14 at 11:54
I dont know maybe we could apply Lagrange's mean value theorem — kdfghdkljfgh, Jan 30 '14 at 11:56
@ kdfghdkljfgh , you did good. You mean y = 2 and -2. Now note that 'a' shifts the graph up or down. You can shift this graph up or down by any number in the interval (-2 , 2) without fear of losing any roots. — neofoxmulder, Jan 30 '14 at 12:18
k i got it now. f(x) is greater than 0 for some range and less than 0 for some range which we found to be a+2 and a-2. Using the theorem, there is a root somewhere in between and hence the solution comes to be |a|<2 or a belongs to (-2,2) — kdfghdkljfgh, Jan 30 '14 at 12:24

score 1 · Answer 1 · answered Jan 30 '14 at 12:12

1

The derivative cancels at $x=1$ and $x=-1$. To these correspond a maximum value of $a+2$ and a minimum value of $a-2$. In order to have three real roots, you need three $x$ intercepts; this means that you must have $a+2>0$ and $a-2<0$. So, the condition is $|a|<2$.

answered Jan 30 '14 at 12:12

Claude Leibovici

260,315

i think i got it now – kdfghdkljfgh Jan 30 '14 at 12:15

score 0 · Answer 2 · answered Jan 30 '14 at 11:58

0

The discriminant of the polynomial is

$$108-27a^2$$

If the discriminant is positive, we have three distint real roots.

So, a must satisfy the condition |a|<2.

answered Jan 30 '14 at 11:58

Peter

84,454

1

what is the discriminant of a polynomial? – kdfghdkljfgh Jan 30 '14 at 11:59
can't we do it using derivative? – kdfghdkljfgh Jan 30 '14 at 12:02
The best known discriminant is $b^2-4ac$ for the polynomial $ax^2+bx+c$, but a discriminant can also be defined for polynomials of higher degree. This discriminant gives informations about the roots. If it is zero, we have multiple roots, for example. – Peter Jan 30 '14 at 12:07
For degree 3, the discriminant tells us immediately, if there is one real root or three. – Peter Jan 30 '14 at 12:08
k read it up on wikipedia but i think derivative would actually give us more answers – kdfghdkljfgh Jan 30 '14 at 12:09
Derivates help to find the bound of the roots. If you want to know the number of real roots in an interval, use the sturm chain of a polynomial. – Peter Jan 30 '14 at 12:11
Yes, actually the answer is given in terms of bound of the roots. – kdfghdkljfgh Jan 30 '14 at 12:13
OK, that's another approach. – Peter Jan 30 '14 at 12:15
1

dont get me wrong, your method gives the answer directly, but I we have not been taught the discriminants of polynomial functions with degree more than 2 – kdfghdkljfgh Jan 30 '14 at 12:18
OK, I understand. – Peter Jan 30 '14 at 12:19

score 0 · Answer 3 · answered Jan 30 '14 at 12:05

0

If you want them to be real and distinct...

Natural choice would be to consider :$$(x-l)(x-m)(x-n)=(x^2-(l+m)x+lm)(x-n)=x^3-(l+m+n)x^2+(lm+n(l+m))x-lmn$$

Equating the coefficients we would have : $$l+m+n=0$$

$$lm+n(l+m)=-3\Rightarrow lm-n^2=-3\Rightarrow -lmn =n(3-n^2)$$

We need to find bound for $-lmn=n(3-n^2)$...

Can you do it by using derivative?

answered Jan 30 '14 at 12:05

if we compare the coefficients we get contradicting values – kdfghdkljfgh Jan 30 '14 at 12:10
who said? what contradicting values are you getting? – Jan 30 '14 at 12:10
i think i made a mistake... correct me if i am wrong... f'(x)=0 for 1 and -1. Doesn't that make them roots of the equation – kdfghdkljfgh Jan 30 '14 at 12:12
what equation are you considering??? you have to make $g'(x)=0$ for $g(x)=x(3-x^2)$ – Jan 30 '14 at 12:13
Just tell me this: f(x)=0 say it is a quadratic equation. Now i find the derivative and put it equal to zero. Say i get two values of x. These would be the roots, right? – kdfghdkljfgh Jan 30 '14 at 12:16
No No.... take $f(x)=x^2+3\Rightarrow f'(x)=2x=0\Rightarrow x=0$ so we should get $f(0)=0$ i.e., $3=0$???? – Jan 30 '14 at 12:25
k got it now. f'(x)=0 gives us some of the critical points where the curve becomes parallel to the x-axis. It is possible that the critical points are roots, but not always. Same goes for the roots. They are not always critical points – kdfghdkljfgh Jan 30 '14 at 12:28
yes but then??? – Jan 30 '14 at 12:29
however the derivative does give us the clue of how the sign of the function changes within a given interval, leading to points of maxima and minima. Between two points of maxima always lies a point of minima. Check above where neofoxmulder tells something like that – kdfghdkljfgh Jan 30 '14 at 12:33
so can you conclude now?? – Jan 30 '14 at 12:34
yeah i did below his comment – kdfghdkljfgh Jan 30 '14 at 12:34

score -1 · Answer 4 · answered Jan 30 '14 at 12:43

-1

Here is a graphical approach. Note that '+ a' shifts the graph up or down.

1) Disregard '+ a'.

2) Graph $ f(x) = x^3 - 3x $

http://www.wolframalpha.com/input/?i=plot%28x%5E3+-+3x%29#

3) from the graph we see that we must find the y values at the local extrema. You did that already getting y = 2 and -2

4) It should be obvious now by inspection that a must be between -2 and 2

Final answer ,

-2 < a < 2

answered Jan 30 '14 at 12:43

neofoxmulder

1,166

Sometimes graphs really put the question to rest. Thanks – kdfghdkljfgh Jan 30 '14 at 12:45

Real and distinct roots of a cubic equation

4 Answers4