Thread: Roots of polynomials

My techer asked me to submit the answer for this question by tomorrow.
I cracked my brain for 2hrs but it did'nt help me.

Question: If the 2 roots of ax³+bx²+cx+d=0 are equal, find the relation between a,b,c,d?

consider the roots x,y,z. x=y. then use the formula. x+y+z=-b/a and xy+yz+zx=c/a and xyz=-d/a. simplify and u will get the answer

thats a third grade equation, it can be up to 3 answers

Originally Posted by anandsatya

My techer asked me to submit the answer for this question by tomorrow.
I cracked my brain for 2hrs but it did'nt help me.

Question: If the 2 roots of ax³+bx²+cx+d=0 are equal, find the relation between a,b,c,d?

Well ashwini9688's solution may be correct but perhaps this is more helpful. Multiply out the followng:

(x - u)(x - u)(v x - w)

then equate the coefficients of the x^3 term with a and of the x^2 term with b and of the x term with c and the constant term with d. That will give what we call a parametric relationship between a, b, c and d. You can eliminate the u, v and w to get the direct relationships between a, b, c and d. But since this involves the algebraic solution of a cubic equation I don't know if this last step is worth the trouble, I would just stick with the parametric relations, or a least eliminate only v and w, but not v.

Unsolvable in my opinion. you need to know more. There are three possible roots, and even if you knew all three, they could all equal any constant, and you probably wouldnt even be able to find a, b, c, and d then.

your's question will be difficult i cant answer this
if u know the answer plese mail the answer to my persional mail

mail id is: reddaiah224@gmail.com

Originally Posted by anandsatya

Question: If the 2 roots of ax³+bx²+cx+d=0 are equal, find the relation between a,b,c,d?

I think you made a mistake (or a typo) there: I think you mean if the 3 roots of the equation are equal. If only two are equal, it's unsolvable - even if you hand-pick a value for the repeated root, you still won't be able to solve it.

Say, suppose the repeated root is 0 and the other root is k. Then you have c = d = 0 and k = -b/a, and that's all you can do. You still can't get any relation between a and b in this particular case.

However, if all three roots are equal, then you can indeed get a relation between the coefficients. You have







and you can eliminate k to get 9ad = bc.

Suppose the roots of the cubic equation are all real and positive.

Prove that .

8)

Originally Posted by JaneBennet

Prove that

I'm not sure the inequality is correct.

I've seen something like that before, you use the fact that the arithmetic mean of a set of positive integers is never less than their geometric mean.

Let the three positive roots be . So









Also







However you can only conclude that and are both greater than or equal to , you can't conclude that .

For you need to use the Cauchy–Schwarz inequality.

How?

Cauchy–Schwarz says that



Now use to complete the job. 8)

Originally Posted by JaneBennet

Cauchy–Schwarz says that



Now use to complete the job. 8)

you can start with 0 =< (a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2)-2(ab+bc+ca)