If (x^4)+(ax^3)+(bx^2)+(cx)+81=0 has 4 real roots, then...

[Akshar]

If the equation (x^4)+(ax^3)+(bx^2)+(cx)+81=0 has four real roots, then

(1) a>= 12
(2) a<= -12
(3) b>= 54
(4) c<= -108

I am a resident of India. I am currently preparing for IIT-JEE. Being in the eleventh grade, in my institute FIITJEE, we have to attempt AITS(All India Test Series). In the examination, I failed to solve this question and that's why I'm asking for help.

Here are my failed attempts:sad: on solving this question:

This is a bi-quadratic equation, so i differentiated it twice to make it a quadratic equation. Then,i thought that since it has real roots, thus D<=0.But, i don't got anything about the answer through that way. Moreover, I also tried to factorize the equation but it was of no help. I also tried to substitute the value of x in terms of some variable y so that i could convert it into quadratic equation. Moreover, I also divided the whole equation by x^2 to see if anything helps. However, I wasn't able to figure out anything meaningful.

I have also got the key of the paper and its says (a)option is the correct answer. But I still couldn't figure out how they've got the answer.

 

[stapel]

If the equation (x^4)+(ax^3)+(bx^2)+(cx)+81=0 has four real roots, then

(1) a>= 12
(2) a<= -12
(3) b>= 54
(4) c<= -108

If this has four real-valued roots (zeroes, solutions), then the graph of the related function crosses the x-axis in four places. Think about that graph. How many turning points must it have? What must be the sign of the y-values for those turning points? How many inflection points must it have?

...the key..says (a) is the correct answer.

Um... not sure which of (1), (2), (3), and (4) you mean by "(a)"...?

If you work with the inflection-points equation, you'll see that (3) results in (1).

 

[JeffM]

If the equation (x^4)+(ax^3)+(bx^2)+(cx)+81=0 has four real roots, then

(1) a>= 12
(2) a<= -12
(3) b>= 54
(4) c<= -108

I am a resident of India. I am currently preparing for IIT-JEE. Being in the eleventh grade, in my institute FIITJEE, we have to attempt AITS(All India Test Series). In the examination, I failed to solve this question and that's why I'm asking for help.

Here are my failed attempts:sad: on solving this question:

This is a bi-quadratic equation, so i differentiated it twice to make it a quadratic equation. Then,i thought that since it has real roots, thus D<=0.But, i don't got anything about the answer through that way. Moreover, I also tried to factorize the equation but it was of no help. I also tried to substitute the value of x in terms of some variable y so that i could convert it into quadratic equation. Moreover, I also divided the whole equation by x^2 to see if anything helps. However, I wasn't able to figure out anything meaningful.

I have also got the key of the paper and its says (a)option is the correct answer. But I still couldn't figure out how they've got the answer.

First, I thank the student for giving information about his background and his thoughts on the problem. At last we have someone who read READ BEFORE POSTING.

Second, someone else needs to give this a try because I have failed to find a proof for the student. This may be my lack of skill, but it may also be that the statement of the problem is not complete and exact. It would be helpful if the student checked that the problem is stated exactly as posed, e.g., four distinct real roots, or four real roots, two positive and two negative. I keep suspecting that more information than was given is required.

Third, some thoughts that may (or may not) be helpful.

\(\displaystyle p,\ q,\ r,\ s \in \mathbb R\ and\ pqrs = 81\ and\ p < q < r < s\ and\ f(x) = (x - p)(x - q)(x - r)(x - s) \implies 0 < p\ or\ s < 0\ or\ q < 0 < r\ and\)

\(\displaystyle f(x) = x^4 - (p + q + r + s)x^3 + (pq + pr + ps + qr + qs + rs)x^2 - (pqr + pqs + prs + qrs)x + pqrs = x^4 + ax^3 + bx^2 + cx + 81.\)

Thus, we can express all the coefficients in terms of the roots.

\(\displaystyle f'(x) = (x - q)(x - r)(x - s) + (x - p)(x - r)(x - s) + (x - p)(x - q)(x - s) +(x - p)(x - q)(x - r) \implies\)

\(\displaystyle f'(p) < 0,\ and\ f'(q) > 0,\ and\ f'(r) < 0,\ and\ f'(s) > 0 \implies\)

\(\displaystyle \exists\ t,\ u,\ v \in \mathbb R\ such\ that\ p < t < q < u < r < v < s\ and\ f'(t) = f'(u) = f'(v) = 0\ and\ f''(t) > 0 < f''(v)\ and\ f''(u) < 0 \implies\)

\(\displaystyle \exists\ m,\ n,\ \in \mathbb R\ such\ that\ p < t < m < u < n < v < s\ and\ f''(m) = f''(n) = 0.\)

\(\displaystyle But\ f'(x) = 4x^3 + 3ax^2 + 2bx + c \implies f''(x) = 12x^2 + 6ax + 2b \implies 36a^2 > 96b \implies 3a^2 > 8b \implies |a| >\sqrt{\dfrac{8b}{3}} \implies b \ge 0.\)

I messed around with some other ideas, but did not see anything that led anywhere.

 

[JeffM]

from expanding it out I found, as you did, that

\(\displaystyle pqrs = 81\mbox{ and }a=-(p + q + r + s)\)

let p=q=r=s=3

then pqrs=81 and a=-12

on the other hand let p=q=r=s=-3

then pqrs=81 and a=12

with the information we're given both of these are valid solutions and correspond to two different choices on the exam.

Yes I found that too, but I have been reading the question to mean four distinct real roots. And it is weird to give a multiple choice exam with two correct answers. But thanks for at least checking my work.

 

[Akshar]

Thanks Everyone!

Thanks to everyone for your support! Sorry but I forgot to tell you that this question was of multi-correct type i.e., one or more options may be correct. So, your deduction was correct. Once, again I thank you all for your support. I can really count on you!