Descartes' Rule Of Sign Is Not Enough!

Algebra Level 4

$\large { x }^{ 6 }-12{ x }^{ 5 }+a{ x }^{ 4 }+b{ x }^{ 3 }+c{ x }^{ 2 }+dx+64 =0$

Let $a,b,c,d$ be all constants.

If all of the roots of the above equation are positive, find $b + c - a$ .

The answer is 20.

2 solutions

Prajwal Krishna
Nov 15, 2016

Let $p,q,r,s,t,u$ be the positive roots . Since degree of polynomial is six it has six positive roots.

By vieta's formula , $p+q+r+s+t+u = 12$ and $pqrstu = 64$

Applying arithmetic mean - geometric mean , we obtain

$\frac { p+q+r+s+t+u }{ 6 } \ge { \quad (pqrstu) }^{ \frac { 1 }{ 6 } }$

Observe that $AM = 2$ and $GM = 2$ , so we are in the equality case. This means that all of the variables are equal, thus $p=q=r=s=t=u=2$ .

Hence $a = 15 * 2 * 2 = 60$ , $b = - 20 * 2 * 2 * 2 = -160$ , $c = 15 * 2 * 2 * 2 * 2 = 240$ .
Therefore $b+c-a = 20$ .

Positive and negative is defined only for real numbers

Positive = Numbers greater than zero

Negative = Numbers lesser than zero

But as you would know complex numbers cannot be compared

Prajwal Krishna - 4 years, 6 months ago

Excellent solution sir. I haven't learned about Vieta's formula (but I learned the AMGM inequality instead haha) but you have presented an elegant solution that is simple to understand. Thankyou.

Rico Lee - 4 years, 6 months ago

Thanks for your appreciation

Prajwal Krishna - 4 years, 6 months ago

FYI To start a new line, leave 3 empty spaces at the end of the sentence. I've edited your solution, so that it is easier to read.

Calvin Lin Staff - 4 years, 7 months ago

Nice solution.

I suggest that you replace the phrase "all of the roots of the above equation are positive" by "all of the roots of the above equation are real and positive"

Abdelhamid Saadi - 4 years, 6 months ago

Positive implies real already.

Calvin Lin Staff - 4 years, 6 months ago

The domain of the variable x is not given. so that: "All of the roots" may signify for some people "All of the real roots" and for others "All of the complex roots".

I mean if this equation has only two positive roots and the others are complex, I can say "all of the roots of the above equation are positive".

Maybe we can say: "the above equation has six positive roots"

Abdelhamid Saadi - 4 years, 6 months ago

@Abdelhamid Saadi – That's where I disagree. You cannot add dependent quantifiers when interpreting a statement. The statement "All of the roots of the above equation are positive" means that all of the roots (real or complex) are positive (and hence real).

If the statement was "All of the real roots of the above equation are positive", then I agree that it allows for complex roots.

Calvin Lin Staff - 4 years, 6 months ago

@Calvin Lin – If this is the convention, just ignore my comment.

I was just thinking about people who didn't know about complex numbers yet, since this problem doesn't need knowledge of complex numbers.

Abdelhamid Saadi - 4 years, 6 months ago

Prakhar Bindal
Nov 15, 2016

Sum of 6 positive numbers = 12

Product = 64

Therefore AM = GM . hence all numbers should be equal.

By vieta's We know a = sum of roots taken two at a time = 6C2 * 2* 2 = 60

b = sum of roots taken 3 at a time = -(6C3* 2* 2* 2) = -160

c = sum of roots taken 4 at a time = 6C4* 2* 2* 2* 2 =240

b+c-a = 80-60 = 20

Descartes' Rule Of Sign Is Not Enough!

The answer is 20.

2 solutions

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