Range of abc (1 Viewer)

[juantheron]

 

[fortune_cookie]

It's fairly obvious it has something to do with p(x)= x^3 -15x^2 +27x +d

We require the above to have three real solutions, which means it's derivative must have two real solutions (for a cubic to cut the x axis three times, it must turn around twice)

p ' (x) = 3x^2 -30x +27 =3x^2 -3x -27x +27 =0

3x(x-1)-27(x-1)=0

(3x-27)(x-1) =0

x= 9 or x=1

Continue from that somehow. I caqn't be bothered to finish it now.

 

[RealiseNothing]

If

Then

Using

We get

But this means

So there isn't any solutions?

I'm sure there is atleast 1 solution though, so I don't know what to do lol.

 

[RealiseNothing]

It's fairly obvious it has something to do with p(x)= x^3 -15x^2 +27x +d

We require the above to have three real solutions, which means it's derivative must have two real solutions (for a cubic to cut the x axis three times, it must turn around twice)

p ' (x) = 3x^2 -30x +27 =3x^2 -3x -27x +27 =0

3x(x-1)-27(x-1)=0

(3x-27)(x-1) =0

x= 9 or x=1

Continue from that somehow. I caqn't be bothered to finish it now.

The third term wouldn't be 27x, it would be 81x.

 

[seanieg89]

Question is vacuous, no triple of real numbers (a, b, c) can satisfy the two given conditions.

 

[Carrotsticks]

Pretty sure OP made a typo as he is infamous for that.

 

[seanieg89]

If the equality signs in the second condition are changed to addition signs, then the question is more interesting. The answer is: -13 =< abc =< 243 using the cubic discriminant.

 

[juantheron]

Sorry friends actually it is

 

[seanieg89]

Then trivially, a=b=c=5, abc=125.

 

[Mr Slick]

Is this in the course? :S

 

[juantheron]

I did not Understand that line

If the equality signs in the second condition are changed to addition signs, then the question is more interesting. The answer is: -13 =< abc =< 243 using the cubic discriminant.

Could you like to explain it to me.

Thanks

 

[seanieg89]

If you change the ab=ac=bc=27 to ab+ac+bc=27 you get a problem that makes sense and is considerably more difficult that the one obtained by replacing 27 with 25.

If we let P(x) be the monic polynomial with roots at a,b,c, then P(x)=x^3-15x^2+27x+c for some real c. The problem amounts to finding the values of c for which P has three real roots. This can be done using a substitution y=(x-5) and the depressed cubic discriminant:

y^3+Ay+B=0 has three real roots <=> 4A^3+27B^2<0

which is easily derived by considering extrema of the function y^3+Ay+B.

Upon applying the cubic discriminant, all that is left is to solve a quadratic inequality.