Polynomials Help (1 Viewer)

[Comeeatmebro]

Hey guys, came across a couple of questions where I had no idea what to do:

1. Find the range of values of k for which f(x) = x^3 -6kx +2 has 3 DISTINCT real roots

2. Prove that x^3 + 3mx + n = 0 has a double root when n^2 = -4m^3

Any ideas how to do these questions? For Q1, i tried finding turning points so product of turning points would be less than 0, but it didn't work out too well.

Thanks for the help chaps

 

[funnytomato]

Hey guys, came across a couple of questions where I had no idea what to do:

1. Find the range of values of k for which f(x) = x^3 -6kx +2 has 3 DISTINCT real roots

2. Prove that x^3 + 3mx + n = 0 has a double root when n^2 = -4m^3

Any ideas how to do these questions? For Q1, i tried finding turning points so product of turning points would be less than 0, but it didn't work out too well.

Thanks for the help chaps

for question1 , find f'(x) and solve for x , you get one -ve and one +ve solution
once you draw the graph of it , you'd realise f(x) is greater than zero at the -ve soln
and smaller than 0 at the +ve soln
Then you solve these two inequalities

 

[funnytomato]

for the next question,
since it's got a double,then for some value of x , p(x) and p'(x) must both be zero

differentiate p(x) to get p'(x), solve p'(x) , you'd get x^2 = -m

x^3 + 3mx = -n
square both sides, then sub in x^2 = -m

 

[kooliskool]

for question1 , find f'(x) and solve for x , you get one -ve and one +ve solution
once you draw the graph of it , you'd realise f(x) is greater than zero at the -ve soln
and smaller than 0 at the +ve soln
Then you solve these two inequalities

Erm, after you solve for x being the turning point, you don't know whether the solution is -ve or +ve.......... It's going to be in terms of k.

Anyway, here is the solution for it:




Then sub this back into , you will get:




and




You want the product of and less than 0, because if there are three distinct roots, the y-coordinates of the turning points should be one positive and one negative, not x.

Hence:
Notice, difference of squares, so




Anyone want to check the answer with me?

 

[Comeeatmebro]

for the next question,
since it's got a double,then for some value of x , p(x) and p'(x) must both be zero

differentiate p(x) to get p'(x), solve p'(x) , you'd get x^2 = -m

x^3 + 3mx = -n
square both sides, then sub in x^2 = -m

brilliant method mate, thanks a lot, what I did was let x=root(-m) then sub it in... etc, yeah pretty messy

 

[Drongoski]

Anyone want to check the answer with me?

Answer correct. Very nice solution.

 

[funnytomato]

Erm, after you solve for x being the turning point, you don't know whether the solution is -ve or +ve.......... It's going to be in terms of k.
Anyway, here is the solution for it:




Then sub this back into , you will get:




and




You want the product of and less than 0, because if there are three distinct roots, the y-coordinates of the turning points should be one positive and one negative, not x.

Hence:
Notice, difference of squares, so




Anyone want to check the answer with me?

I basically did it the same way as you

what I was thinking is that if one were to draw the graph of this function, since its leading coefficient is +ve , has 3 distinct roots. it could be seen that f(a)>0 and f(b)<0, where a<b, f'(a)=0, f'(b)=0
( i.e. the "extreme point" on the left is +ve, the one on the right is -ve)

so after evaluating the function values at these 2 points, which are line 4 and 6 in your working, since -sqrt(2k) < sqrt(2k)

f( -sqrt(2k) ) > 0
f( sqrt(2k) ) < 0

then I solved these 2 inequalities and got the same answer, but your way is easier since you only had to solve 1 inequality

 

[funnytomato]

it was meant to be a < b before the brackets