4U Polynomial question HELP !! (1 Viewer)

[_mysteryatar_]

If the polynomial x^3+3ax^2+3bx+c=0 has a double root, show that the double root is (c-ab)/(2(a^2-b)) , given that a^2 doesnt equal to b

Please help Im struggling with this question

 

[HeroWise]

I am gonna do it a bit later, a bit busy, but have you tried 3rd derivative getting it to be 0 then testing against second derivative and subbing in the original function?

 

[_mysteryatar_]

But since the root has a multiplicity of 2 can it be found in the third derivative?

 

[HeroWise]

I read third for some reason im blind

 

[_mysteryatar_]

yeah okay can you do it now?

 

[fluffchuck]

can't do the question right now but i assume that this is how you do it:

since x^3 + 3ax^2 + 3bx + c = 0 has a double root then this means that p(x) = x^3 + 3ax^2 + 3bx + c has a double zero. let this double zero be y. then p(y) = 0 and p'(y) = 0. this means that y^3 + 3ay^2 + 3by + c = 0 and 3y^2 + 6ay + 3b = 0 -> y^2 + 2ay + b = 0. now we solve this last equation for the double root, y. u will get 2 solutions (since quadratic), then sub each presumed double root into the original equation (p(y)). whichever results in p(y) = 0 is the double root. good luck!

 

[fan96]

Notice that the required answer includes the constant term . That should be a hint - if we were to differentiate the polynomial, we would lose , so not only would we have to expand the cube of a binomial (which can be really messy), we would also have to rewrite the result to include in order to find the double root, so let's try a different approach.

Let and let the roots be .

By the relationships between the roots and co-efficients,







Rearranging,







Now:





Therefore,

 

[BChen65536]

While the method above is indeed valid, one could also solve the problem using calculus:

Let P(x)=x³+3ax²+3bx+c.

Then P'(x)=3x²+6ax+3b.

For there to be a double root, there must exist some x such that P(x)=0 and P'(x)=0.

P'(x)=0 iff 3x²+6ax+3b=0, i.e. x²+2ax+b=0.

Now, P(x)=0 for the same value of x,

so x³+3ax²+3bx+c=0,

i.e. x(x²+2ax+b)+ax²+2bx+c=0.

x(x²+2ax+b)+a(x²+2ax+b)-2a²x-ab+2bx+c=0.

But x²+2ax+b=0 (proven above),

so -2a²x-ab+2bx+c=0,

i.e. x(2b-2a²)+c-ab=0.

⇒x=(c-ab)/(2(a²-b)) as a²-b≠0.

Hence, the double root must be equal to (c-ab)/(2(a²-b)) as required.