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?
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
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?
But since the root has a multiplicity of 2 can it be found in the third derivative?
I read third for some reason im blind
yeah okay can you do it now?
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!
HSC 2017
Chemistry 89 - Physics 91 - English (Standard) 87 - Maths Ext 1 99 - Maths Ext 2 98
98.7
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,
Last edited by fan96; 11 Oct 2018 at 10:25 PM.
While the method above is indeed valid, one could also solve the problem using calculus:
Let P(x)=x^{3}+3ax^{2}+3bx+c.
Then P'(x)=3x^{2}+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 3x^{2}+6ax+3b=0, i.e. x^{2}+2ax+b=0.
Now, P(x)=0 for the same value of x,
so x^{3}+3ax^{2}+3bx+c=0,
i.e. x(x^{2}+2ax+b)+ax^{2}+2bx+c=0.
x(x^{2}+2ax+b)+a(x^{2}+2ax+b)-2a^{2}x-ab+2bx+c=0.
But x^{2}+2ax+b=0 (proven above),
so -2a^{2}x-ab+2bx+c=0,
i.e. x(2b-2a^{2})+c-ab=0.
⇒x=(c-ab)/(2(a^{2}-b)) as a^{2}-b≠0.
Hence, the double root must be equal to (c-ab)/(2(a^{2}-b)) as required.
Last edited by BChen65536; 11 Oct 2018 at 10:47 PM.
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