# 4U Polynomial question HELP !! (1 Viewer)

#### _mysteryatar_

##### New Member
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

#### HeroWise

##### Active Member
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_

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

#### _mysteryatar_

##### New Member
yeah okay can you do it now?

#### fluffchuck

##### Active Member
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

##### 617 pages
Notice that the required answer includes the constant term $c$. That should be a hint - if we were to differentiate the polynomial, we would lose $c$, 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 $c$ in order to find the double root, so let's try a different approach.

Let $P(x) = x^3 + 3ax^2+3bx+c$ and let the roots be $\alpha, \, \alpha , \,\beta$.

By the relationships between the roots and co-efficients,

$2\alpha + \beta = -3a$

$\alpha^2 + 2\alpha\beta = 3b$

$\alpha^2 \beta = -c$

Rearranging,

$a = - \frac{2\alpha + \beta}{3}$

$b = \frac{\alpha^2 + 2\alpha\beta}{3}$

$c=-\alpha^2 \beta$

Now:

$c - ab = \frac{(2\alpha+\beta)(\alpha^2 + 2\alpha\beta)-9\alpha^2\beta}{9} = \frac{2\alpha(\alpha - \beta)^2}{9}$

$2(a^2-b) = 2 \cdot \frac{(2\alpha + \beta)^2-3(\alpha^2 + 2\alpha\beta)}{9} = \frac{2(\alpha - \beta)^2}{9}$

Therefore,

$\frac{c-ab}{2(a^2-b)} = \alpha, \quad (a^2 - b \neq 0)$

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#### BChen65536

##### New Member
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.

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