$\displaystyle \text{If }1,\,2,\,3 \text{ are three of the roots of }\,x^4 + ax^2 + bx + c \:=\:0,\;\text{find }a + c$.

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- Feb 1st 2007, 10:29 AMSorobanQuickie #18

$\displaystyle \text{If }1,\,2,\,3 \text{ are three of the roots of }\,x^4 + ax^2 + bx + c \:=\:0,\;\text{find }a + c$.

- Feb 1st 2007, 11:37 AMtopsquark
Again, "bulling" my way through the problem:

Let the fourth root be "d". Then

$\displaystyle (x - 1)(x - 2)(x - 3)(x - d) = x^4 + (-d - 6)x^3 + (6d + 11)x^2 + (-11d - 6)x + 6d$

So

$\displaystyle x^4 + (-d - 6)x^3 + (6d + 11)x^2 + (-11d - 6)x + 6d = x^4 + ax^2 + bx + c$

Comparing the cubic terms gives:

$\displaystyle -d - 6 = 0$

Thus

$\displaystyle d = -6$

Thus

$\displaystyle x^4 + (-d - 6)x^3 + (6d + 11)x^2 + (-11d - 6)x + 6d = x^4 - 25x^2 + 60x - 36$

Thus $\displaystyle a + c = -25 - 36 = -61$

-Dan - Feb 2nd 2007, 12:53 AMearboth
Hello, Soroban,

I'm not quite certain if I remember correctly, but I believe that there exist a "coefficient theorem of Vieta" (that's the name I learned long, long time ago) which says:

If $\displaystyle x^n+a_1 x^{n-1}+...+a_n=(x-x_1)(x-x_2)...(x-x_n)$ then

$\displaystyle \begin{array}{l} a_1=-(x_1+x_2+...+x_n) \\ a_2=x_1 x_2+x_1 x_3+...+x_2 x_3+ ... +x_{n-1} x_n \\ a_3=-\sum_{i,k,l=1}^{n}{x_i x_k x_l} , (i<k<l) \\ ... \\ a_n=(-1)^n x_1 x_2 x_3 ...x_n \end{array}$

Let r be the missing root of the equation and using this theorem I get:

a = -(1 + 2 + 3 + r) = -6 - r which is zero here: thus r = -6

b = 1*2 + 1*3 + 1*r + 2*3 + 2*r + 3*r = 11 + 6r

c = -(1*2*3 + 1*2*r + 2*3*r) = -6 - 8r

d = (-1)^4 * (1*2*3*r) = 6r

According to my results b + d (these are your a+c) = 11+6r + 6r = 11 +12 r = 11 - 72 = -61

But - as I mentioned before - I'm not sure that this attempt is correct.

EB - Feb 2nd 2007, 08:17 AMSoroban
A great algebraic solution, Dan . . . Nice!

EB, Vieta's theorem was used in my book's solution.

. . . Your work was absolutely correct.

Let $\displaystyle r$ be the fourth solution.

Vieta says: .$\displaystyle 1 + 2 + 3 + r \:=\:0\quad\Rightarrow\quad r \,= \,\text{-}6$

And he says: .$\displaystyle 1\cdot2 + 1\cdot3 + 1\cdot r + 2\cdot 3 + 2\cdot r + 3\cdot r \:=\:a$

Hence: .$\displaystyle a \:=\:6r + 11\:=\:6(\text{-}6) + 11\quad\Rightarrow\quad\boxed{ a\,=\,\text{-}25}$

He also says: .$\displaystyle 1\cdot2\cdot3\cdot r\:=\:c\quad\Rightarrow\quad c \,= \,6r \,=\,6(\text{-}6)\quad\Rightarrow\quad\boxed{c \,=\,\text{-}36}$

Therefore: .$\displaystyle a + c \;=\;(\text{-}25) + (\text{-}36)\;=\;\boxed{\text{-}61}$