# Product + Sum of the Roots

• Jan 22nd 2010, 12:24 PM
Sunyata
Product + Sum of the Roots
Hey guys

Solve the equation$\displaystyle 8x^4-2x^3-27x^2+6x+9=0$ given that the sum of two of its roots is zero.

I have no idea how to solve this. I have never dealt with a quartic before.

• Jan 22nd 2010, 01:27 PM
Dinkydoe
I admit this is not an easy problem.

Let $\displaystyle f(x) = 8x^4-2x^3-27x^2+6x+9$

Suppose $\displaystyle \alpha,\beta$ are roots such that $\displaystyle \alpha+\beta = 0$. This means $\displaystyle \beta = -\alpha$.

Thus we can write $\displaystyle f(x) = q(x)(x-\alpha)(x+\alpha) = q(x)(x^2-\alpha^2)$.

If we let $\displaystyle q(x) = ax^2+bx+c$.

we have $\displaystyle 8x^4-2x^3-27x^2+6x+9 = (ax^2+bx+c)(x^2-\alpha^2)\Rightarrow a = 8, b= -2$

And since $\displaystyle -\alpha^2b = 2\alpha^2 = 6 \Rightarrow \alpha^2 = 3$
And since $\displaystyle -\alpha^2c = -3c = 9 \Rightarrow c = -3$.

Thus we found $\displaystyle f(x) = (8x^2-2x-3)(x^2-3)$

Now you can find the roots of the 2 quadratic factors yourself.
• Jan 22nd 2010, 04:44 PM
Soroban
Hello, Sunyata!

Quote:

Solve the equation: $\displaystyle f(x) \:=\: 8x^4-2x^3-27x^2+6x+9\:=\:0$
given that the sum of two of its roots is zero.

Let the two roots be: .$\displaystyle a\text{ and }-a.$

Then: .$\displaystyle \begin{array}{ccccccc}f(a) = 0\!: & a^4 - 2a^3 - 27a^2 + 6a + 9 &= & 0 & [1] \\ f(\text{-}a) = 0\!: & a^4 + 2a^3 - 27a^2 - 6a + 9 &=& 0 & [2] \end{array}$

Subtract [2] - [1]: .$\displaystyle 4a^3 - 12a \:=\:0 \quad\Rightarrow\quad 4a(a^2-3) \:=\:0$

. . and we have three roots: .$\displaystyle a \:=\:0,\:\pm\sqrt{3}$

. . but $\displaystyle a = 0$ is an exraneous root.

Hence, two of the factors are: .$\displaystyle (x - \sqrt{3})(x + \sqrt{3})$

Dividing $\displaystyle f(x)\text{ by }x^2-3$, we get: .$\displaystyle 8x^2 - 2x - 3$

The equation becomes: .$\displaystyle (x^2-3)(2x+1)(4x-3) \:=\:0$

Therefore, the roots are: .$\displaystyle x \;=\;\pm\sqrt{3},\;-\frac{1}{2},\;\frac{3}{4}$