# Thread: simulatenous equations from quartic root polynomial

1. ## simulatenous equations from quartic root polynomial

the question was to find the roots of this equation
$8x^4-2x^3-27x^2+6x+9$
when two of the roots equalled zero.
so i let the roots be $\alpha, \beta, \gamma, \delta$
therefore $\alpha+\beta = 0$
i used the standard sum of roots, product of roots etc technique to work out simultaneous equations and i ended up with this result
$-2\alpha^2\gamma + 8\alpha^2\gamma^2 = 9$
$-8\alpha^2+2\gamma-8\gamma^2=-27$
$-4\alpha^2\gamma-\alpha^2+4\alpha\gamma^2=-3$

i'm not realli sure if those equations are correct but if they are, i dont realli no how to proceed to figure out gamma and alpha
if anyone could help me, i would realli realli appreciate it !!! thankyou !!

2. Originally Posted by THSKluv
the question was to find the roots of this equation
$8x^4-2x^3-27x^2+6x+9$
when two of the roots equalled zero.
so i let the roots be $\alpha, \beta, \gamma, \delta$
therefore $\alpha+\beta = 0$
i used the standard sum of roots, product of roots etc technique to work out simultaneous equations and i ended up with this result
$-2\alpha^2\gamma + 8\alpha^2\gamma^2 = 9$
$-8\alpha^2+2\gamma-8\gamma^2=-27$
$-4\alpha^2\gamma-\alpha^2+4\alpha\gamma^2=-3$

i'm not realli sure if those equations are correct but if they are, i dont realli no how to proceed to figure out gamma and alpha
if anyone could help me, i would realli realli appreciate it !!! thankyou !!
The rational root theorem suggests you try +/-1/2, +/-1/4, +/-1/8, +/-3/2, +/-3/4, +/-3/8, +/-9, +/-9/2, +/-9/4 and +/-9/8 as possible roots.

Of these -1/2 and 3/4 are roots.

CB

3. thankyou but i actually havent learnt the rational root theorem and im supposed
to be solving these equations through the equations for sum of roots, sum of product of roots two at a time, sum of product of roots three at a time and product of roots
thanks though
if anyone can help me with my above equations i would realli appreciate it !!

4. Hi

You know
a + b + c + d = 1/4
ab + ac + ad + bc + bd + cd = -27/8
abc + abd + acd + bcd = -3/4
abcd = 9/8
a + b = 0

This means
c + d = 1/4
ab + cd = -27/8 (since ac + ad + bc + bd = (a+b)(c+d) = 0)
ab(c + d) = -3/4 (since acd + bcd = (a+b)cd = 0)
abcd = 9/8
a + b = 0

You can find
ab = -3 (which gives you a and b using a+b=0)
cd = -3/8 and c+d = 1/4 which gives you c and d

5. wow thankyou SO much !!!
u made the qs so much easier for me
thanks again !! realli appreciate it =DD