Question 412586
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Hi
Determine the sum and product of the roots of  
3y^2– 2y + 12 = 0
{{{x = (2 +- sqrt(-140))/(6) }}}
{{{x = (2 +- sqrt(-4*35))/(6) }}}
{{{x = 1/3 +- 2isqrt(35)/6 }}}
{{{x = 1/3 +- isqrt(35)/3 }}}
Roots are {1/3 + isqrt(35), 1/3 -isqrt(35)/3}
Sum of roots is 2/3
Product of roots is [1/3 + isqrt(35)][1/3 - isqrt(35)] = 1/9 + 35 = 35 1/9
                       Note i^2 = -1