It is known that x1 and x2 are roots of the equation 3x^2+2x+k=0, where 2x1=−3x2.

Question

Please find k.

Answer 1

3x^2+2x+k=0

use the quadratic formula

(-2±√(4-12k))/6

(-2+√(4-12k))/6 and (-2-√(4-12k))/6 are the roots

2(-2+√(4-12k))/6=(-3)(-2-√(4-12k))/6 because 2x1=-3x2

simplify

(-2+√(4-12k))/3=(2+√(4-12k))/2

cross multiply

-4+2√(4-12k)=6+3√(4-12k)

simplify

-4-6=3√(4-12k)-2√(4-12k)

-10=√(4-12k)

square both sides

100=4-12k

100-4=-12k

96=-12k

k=96/(-12)

k=-8 answer

3x^2+2x-8=0 is the equation

roots are (-2±√100)/6

(-2±10)/6

-12/6 and 8/6

x=-2 and x=4/3

check: 2(-2)=-3(4/3), -4=-4