If the roots are m and 2m, then
(x-m)(x-2m) = x^2-3m+2m^2
so 2m^2 = 12, and m^2 = 6
(x-√6)(x-2√6) = x^2 - 3√6 x + 12
Looks like k = ±3√6
In the equation x^2+kx+12 one root is twice the other root. What are the value(s) of k?
2 answers
or
using x^ - (sum of roots)x + (product of roots) = 0 property
using ooblecks roots of m and 2m
sum of roots = m+2m = 3m
product of the roots = 2m^2
2m^2 = 12 or m = ±√6
k = -3m = ±3√6
using x^ - (sum of roots)x + (product of roots) = 0 property
using ooblecks roots of m and 2m
sum of roots = m+2m = 3m
product of the roots = 2m^2
2m^2 = 12 or m = ±√6
k = -3m = ±3√6
1 answer