Question

For what values of k does the equation x^2 - (5k+1)x + 9k^2 = 0 have real roots?

Answer 1

real roots occur when the determinant b^2-4ac >= 0

so substituting:-

(5k + 1)^2 - 4*1* (9k^2) >= 0
25k^2 + 10k + 1 - 36k^2 >= 0
-11k^2 +10k + 1 >= 0
11k^2 - 10k - 1 <= 0
(11k + 1)(k - 1) <= 0

the equation will have real roots for -1/11 =< k <= 1

Answer 2

K has a range from[ -1/11 to 1]