Algebra help?

Question

For what value of k are the roots of 3x^2-6x+k=0

Answer 1

For what value of k are the roots what? You didn't finish the question.

3x² - 6x + k = 0
3(x² - 2x) = -k
3(x² - 2x + 1) = -k + 3
3(x - 1)² = 3 - k
(x - 1)² = (3 - k)/3
x - 1 = ±√[(3-k)/3]
x = 1 ± [√(9 - 3k)] / 3
x is real for 9 - 3k ≥ 0, 9 ≥ 3k, 3 ≥ k
if you want x rational, you have to analyze what perfect squares have the form 9 - 3k = 3(3-k)