Quadratics: x^2 - 3kx + 1 = 0, complete the square?

Question

Also: 3x^2 - 6x + 12, change to turning point form?


Can you explain it in detail, thanks!!

Answer 1

(x-3k/2)^2 - 9k^2/4 + 1 = 0

x = 3k/2 +- sqrt (9k^2/4 -1) = 3k/2 +- sqrt ([9k^2 - 4]/4))

I don't understand what you mean by 'turning point form'. If you just want to get your coefficient in x^2 term = 1 then complete the square again here we go:

3x^2 - 6x + 12 = 0
x^2 - 2x + 4 = 0
(x-1)^2 + 3 = 0
x= 1 +- sqrt (-3) = 1+- i*sqrt(3)
i.e. no real solutions

Answer 2

x^2 - 3kx + 1 = 0
x^2 - 3kx = -1

Divide the coefficient of x by 2, square it and add it to both sides

-3/2 = -1.5 ==> (1.5)^2 = 2.25

x^2 - 3kx + 2.25 = -1 + 2.25

Factor the left side and simplify the right side

(x - 1.5k) (x - 1.5k) = 1.25
(x - 1.5k)^2 = 1.25
Take the square root of both sides

x - 1.5 = +/- sqrt 1.25
x = 1.5 + sqrt 1.25 and
x = 1.5 - sqrt 1.25

sqrt 1.25 = 1.118 (approximately)

x = 1.5 + 1.118 = 2.618 and
x = 1.5 - 1.118 = 0.382

Answer 3

Question 1
[ x² - 3k x + (3k/2)² ] - (3k/2)² = - 1
(x - 3k/4)² = (3k/2)² - 1
(x - 3k/4)² = 9k²/4 - 1
(x - 3k/4)² = (9k² - 4) / 4
(x - 3k/4) = ±√((3k - 2).(3k + 2)) / 2
x = 3k/4 ±√((3k - 2).(3k + 2) / 2

Question 2
= 3.(x² - 2x + 12)
= 3.(x² - 2x + 1 - 1 + 12)
= 3.((x - 1)² + 11)
= 3.(x - 1)² + 33