Im very new to further maths and answered the following question -

The quadratic equation3x²-9x+1=0has roots α and β. Find a quadratic equation with integer coefficients and with roots3/α , 3/β

This is how I answered it -

αβ = 1/3

α+β = 9/3 = 3

new sum = 3/α + 3/β

new sum = (3β + 3α) / (1/3)

new sum = 9 / (1/3) =27

new product = 3/α * 3/β

new product = 9/αβ

= 9 / (1/3) =27

x² - 27x + 27 = 0

Worked out right but now I'm stuck with this one. Its the same equation but with different roots -

The quadratic equation3x²-9x+1=0has roots α and β. Find a quadratic equation with integer coefficients and with roots2α , 2β

I have started it out like this but getting the wrong answer -

αβ = 1/3

α + β = 9/3 = 3

new sum = 2α + 2β

new sum = 2(α+β)

= 2 * 3 =6

However this is wrong :|

Could anybody give me any tips on how i answered the first question and help with the second question please

thanks