OK, that's what I thought. I really like @ibad's approach on the first question, so let me see if I can apply that here as well.
From the polynomial that we started with, we had roots of a & b. So we can say:
\(\Large 2(x^2-\dfrac{3}{2}x-2)=0\) has roots a & b, and so:
\(\Large x^2-\dfrac{3}{2}x-2=(x-a)(x-b)=x^2-x(a+b)+ab\)
Therefore: \(\Large a+b=\dfrac{3}{2}\) and \(\Large ab=-2\)
Now save those, we'll come back to use them later.
Now we want to cook up a polynomial with the roots:
\(\large m=b+\dfrac{ 1 }{a }~and~n=a+\dfrac{ 1 }{b}\)
so that polynomial has, as factors, (x-m) and (x-n)
So we get:
\(\Large (x-m)(x-n)=x^2-x(m+n)+mn\)
so using \(\large m=b+\dfrac{ 1 }{a }~and~n=a+\dfrac{ 1 }{b}\)
we get:
\(\large m+n=b+\dfrac{ 1 }{a }+a+\dfrac{ 1 }{b}=\dfrac{ 2ab+(b+a) }{ ab }\)
and
\(\large mn=(b+\dfrac{ 1 }{a })(a+\dfrac{ 1 }{b})=ab+2+\dfrac{1}{ab}=\dfrac{ (ab)^2+2ab+1 }{ ab }\)
OK now to put it all together:
\(\large m+n=\dfrac{ 2ab+(b+a) }{ ab }=\dfrac{ 2(-2)+\dfrac{3}{2} }{ -2 }=2-\dfrac{3}{4}=\dfrac{5}{4}\)
and
\(\large mn=\dfrac{ (ab)^2+2ab+1 }{ ab }=\dfrac{ 4+2(-2)+1 }{ -2 }=-2+2-\dfrac{1 }{ 2 }=-\dfrac{1 }{ 2 }\)
So putting that into \(\Large x^2-x(m+n)+mn\) yields:
\(\Large x^2-x(\dfrac{5}{4})-\dfrac{1 }{ 2 }\) So we can get:
\(\Large y=4\cdot[x^2-x(\dfrac{5}{4})-\dfrac{1 }{ 2 }]\)
\(\Large y=4x^2-5x-2\)
OK, I'm trying to confirm that this function has the right roots (related to the roots of the original equation), using Wolfram, and it is NOT WORKING so I guess I have an error somewhere. :( But I think the method is sound, I'll look at it and see if I can figure out why it isn't working.