2x^2 âˆ’ax + 2 > 0

2{ (x-a/4)^2 – a^2/16+1} > 0 âˆ€ x âˆˆR

-a^2/16+1 > 0

a âˆˆ{ -3,-2,-1,0,1,2,3}

x^2 âˆ’bx + 8 â‰¥ 0

(x-b/2)^2 – b^2 /4 + 8>0 âˆ€ x âˆˆR -b^2 /4 + 8>0

b âˆˆ{-5,-4,-3,-2,-1,0,1,2,3,4,5 }

So largest possible value of 2a â€“ 6b = 3*2 â€“ 6(-5) = 36

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