a)

3x²-5x+m=0,

x=(5±√(25-12m))/6.

25-12m≥0 for non-complex roots.

Therefore 12m≤25, m≤25/12.

EXAMPLE

__When m=0__, x(3x-5)=0, x=0 or 5/3.

__When m=2__, x=(5±√(25-24))/6=(5±1)/6=1 or ⅔.

3x²-5x+2=(3x-2)(x-1).

So the solution is m≤25/12 (including, for example, m=0 and m=2).

b)

2(x²+mx)+6=m(x-1),

2x²+2mx-mx+6+m=0,

2x²+mx+6+m=0.

x=(-m±√(m²-8(6+m)))/4,

x=(-m±√(m²-8m-48))/4=(-m±√((m-12)(m+4)))/4.

When m=-4 or 12, the quadratic is a perfect square, so has only one distinct root (x=1 when m=-4 and the quadratic is 2x²-4x+2=2(x-1)²; x=-3 when m=12 and the quadratic is 2x²+12x+18=2(x+3)²).

(m-12)(m+4)>0 when m>12 or m<-4.

So the solution to (b) is m>12 or m<-4 (for example, m=14 gives x=-2, -5 and the quadratic is 2x²+14x+20=2(x+2)(x+5); m=-6 gives the quadratic equation 2x²-6x=0=2x(x-3), x=0, 3).

If the same m is to satisfy both (a) and (b), m<-4.