The number of integral values of m for which the quadratic expression (1+2m)x2−2(1+3m)x+4(1+m), x∈R, is always positive, is:
Solutions · 1
Official solution
For the expression to be always positive: 1+2m>0 ...(i), and D<0:
\[4(1+3m)^2-4(1+2m)\cdot4(1+m)<0 \implies (1+3m)^2-4(1+2m)(1+m)<0\]
\[1+6m+9m^2 - 4(1+3m+2m^2) < 0 \implies m^2-6m-3<0 \implies m\in(3-2\sqrt{3},3+2\sqrt{3})\]
Since 3−23≈−0.46>−21, condition (i) is automatically satisfied within this range.
Integral values of m: 0,1,2,3,4,5,6 --- 7 values.