
Bobosharif S. answered 02/27/18
Tutor
4.4
(32)
PhD in Math, MS's in Calulus
n2 - b ≥0 ≤
n2 ≥b, assuming b ≥0.
|n|≥b
n ≤-b and/or n≥b
3x2≤7x-2
3x2-7x+2≤0 (factor LHS)
(x-2)(3x-1)≤0
Now, product of two factor is ≤0 is they have opposite signs:
a) x-2≥0 ---> x≥2
3x-1≤0 ---> x≤1/3
Both inequalities here cannot be satisfied simultaneously. So x belongs to empty set.
b) x-2≤0 -->x≤2
3x-1≥0 ---> x≥1/3
From these it follows that 1/3 ≤ x≤2.