calc question... solve for x: Ix-2I/3<4?

Answer 1

Ix-2I/3<4

time 3 both sides

Ix-2I < 12
so
x-2 < 12 or -(x-2)< 12
x<14 or -x+2<12
x<14 or -x< 10
x<14 or x> -10

that is not calculus , precal maybe

Answer 2

a million. Subtract 4x^2 from the two facets x^4-4x^2+3=0 2. Say that y =x^2, and replace that fee interior the equation (x^2)^2-4(x^2)+3=0 y^2-4y+3=0 3. resolve like familiar quadratic (y-a million)(y-3) 4. resolve for the two y values y=a million y=3 5. If y=^2, then x=sqrt(a million) x=sqrt(3)

Answer 3

| x - 2 | < 12

x - 2 < 12
x < 14

x - 2 > - 12
x > - 10

- 10 < x < 14