Answer:
The answer in interval notation is [tex]\left(-\frac{1}{2},\:\frac{2}{3}\right)[/tex]
Step-by-step explanation:
Consider the provided inequality.
[tex]6x^2 - x < 2[/tex]
Subtract 2 from both sides.
[tex]6x^2-x-2<0[/tex]
[tex]6x^2+3x-4x-2<0[/tex]
[tex]3x(2x+1)-2(2x+1)<0[/tex]
[tex](3x-2)(2x+1)<0[/tex]
[tex]x<\frac{2}{3}\ or\ x<\frac{-1}{2}[/tex]
The value of function [tex](3x-2)(2x+1)<0[/tex] is positive for [tex]x<-\frac{1}{2}[/tex]
The value function [tex](3x-2)(2x+1)<0[/tex] is zero at [tex]x=-\frac{1}{2}[/tex]
The value function [tex](3x-2)(2x+1)<0[/tex] is negative for [tex]-\frac{1}{2}<x<\frac{2}{3}[/tex]
The value function [tex](3x-2)(2x+1)<0[/tex] is zero at [tex]x=\frac{2}{3}[/tex]
The value function [tex](3x-2)(2x+1)<0[/tex] is positive for [tex]x>\frac{2}{3}[/tex]
Since we want the value of function less than 0, so the required interval which satisfy the condition < 0 is:
[tex]-\frac{1}{2}<x<\frac{2}{3}[/tex]
The required number line is shown below:
The answer in interval notation is [tex]\left(-\frac{1}{2},\:\frac{2}{3}\right)[/tex]
