Solve Algebra Inequality: $\frac 1{3x^2-x-2}<0$

[Air]

$$\frac 1{3x^2-x-2}<0$$

I understand that in algebra inequalities, one side has to be made zero but in the equation shown above, I am unable to find the set of values that satisfies the equation.

 

[genneth]

For what values of x is 1/x < 0 ?

 

[JonF]

one side needs to be "made zero" for a continuous function to change signs, This function isn’t continuous where it’s undefined: I suggest you factor the bottom and check the intermediate values

 

[Air]

For what values of x is 1/x < 0 ?

Yes, that is what needs to be worked out. I know the answer but when I do the working it doesn't come to the answer so I want to know how to do working out.

The answer is -2/3 < x < 1[/color].

 

[genneth]

Yes, that is what needs to be worked out. I know the answer but when I do the working it doesn't come to the answer so I want to know how to do working out.

The answer is -2/3 < x < 1[/color].

No -- I mean if I gave you the (in)equation 1/x < 0, what's the solution in terms of x? Then I was going to ask you to solve x^2 - 1 < 0, followed by (x-a)(x-b) < 0. Then hoping that you'd managed the rest...

 

[arildno]

Use the quadratic formula to factorize the denominator.

 

[symbolipoint]

$$\frac 1{3x^2-x-2}<0$$

I understand that in algebra inequalities, one side has to be made zero but in the equation shown above, I am unable to find the set of values that satisfies the equation.

The denominator factors to two binomials. In order for the original rational left side to be less than zero, exactly ONE of the binomials must equal zero but not both equal to zero.

The binomials for the denominator are (3x+2) and (x-1). Additionally, x must NOT equal -2/3, and x must NOT equal +1.

Critical points to use for boundary between solution ranges might be x at -2/3 and x at +1.

 

[symbolipoint]

... in fact, the solution appears to be disjoint , but actually checking each section of the number line tells you something more specific.

Check a point less than -2/3;
Check a point more than -2/3 and less than +1;
check a point more than +1 (should not really be necessary).

The values of x between -2/3 and +1 exclusive should be the solution.

 

[murshid_islam]

$$\frac 1{3x^2-x-2}<0$$

I understand that in algebra inequalities, one side has to be made zero but in the equation shown above, I am unable to find the set of values that satisfies the equation.

$$\frac 1{3x^2-x-2}<0$$

$$\Rightarrow3x^2-x-2<0$$

$$\Rightarrow(3x+2)(x-1)<0$$

i think it's easy from here.

 

[symbolipoint]

from murshid_islam

$$\Rightarrow(3x+2)(x-1)<0$$

With that, you need just to look at two situations:
3x+2<0 AND x-1>0
OR (which you must check carefully to see whether conjoint or disjoint)
3x+1>0 AND x-1<0

Important is that one binomial must be positive and the other binomial must be negative.