Find the range of values of x which satisfies the inequality 12\(x^2\)< x+1
The correct answer is A. -\(\frac{1}{4}\) < x < \(\frac{1}{3}\)
To find the range of values of x that satisfy the inequality 12\(x^2\) < x + 1, we can start by rearranging the inequality to bring all terms to one side:
12\(x^2\) - x - 1 < 0
Next, we need to solve this quadratic inequality. One approach is to factorize the quadratic expression:
(4x - 1)(3x + 1) < 0
Now we have two factors: (4x - 1) and (3x + 1). To determine the sign of the expression, we can use a sign chart:
```
-1/3 1/4
| |
+----------+
(3x + 1) (4x - 1)
| |
- +
```
From the sign chart, we can see that the inequality is true when (3x + 1) is negative and (4x - 1) is positive, or when (3x + 1) is positive and (4x - 1) is negative.
Case 1: (3x + 1) is negative and (4x - 1) is positive
-1/3 < x < 1/4
Case 2: (3x + 1) is positive and (4x - 1) is negative
No solution in this case.
Therefore, the range of values of x that satisfy the inequality is: -1/3 < x < 1/4
