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# Find the range of values of x which satisfies the inequality 12x2 < x + 1....

### Question

Find the range of values of x which satisfies the inequality 12x2 < x + 1.

### Options

A) -$$\frac{1}{4}$$ < x < $$\frac{1}{3}$$

B) $$\frac{1}{4}$$ < x < -$$\frac{1}{3}$$

C) -$$\frac{1}{3}$$ < x < $$\frac{1}{4}$$

D) -$$\frac{1}{4}$$ < x < - $$\frac{1}{3}$$

### Explanation:

12x2< x + 1

12 - x - 1 < 0

12x2 - 4x + 3x - 1 < 0

4x(3x - 1) + (3x - 1) < 0

Case 1 (+, -)
4x + 1 > 0, 3x - 1 < 0

x > -$$\frac{1}{4}$$x < $$\frac{1}{3}$$

Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0
-$$\frac{1}{4}$$ < x < - $$\frac{1}{3}$$

## Dicussion (1)

• 12x2< x + 1

12 - x - 1 < 0

12x2 - 4x + 3x - 1 < 0

4x(3x - 1) + (3x - 1) < 0

Case 1 (+, -)
4x + 1 > 0, 3x - 1 < 0

x > -$$\frac{1}{4}$$x < $$\frac{1}{3}$$

Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0
-$$\frac{1}{4}$$ < x < - $$\frac{1}{3}$$