MCQ
The complete solution of the inequation ${x^2} - 4x < 12\,{\rm{ is}}$
- A$x < - \,2$ or $x > 6$
- B$ - \,6 < x < 2$
- C$2 < x < 6$
- ✓$ - \,2 < x < 6$
==> ${x^2} - 4x - 12 < 0$ ==> ${x^2} - 6x + 2x - 12 < 0$
==> $(x - 6)(x + 2) < 0$ ==> $ - 2 < x < 6$.
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