Answer: (a) \(x\leqslant-2\) or \(x\geqslant3\). (b) \(\frac12\lt x\lt 3\).
(a) Factorise the quadratic:
\(x^2-x-6=(x-3)(x+2).\)
The critical values are
\(x=-2\quad\text{and}\quad x=3.\)
Since the coefficient of \(x^2\) is positive, the parabola opens upwards. Therefore the expression is non-negative outside the two roots, including the roots themselves.
Hence
\(x\leqslant-2\quad\text{or}\quad x\geqslant3.\)
(b) For
\(|3x-4|\lt x+2,\)
the right-hand side must be positive, and the inequality is equivalent to
\(-(x+2)\lt3x-4\lt x+2.\)
From the left inequality,
\(-x-2\lt3x-4.\)
Adding \(x+4\) to both sides gives
\(2\lt4x,\)
so
\(x\gt\dfrac12.\)
From the right inequality,
\(3x-4\lt x+2.\)
Therefore
\(2x\lt6,\)
so
\(x\lt3.\)
Combining the two inequalities gives
\(\dfrac12\lt x\lt3.\)
