Solution.

1.

By Graph

Let \(y=x^2+x-6\). The graph of this function is

\(x^2+x-6<0\implies y<0\)

Here we see that values of \(y\) are negative between \(-3\) and 2. Therefore the solution set is \[SS: \{x\in \mathbb {R}:-3<x<2\}\]

2.

By Tables

Here we first find critical values. Critical values of \(x\) where the factors in the given expression are zeros. \(x^2+x-6=0\)

Now, \((x+3)(x-2)<0\)

critical values: \(x=-3\) and \(x=2\)

	\((-5)\)	\(-3\)	0	2	\((3)\)
\(x+3\)	\(-ve\)	\(|\)	\(+ve\)	\(|\)	\(+ve\)
\(x-2\)	\(-ve\)	\(|\)	\(-ve\)	\(|\)	\(+ve\)
\((x+3)(x-2)\)	\(+ve\)	\(|\)	\(-ve\)	\(|\)	\(+ve\)

Thus \(x^2+x-6<0\) if \(-3<x<2\). Therefore the solution set is \(SS:(-3,2)\).

Solution.

1.

By Table: First we find critical values.

Now \begin {align*} 2x-1 & < x^2-4\\ \implies 0 &< x^2-4-2x+1\\ \implies x^2-2x-3 &>0\\ \implies (x-3)(x+1)&>0 \end {align*}

Critical values \(x-3=0\implies x=3\) and \(x+1=0\implies x=-1\)

		\((-2)\)	\(-1\)	0	3	\((4)\)
\((x-3)\)		\(-ve\)	\(|\)	\(-ve\)	\(|\)	\(+ve\)
\((x+1)\)		\(-ve\)	\(|\)	\(+ve\)	\(|\)	\(+ve\)
\((x-3)(x+1)\)		\(+ve\)	\(|\)	\(-ve\)	\(|\)	\(+ve\)

The product is positive in the interval \((-\infty ,-1)\) and \((3,\infty )\). Therefore \[SS:(-\infty ,-1)\cup (3,\infty )\]

2.

Analytically
\((x-3)(x+1)>0\)
Since the product of two factors is positive then both factors must have the same sign.

case 1
\(x-3>0\) and \(x+1>0\)
\(\implies x>3\) and \(x>-1\)
\((3,\infty )\) and \((-1,\infty )\)

Thus we are looking for the intersection. i.e \((3,\infty )\cap (-1,\infty )=(3,\infty )\).

case 2
\(x-3<0\) and \(x+1<0\)
\(\implies x<3\) and \(x<-1\)
\(\implies (-\infty ,3)\cap (-\infty ,-1)=(-\infty ,-1)\)

Hence, \(SS:(-\infty ,-1)\cup (3,\infty )\)