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To solve an equation we have to find the one (or many) values of $x$ which satisfy the equation. The solution set for an equation consists of a discrete set of values. For example, the solutions to $(x-3)^2=4$ are $x=1$ and $x=5$.
In this section, we will learn how to solve equations which involve inequalities. The solution to an inequality is usually an entire range of numbers. For example the inequality $(x-3)^2 \leq 4$ is equivalent to asking the question “for which values of $x$ is $(x-3)^2$ less than or equal to $4$.” The answer is the interval $[1,5] \equiv \{ x\in \mathbb{R}\ | \ 1 \leq x \leq 5 \}$.
The techniques used to deal with inequalities are roughly the same as the techniques which we learned for dealing with equations: we have to perform simplifying steps to both sides of the inequality until we obtain the answer.
The different type of inequality conditions are:
The solutions to an inequality correspond to subsets of the real line. Depending on the type of inequality we are dealing with, the answer will be either a closed or open interval:
Sometimes the we will have to deal with intervals which consists of two disjoint parts:
The main idea for solving inequalities is the same as solving equations except for one small special step. When multiplying by a negative number on both sides, the direction of the inequality must be flipped: \[ f(x) \leq g(x) \qquad \Rightarrow \qquad -f(x) \geq -g(x). \]
To solve $(x-3)^2\leq 4$ we must dig towards the $x$ and undo all the operations that stand in our way: \[ \begin{align*} & \ (x-3)^2 \leq 4, \nl -2 \leq & \ (x-3) \leq 2, \nl 1 \leq & \ \ \ \ \ x \ \ \ \ \leq 5. \end{align*} \] where in the first step we took the square root operation (the inverse of the quadratic function) and then we added $3$ to both sides. The final answer is $x\in[1,5]$.
As you can see, solving inequalities is not more complicated than solving equations. Indeed, the best way to think about an inequality is in terms of the end points – which correspond to the equality condition.