Question

How do you solve $\left( {x + 1} \right)\left( {x - 3} \right) > 0$?

Answer 1

Hint: First off, use the fact that either (x + 1) > 0 and (x – 3) > 0 or (x + 1) < 0 and (x – 3) < 0. Using both of these, we get the possible values of x and thus have the answer.

Complete step-by-step solution:
We are given that we are required to solve $\left( {x + 1} \right)\left( {x - 3} \right) > 0$.
Now, we know that if we have a.b > 0, then either a > 0 and b > 0 or a < 0 and b < 0.
Replacing a by (x + 1) and b by (x – 3), then since $\left( {x + 1} \right)\left( {x - 3} \right) > 0$, therefore, we have:-
Case 1: $(x + 1) > 0$ and $(x - 3) > 0$
This implies that $x > - 1$ and $x > 3$.
This implies that $x > 3$.
Case 2: $(x + 1) < 0$ and $(x - 3) < 0$
This implies that $x < - 1$ and $x < 3$.
This implies that $x < - 1$ .
Thus, the solution set is $\left( { - \infty ,\left. { - 1} \right)} \right. \cup \left( {3,\infty } \right)$.

Note: The students must note that we can also prove the fact that we used above which states that: If a.b > 0, then either a > 0 and b > 0 or a < 0 and b < 0.
Given that a.b > 0.
Case 1: Let us suppose that a > 0 is given then we need to prove that b > 0.
Let us assume that on contrary b < 0, then – b > 0.
So, let us consider a.b > 0 which is already given.
Multiplying both sides by – sign, we have: - a.b < 0
We can write this as: a. (- b) < 0
Since, - b is positive, taking it on the right hand side, we have: a < 0 which is a contradiction.
Case 2: Let us suppose that a < 0 is given then we need to prove that b < 0.
Let us assume that on contrary b > 0, then – b < 0.
So, let us consider a.b > 0 which is already given.
Multiplying both sides by – sign, we have: - a.b < 0
We can write this as: a. (- b) < 0
Since, - b is negative, taking it on the right hand side, we have: a > 0 which is a contradiction.