Question

How do you solve the following Quadratic inequality $ {x^2} + 2x - 15 < 0 $ ?

Answer 1

Hint: In order to solve the inequation given in the above question, we will first factorize the quadratic part of the inequation by using the method of Splitting the middle term in order to obtain the critical points. Use the reasoning that the product of two real numbers can be negative when one of them is positive and another be negative to find the different cases of the solution. Check each one of the solution cases to get the possible and feasible set of solutions.

Complete step-by-step answer:
We are given with the inequation ( that do not has only ‘ = ’ sign in it . ) as $ {x^2} + 2x - 15 < 0 $
Since this is a inequation, the solution will be in a range of set . To obtain the sets of solutions over the number line we have to first find the critical points of the quadratic inequation.
In order to find the critical points we will factorize the quadratic part of the inequation
Factoring the quadratic part of the inequation using the splitting up the middle term method.
 $ \Rightarrow {x^2} + 2x - 15 < 0 $
As we can see the middle term $ 2x $ can be written as $ 5x - 3x $ . Rewriting the above expression we get
 $ \Rightarrow {x^2} + 5x - 3x - 15 < 0 $
Now taking common from the first 2 terms and last 2 terms
 $ \Rightarrow x\left( {x + 5} \right) - 3\left( {x + 5} \right) < 0 $
Finding the common binomial parenthesis, the inequation now becomes
 $ \Rightarrow \left( {x - 3} \right)\left( {x + 5} \right) < 0 $
As we know the product of two real numbers can be negative when one of them is positive and another is negative. By using this reasoning, we obtain the 2 solution sets for the above inequation
  $ {A_1}. $ $ x + 5 > 0\,and\,x - 3 < 0 $
\[{A_2}\]\[\,x + 50\,and\,x - 3 > 0\]
Let see each of the case one by one,
So, the case $ {A_1} $ defines the variable $ x $ as
 $ x > - 5\,and\,x < 3 $ , which defines the interval of $ x $ as: $ - 5 < x < 3 $
Now In the case $ {A_2} $ the variable $ x $ is defined as
 $ x < - 5\,and\,x > 3 $ which is completely impossible as this will have null value.
Hence the solution of the inequation will be the case $ {A_1} $ i.e. $ - 5 < x < 3 $
Therefore, the solution of the inequation $ {x^2} + 2x - 15 < 0 $ is $ - 5 < x < 3 $
So, the correct answer is “$ - 5 < x < 3 $”.

Note: 1.The points on the graph are called to be critical points where the function's rate of change is differentiable or changeable —either a change from increasing to decreasing or in some uncertain fashion .
2.Write the factors while splitting the middle term using the proper sign as it may lead to calculation error.
3. The inequality symbols ( $ < , > $ )are used to express strict inequalities and on the other hand symbols $ ( \geqslant , \leqslant ) $ are slack inequalities.