Question

How do you solve the inequality \[\left( x-3 \right)\left( x-5 \right)>15\] ?

Answer 1

Hint: In the above question we are provided with the inequality of the form\[\left( x-a \right)\left( x-b \right) > c\]. So, to solve it we will follow a few steps. First, we will perform a multiplication operation on the left-hand side of the inequality, so that we can simplify the given expression. After that we will simplify further by combining like terms and then perform simple addition or subtraction to get a variable on one side and the constant on the other side. Solve the simplified inequality to get the answer.

Complete step by step answer:
Now before coming to the question let us get familiar with inequalities and rules to be followed while solving it.
Inequalities are the mathematical relations which we use to make a non-equal comparison between two expressions. To solve an inequality, we have to follow some rules. If you add, subtract, divide or multiply any number on both sides, the inequality remains the same. If we multiply both sides by the same negative number then we have to reverse the sign of equality. If we divide both sides by the same negative number in this case also, we have to reverse the sign of equality.
Now coming to the question, we have \[\left( x-3 \right)\left( x-5 \right) > 15\]
 First, we will remove the parentheses
 \[\left( x-3 \right)\left( x-5 \right) > 15\]
\[\Rightarrow {{x}^{2}}-5x-3x+15 > 15\]
Here, we can see that we both have the same constant term on both sides. Thus, cancel the constant term and add the like terms.
\[\Rightarrow {{x}^{2}}-8x > 0\]
\[\Rightarrow x(x-8) > 0\]
\[\Rightarrow x > 0\] or \[x-8 > 0\]
 Thus, \[x > 0\] or \[x > 8\] is the required answer.

Note: Remember the approach used above to solve the question for future use. Since, whenever you are dealing with inequalities, keep in mind the rules to solve the inequality and by which operation you have to reverse the sign of the inequality. Do not get confused by signs while finding the range of the variable.