Find the number of integer values of x that satisfies the inequation:
−x2+2x+15>0;y∈R.
- 7
−x2+2x+15>0;y∈R.
- 7
The correct option is A 7
Given: −x2+2x+15>0
Rule: If both the sides of an inequation are multiplied or divided by the same negative number, then the sign of the inequality will get reversed.
Again, multiplying by -1 on both the sides, we get;
⇒x2−2x−15<0
⇒(x+3)(x−5)<0
For this to happen, there will be two cases.
Case 1: x+3>0 and x−5<0
Rule: If a term of an inequation is transferred from one side to the other side of the inequation, the sign of the term gets changed. Let's apply this rule in the following inequations.
Which gives x>−3 and x<5
Case 2: x+3<0 and x−5>0
Which gives x<−3 and x>5. This is never possible.
So, the solution set is {x:x>−3 and x<5x∈R}
The same solution can be represented on the number line as shown below:
The integers present in the solutions set are: -2, -1, 0, 1, 2, 3, 4. Therefore, the total number of integer values that satisfies the above inequation are 7.
Given: −x2+2x+15>0
Rule: If both the sides of an inequation are multiplied or divided by the same negative number, then the sign of the inequality will get reversed.
Again, multiplying by -1 on both the sides, we get;
⇒x2−2x−15<0
⇒(x+3)(x−5)<0
For this to happen, there will be two cases.
Case 1: x+3>0 and x−5<0
Rule: If a term of an inequation is transferred from one side to the other side of the inequation, the sign of the term gets changed. Let's apply this rule in the following inequations.
Which gives x>−3 and x<5
Case 2: x+3<0 and x−5>0
Which gives x<−3 and x>5. This is never possible.
So, the solution set is {x:x>−3 and x<5x∈R}
The same solution can be represented on the number line as shown below:
The integers present in the solutions set are: -2, -1, 0, 1, 2, 3, 4. Therefore, the total number of integer values that satisfies the above inequation are 7.
