Algebra
Calculus
Trigonometry
Matrix
Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
Solve for x
x∈(−∞,−1)∪(2,+∞)
Evaluate
x2−x−2>0
Rewrite the expression
x2−x−2=0
Factor the expression
More Steps
Evaluate
x2−x−2
Rewrite the expression
x2+(1−2)x−2
Calculate
x2+x−2x−2
Rewrite the expression
x×x+x−2x−2
Factor out x from the expression
x(x+1)−2x−2
Factor out −2 from the expression
x(x+1)−2(x+1)
Factor out x+1 from the expression
(x−2)(x+1)
(x−2)(x+1)=0
When the product of factors equals 0,at least one factor is 0
x−2=0x+1=0
Solve the equation for x
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Evaluate
x−2=0
Move the constant to the right-hand side and change its sign
x=0+2
Removing 0 doesn't change the value,so remove it from the expression
x=2
x=2x+1=0
Solve the equation for x
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Evaluate
x+1=0
Move the constant to the right-hand side and change its sign
x=0−1
Removing 0 doesn't change the value,so remove it from the expression
x=−1
x=2x=−1
Determine the test intervals using the critical values
x<−1−1<x<2x>2
Choose a value form each interval
x1=−2x2=1x3=3
To determine if x<−1 is the solution to the inequality,test if the chosen value x=−2 satisfies the initial inequality
More Steps
Evaluate
(−2)2−(−2)−2>0
Subtract the numbers
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Evaluate
(−2)2−(−2)−2
Evaluate the power
4−(−2)−2
Subtract the terms
6−2
Subtract the numbers
4
4>0
Check the inequality
true
x<−1 is the solutionx2=1x3=3
To determine if −1<x<2 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
More Steps
Evaluate
12−1−2>0
Simplify
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Evaluate
12−1−2
1 raised to any power equals to 1
1−1−2
Apply the inverse property of addition
−2
−2>0
Check the inequality
false
x<−1 is the solution−1<x<2 is not a solutionx3=3
To determine if x>2 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
More Steps
Evaluate
32−3−2>0
Subtract the numbers
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Evaluate
32−3−2
Evaluate the power
9−3−2
Subtract the numbers
4
4>0
Check the inequality
true
x<−1 is the solution−1<x<2 is not a solutionx>2 is the solution
Solution
x∈(−∞,−1)∪(2,+∞)
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