Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
x∈[2,+∞)∪{0}
Evaluate
(x−2)(x×1)x5≥0
Remove the parentheses
(x−2)x×1×x5≥0
Multiply the terms
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Evaluate
(x−2)x×1×x5
Rewrite the expression
(x−2)x×x5
Multiply the terms with the same base by adding their exponents
(x−2)x1+5
Add the numbers
(x−2)x6
Multiply the terms
x6(x−2)
x6(x−2)≥0
Rewrite the expression
x6(x−2)=0
Separate the equation into 2 possible cases
x6=0x−2=0
The only way a power can be 0 is when the base equals 0
x=0x−2=0
Solve the equation
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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=0x=2
Determine the test intervals using the critical values
x<00<x<2x>2
Choose a value form each interval
x1=−1x2=1x3=3
To determine if x<0 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
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Evaluate
(−1)6(−1−2)≥0
Simplify
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Evaluate
(−1)6(−1−2)
Subtract the numbers
(−1)6(−3)
Evaluate the power
1×(−3)
Any expression multiplied by 1 remains the same
−3
−3≥0
Check the inequality
false
x<0 is not a solutionx2=1x3=3
To determine if 0<x<2 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
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Evaluate
16×(1−2)≥0
Simplify
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Evaluate
16×(1−2)
Subtract the numbers
16×(−1)
1 raised to any power equals to 1
1×(−1)
Any expression multiplied by 1 remains the same
−1
−1≥0
Check the inequality
false
x<0 is not a solution0<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
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Evaluate
36(3−2)≥0
Simplify
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Evaluate
36(3−2)
Subtract the numbers
36×1
Any expression multiplied by 1 remains the same
36
36≥0
Calculate
729≥0
Check the inequality
true
x<0 is not a solution0<x<2 is not a solutionx>2 is the solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
x≥2 is the solutionx=0
Solution
x∈[2,+∞)∪{0}
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