Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
a∈(−∞,−1)∪(0,1)∪(2,+∞)
Evaluate
(1−a2)(2a−a2)>0
Rewrite the expression
(1−a2)(2a−a2)=0
Separate the equation into 2 possible cases
1−a2=02a−a2=0
Solve the equation
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Evaluate
1−a2=0
Move the constant to the right-hand side and change its sign
−a2=0−1
Removing 0 doesn't change the value,so remove it from the expression
−a2=−1
Change the signs on both sides of the equation
a2=1
Take the root of both sides of the equation and remember to use both positive and negative roots
a=±1
Simplify the expression
a=±1
Separate the equation into 2 possible cases
a=1a=−1
a=1a=−12a−a2=0
Solve the equation
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Evaluate
2a−a2=0
Factor the expression
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Evaluate
2a−a2
Rewrite the expression
a×2−a×a
Factor out a from the expression
a(2−a)
a(2−a)=0
When the product of factors equals 0,at least one factor is 0
a=02−a=0
Solve the equation for a
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Evaluate
2−a=0
Move the constant to the right-hand side and change its sign
−a=0−2
Removing 0 doesn't change the value,so remove it from the expression
−a=−2
Change the signs on both sides of the equation
a=2
a=0a=2
a=1a=−1a=0a=2
Determine the test intervals using the critical values
a<−1−1<a<00<a<11<a<2a>2
Choose a value form each interval
a1=−2a2=−21a3=21a4=23a5=3
To determine if a<−1 is the solution to the inequality,test if the chosen value a=−2 satisfies the initial inequality
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Evaluate
(1−(−2)2)(2(−2)−(−2)2)>0
Simplify
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Evaluate
(1−(−2)2)(2(−2)−(−2)2)
Subtract the numbers
(−3)(2(−2)−(−2)2)
Remove the parentheses
−3(2(−2)−(−2)2)
Multiply the numbers
−3(−4−(−2)2)
Subtract the numbers
−3(−8)
Multiplying or dividing an even number of negative terms equals a positive
3×8
Multiply the numbers
24
24>0
Check the inequality
true
a<−1 is the solutiona2=−21a3=21a4=23a5=3
To determine if −1<a<0 is the solution to the inequality,test if the chosen value a=−21 satisfies the initial inequality
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Evaluate
(1−(−21)2)(2(−21)−(−21)2)>0
Simplify
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Evaluate
(1−(−21)2)(2(−21)−(−21)2)
Subtract the numbers
43(2(−21)−(−21)2)
Multiply the numbers
43(−1−(−21)2)
Subtract the numbers
43(−45)
Multiplying or dividing an odd number of negative terms equals a negative
−43×45
To multiply the fractions,multiply the numerators and denominators separately
−4×43×5
Multiply the numbers
−4×415
Multiply the numbers
−1615
−1615>0
Calculate
−0.9375>0
Check the inequality
false
a<−1 is the solution−1<a<0 is not a solutiona3=21a4=23a5=3
To determine if 0<a<1 is the solution to the inequality,test if the chosen value a=21 satisfies the initial inequality
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Evaluate
(1−(21)2)(2×21−(21)2)>0
Simplify
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Evaluate
(1−(21)2)(2×21−(21)2)
Subtract the numbers
43(2×21−(21)2)
Multiply the numbers
43(1−(21)2)
Subtract the numbers
43×43
To multiply the fractions,multiply the numerators and denominators separately
4×43×3
Multiply the numbers
4×49
Multiply the numbers
169
169>0
Calculate
0.5625>0
Check the inequality
true
a<−1 is the solution−1<a<0 is not a solution0<a<1 is the solutiona4=23a5=3
To determine if 1<a<2 is the solution to the inequality,test if the chosen value a=23 satisfies the initial inequality
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Evaluate
(1−(23)2)(2×23−(23)2)>0
Simplify
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Evaluate
(1−(23)2)(2×23−(23)2)
Subtract the numbers
(−45)(2×23−(23)2)
Remove the parentheses
−45(2×23−(23)2)
Multiply the numbers
−45(3−(23)2)
Subtract the numbers
−45×43
To multiply the fractions,multiply the numerators and denominators separately
−4×45×3
Multiply the numbers
−4×415
Multiply the numbers
−1615
−1615>0
Calculate
−0.9375>0
Check the inequality
false
a<−1 is the solution−1<a<0 is not a solution0<a<1 is the solution1<a<2 is not a solutiona5=3
To determine if a>2 is the solution to the inequality,test if the chosen value a=3 satisfies the initial inequality
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Evaluate
(1−32)(2×3−32)>0
Simplify
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Evaluate
(1−32)(2×3−32)
Subtract the numbers
(−8)(2×3−32)
Remove the parentheses
−8(2×3−32)
Multiply the numbers
−8(6−32)
Subtract the numbers
−8(−3)
Multiplying or dividing an even number of negative terms equals a positive
8×3
Multiply the numbers
24
24>0
Check the inequality
true
a<−1 is the solution−1<a<0 is not a solution0<a<1 is the solution1<a<2 is not a solutiona>2 is the solution
Solution
a∈(−∞,−1)∪(0,1)∪(2,+∞)
Show Solution
