Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
x∈[3,5]∪{0}
Evaluate
8(x−5)×2x2(x−3)≤0
Multiply
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Evaluate
8(x−5)×2x2(x−3)
Multiply the terms
16(x−5)x2(x−3)
Multiply the terms
16x2(x−5)(x−3)
16x2(x−5)(x−3)≤0
Rewrite the expression
16x2(x−5)(x−3)=0
Elimination the left coefficient
x2(x−5)(x−3)=0
Separate the equation into 3 possible cases
x2=0x−5=0x−3=0
The only way a power can be 0 is when the base equals 0
x=0x−5=0x−3=0
Solve the equation
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Evaluate
x−5=0
Move the constant to the right-hand side and change its sign
x=0+5
Removing 0 doesn't change the value,so remove it from the expression
x=5
x=0x=5x−3=0
Solve the equation
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Evaluate
x−3=0
Move the constant to the right-hand side and change its sign
x=0+3
Removing 0 doesn't change the value,so remove it from the expression
x=3
x=0x=5x=3
Determine the test intervals using the critical values
x<00<x<33<x<5x>5
Choose a value form each interval
x1=−1x2=2x3=4x4=6
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
16(−1)2(−1−5)(−1−3)≤0
Simplify
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Evaluate
16(−1)2(−1−5)(−1−3)
Subtract the numbers
16(−1)2(−6)(−1−3)
Subtract the numbers
16(−1)2(−6)(−4)
Evaluate the power
16×1×(−6)(−4)
Rewrite the expression
16(−6)(−4)
Rewrite the expression
16×6×4
Multiply the terms
96×4
Multiply the numbers
384
384≤0
Check the inequality
false
x<0 is not a solutionx2=2x3=4x4=6
To determine if 0<x<3 is the solution to the inequality,test if the chosen value x=2 satisfies the initial inequality
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Evaluate
16×22(2−5)(2−3)≤0
Simplify
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Evaluate
16×22(2−5)(2−3)
Subtract the numbers
16×22(−3)(2−3)
Subtract the numbers
16×22(−3)(−1)
Any expression multiplied by 1 remains the same
16×22×3
Transform the expression
24×22×3
Multiply the terms with the same base by adding their exponents
24+2×3
Add the numbers
26×3
Evaluate the power
64×3
Multiply the numbers
192
192≤0
Check the inequality
false
x<0 is not a solution0<x<3 is not a solutionx3=4x4=6
To determine if 3<x<5 is the solution to the inequality,test if the chosen value x=4 satisfies the initial inequality
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Evaluate
16×42(4−5)(4−3)≤0
Simplify
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Evaluate
16×42(4−5)(4−3)
Subtract the numbers
16×42(−1)(4−3)
Subtract the numbers
16×42(−1)×1
Rewrite the expression
16×42(−1)
Any expression multiplied by 1 remains the same
−16×42
Transform the expression
−42×42
Multiply the terms with the same base by adding their exponents
−42+2
Add the numbers
−44
−44≤0
Calculate
−256≤0
Check the inequality
true
x<0 is not a solution0<x<3 is not a solution3<x<5 is the solutionx4=6
To determine if x>5 is the solution to the inequality,test if the chosen value x=6 satisfies the initial inequality
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Evaluate
16×62(6−5)(6−3)≤0
Simplify
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Evaluate
16×62(6−5)(6−3)
Subtract the numbers
16×62×1×(6−3)
Subtract the numbers
16×62×1×3
Rewrite the expression
16×62×3
Multiply the terms
48×62
Evaluate the power
48×36
Multiply the numbers
1728
1728≤0
Check the inequality
false
x<0 is not a solution0<x<3 is not a solution3<x<5 is the solutionx>5 is not a solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
3≤x≤5 is the solutionx=0
Solution
x∈[3,5]∪{0}
Show Solution
