Algebra
Calculus
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Matrix
Question
Solve the inequality
x∈[−1.517113,2.403576]∪[4.113538,5.4965]
Evaluate
x2∣x−5∣≤15
When the expression in absolute value bars is not negative, remove the bars
x2∣x−5∣≤15
Transform the expression
x2(x−5)≤15,x−5≥0x2(−x+5)≤15,x−5<0
Calculate
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Calculate
x2(x−5)≤15
Expand the expression
More Steps
Evaluate
x2(x−5)
Apply the distributive property
x2×x−x2×5
Multiply the terms
x3−x2×5
Use the commutative property to reorder the terms
x3−5x2
x3−5x2≤15
Move the expression to the left side
x3−5x2−15≤0
Rewrite the expression
x3−5x2−15=0
Find the critical values by solving the corresponding equation
x≈5.4965
Determine the test intervals using the critical values
x<5.4965x>5.4965
Choose a value form each interval
x1=4x2=6
To determine if x<5.4965 is the solution to the inequality,test if the chosen value x=4 satisfies the initial inequality
More Steps
Evaluate
43−5×42−15≤0
Simplify
−31≤0
Check the inequality
true
x<5.4965 is the solutionx2=6
To determine if x>5.4965 is the solution to the inequality,test if the chosen value x=6 satisfies the initial inequality
More Steps
Evaluate
63−5×62−15≤0
Simplify
21≤0
Check the inequality
false
x<5.4965 is the solutionx>5.4965 is not a solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
x≤5.4965 is the solution
The final solution of the original inequality is x≤5.4965
x≤5.4965
x≤5.4965,x−5≥0x2(−x+5)≤15,x−5<0
Calculate
More Steps
Calculate
x−5≥0
Move the constant to the right side
x≥0+5
Removing 0 doesn't change the value,so remove it from the expression
x≥5
x≤5.4965,x≥5x2(−x+5)≤15,x−5<0
Calculate
More Steps
Calculate
x2(−x+5)≤15
Expand the expression
More Steps
Evaluate
x2(−x+5)
Apply the distributive property
x2(−x)+x2×5
Multiply the terms
−x3+x2×5
Use the commutative property to reorder the terms
−x3+5x2
−x3+5x2≤15
Move the expression to the left side
−x3+5x2−15≤0
Rewrite the expression
−x3+5x2−15=0
Find the critical values by solving the corresponding equation
x≈−1.517113x≈2.403576x≈4.113538
Determine the test intervals using the critical values
x<−1.517113−1.517113<x<2.4035762.403576<x<4.113538x>4.113538
Choose a value form each interval
x1=−3x2=0x3=3x4=5
To determine if x<−1.517113 is the solution to the inequality,test if the chosen value x=−3 satisfies the initial inequality
More Steps
Evaluate
−(−3)3+5(−3)2−15≤0
Simplify
57≤0
Check the inequality
false
x<−1.517113 is not a solutionx2=0x3=3x4=5
To determine if −1.517113<x<2.403576 is the solution to the inequality,test if the chosen value x=0 satisfies the initial inequality
More Steps
Evaluate
−03+5×02−15≤0
Simplify
−15≤0
Check the inequality
true
x<−1.517113 is not a solution−1.517113<x<2.403576 is the solutionx3=3x4=5
To determine if 2.403576<x<4.113538 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
More Steps
Evaluate
−33+5×32−15≤0
Simplify
3≤0
Check the inequality
false
x<−1.517113 is not a solution−1.517113<x<2.403576 is the solution2.403576<x<4.113538 is not a solutionx4=5
To determine if x>4.113538 is the solution to the inequality,test if the chosen value x=5 satisfies the initial inequality
More Steps
Evaluate
−53+5×52−15≤0
Simplify
−15≤0
Check the inequality
true
x<−1.517113 is not a solution−1.517113<x<2.403576 is the solution2.403576<x<4.113538 is not a solutionx>4.113538 is the solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
−1.517113≤x≤2.403576 is the solutionx≥4.113538 is the solution
The final solution of the original inequality is x∈[−1.517113,2.403576]∪[4.113538,+∞)
x∈[−1.517113,2.403576]∪[4.113538,+∞)
x≤5.4965,x≥5x∈[−1.517113,2.403576]∪[4.113538,+∞),x−5<0
Calculate
More Steps
Calculate
x−5<0
Move the constant to the right side
x<0+5
Removing 0 doesn't change the value,so remove it from the expression
x<5
x≤5.4965,x≥5x∈[−1.517113,2.403576]∪[4.113538,+∞),x<5
Calculate
5≤x≤5.4965x∈[−1.517113,2.403576]∪[4.113538,+∞),x<5
Solution
x∈[−1.517113,2.403576]∪[4.113538,5.4965]
Show Solution
