Algebra
Calculus
Trigonometry
Matrix
Question
Solve the inequality
x≤1+34+15+34−15
Alternative Form
x∈(−∞,1+34+15+34−15]
Evaluate
2−(x−3)x2≥−4
Multiply the terms
2−x2(x−3)≥−4
Move the expression to the left side
2−x2(x−3)−(−4)≥0
Subtract the terms
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Evaluate
2−x2(x−3)−(−4)
Expand the expression
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Calculate
−x2(x−3)
Apply the distributive property
−x2×x−(−x2×3)
Multiply the terms
−x3−(−x2×3)
Use the commutative property to reorder the terms
−x3−(−3x2)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
−x3+3x2
2−x3+3x2−(−4)
Subtract the numbers
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Evaluate
2−(−4)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
2+4
Add the numbers
6
6−x3+3x2
6−x3+3x2≥0
Rewrite the expression
6−x3+3x2=0
Find the critical values by solving the corresponding equation
x=1+34+15+34−15x=22−34+15−34−15
Determine the test intervals using the critical values
x<22−34+15−34−1522−34+15−34−15<x<1+34+15+34−15x>1+34+15+34−15
Choose a value form each interval
x1=−1x2=2x3=4
To determine if x<22−34+15−34−15 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
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Evaluate
2−(−1)2(−1−3)≥−4
Simplify
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Evaluate
2−(−1)2(−1−3)
Subtract the numbers
2−(−1)2(−4)
Evaluate the power
2−1×(−4)
Any expression multiplied by 1 remains the same
2−(−4)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
2+4
Add the numbers
6
6≥−4
Check the inequality
true
x<22−34+15−34−15 is the solutionx2=2x3=4
To determine if 22−34+15−34−15<x<1+34+15+34−15 is the solution to the inequality,test if the chosen value x=2 satisfies the initial inequality
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Evaluate
2−22(2−3)≥−4
Simplify
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Evaluate
2−22(2−3)
Subtract the numbers
2−22(−1)
Multiply the terms
2−(−22)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
2+22
Evaluate the power
2+4
Add the numbers
6
6≥−4
Check the inequality
true
x<22−34+15−34−15 is the solution22−34+15−34−15<x<1+34+15+34−15 is the solutionx3=4
To determine if x>1+34+15+34−15 is the solution to the inequality,test if the chosen value x=4 satisfies the initial inequality
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Evaluate
2−42(4−3)≥−4
Simplify
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Evaluate
2−42(4−3)
Subtract the numbers
2−42×1
Any expression multiplied by 1 remains the same
2−42
Evaluate the power
2−16
Subtract the numbers
−14
−14≥−4
Check the inequality
false
x<22−34+15−34−15 is the solution22−34+15−34−15<x<1+34+15+34−15 is the solutionx>1+34+15+34−15 is not a solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
x≤22−34+15−34−15 is the solution22−34+15−34−15≤x≤1+34+15+34−15 is the solution
Solution
x≤1+34+15+34−15
Alternative Form
x∈(−∞,1+34+15+34−15]
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