Algebra
Calculus
Trigonometry
Matrix
Question
Solve the inequality
x∈[−2,0)∪[2−3,2+3]
Evaluate
4−x2≥x1
Find the domain
More Steps
Evaluate
{4−x2≥0x=0
Calculate
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Evaluate
4−x2≥0
Rewrite the expression
−x2≥−4
Change the signs on both sides of the inequality and flip the inequality sign
x2≤4
Take the 2-th root on both sides of the inequality
x2≤4
Calculate
∣x∣≤2
Separate the inequality into 2 possible cases
{x≤2x≥−2
Find the intersection
−2≤x≤2
{−2≤x≤2x=0
Find the intersection
x∈[−2,0)∪(0,2]
4−x2≥x1,x∈[−2,0)∪(0,2]
Separate the inequality into 2 possible cases
4−x2≥x1,x1≥04−x2≥x1,x1<0
Solve the inequality
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Solve the inequality
4−x2≥x1
Square both sides of the inequality
4−x2≥(x1)2
Evaluate the power
4−x2≥x−2
Move the expression to the left side
4−x2−x−2≥0
Rewrite the expression
4−x2−x21≥0
Convert the expressions
x24x2−x4−1≥0
Separate the inequality into 2 possible cases
{4x2−x4−1≥0x2>0{4x2−x4−1≤0x2<0
Solve the inequality
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Evaluate
4x2−x4−1≥0
Rewrite the expression
4x2−x4−1=0
Solve the equation using substitution t=x2
4t−t2−1=0
Rewrite in standard form
−t2+4t−1=0
Multiply both sides
t2−4t+1=0
Substitute a=1,b=−4 and c=1 into the quadratic formula t=2a−b±b2−4ac
t=24±(−4)2−4
Simplify the expression
t=24±12
Simplify the radical expression
t=24±23
Separate the equation into 2 possible cases
t=24+23t=24−23
Simplify the expression
t=2+3t=24−23
Simplify the expression
t=2+3t=2−3
Substitute back
x2=2+3x2=2−3
Solve the equation for x
x=2+3x=−2+3x2=2−3
Solve the equation for x
x=2+3x=−2+3x=2−3x=−2−3
Determine the test intervals using the critical values
x<−2+3−2+3<x<−2−3−2−3<x<2−32−3<x<2+3x>2+3
Choose a value form each interval
x1=−3x2=−1x3=0x4=1x5=3
To determine if x<−2+3 is the solution to the inequality,test if the chosen value x=−3 satisfies the initial inequality
x<−2+3 is not a solutionx2=−1x3=0x4=1x5=3
To determine if −2+3<x<−2−3 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
x<−2+3 is not a solution−2+3<x<−2−3 is the solutionx3=0x4=1x5=3
To determine if −2−3<x<2−3 is the solution to the inequality,test if the chosen value x=0 satisfies the initial inequality
x<−2+3 is not a solution−2+3<x<−2−3 is the solution−2−3<x<2−3 is not a solutionx4=1x5=3
To determine if 2−3<x<2+3 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
x<−2+3 is not a solution−2+3<x<−2−3 is the solution−2−3<x<2−3 is not a solution2−3<x<2+3 is the solutionx5=3
To determine if x>2+3 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
x<−2+3 is not a solution−2+3<x<−2−3 is the solution−2−3<x<2−3 is not a solution2−3<x<2+3 is the solutionx>2+3 is not a solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
−2+3≤x≤−2−3 is the solution2−3≤x≤2+3 is the solution
The final solution of the original inequality is x∈[−2+3,−2−3]∪[2−3,2+3]
x∈[−2+3,−2−3]∪[2−3,2+3]
⎩⎨⎧x∈[−2+3,−2−3]∪[2−3,2+3]x2>0{4x2−x4−1≤0x2<0
Solve the inequality
More Steps
Evaluate
x2>0
Since the left-hand side is always positive or 0,and the right-hand side is always 0,the statement is true for any value of x,except when x2=0
x2=0
The only way a power can be 0 is when the base equals 0
x=0
Exclude the impossible values of x
x=0
⎩⎨⎧x∈[−2+3,−2−3]∪[2−3,2+3]x=0{4x2−x4−1≤0x2<0
Solve the inequality
More Steps
Evaluate
4x2−x4−1≤0
Rewrite the expression
4x2−x4−1=0
Solve the equation using substitution t=x2
4t−t2−1=0
Rewrite in standard form
−t2+4t−1=0
Multiply both sides
t2−4t+1=0
Substitute a=1,b=−4 and c=1 into the quadratic formula t=2a−b±b2−4ac
t=24±(−4)2−4
Simplify the expression
t=24±12
Simplify the radical expression
t=24±23
Separate the equation into 2 possible cases
t=24+23t=24−23
Simplify the expression
t=2+3t=24−23
Simplify the expression
t=2+3t=2−3
Substitute back
x2=2+3x2=2−3
Solve the equation for x
x=2+3x=−2+3x2=2−3
Solve the equation for x
x=2+3x=−2+3x=2−3x=−2−3
Determine the test intervals using the critical values
x<−2+3−2+3<x<−2−3−2−3<x<2−32−3<x<2+3x>2+3
Choose a value form each interval
x1=−3x2=−1x3=0x4=1x5=3
To determine if x<−2+3 is the solution to the inequality,test if the chosen value x=−3 satisfies the initial inequality
x<−2+3 is the solutionx2=−1x3=0x4=1x5=3
To determine if −2+3<x<−2−3 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
x<−2+3 is the solution−2+3<x<−2−3 is not a solutionx3=0x4=1x5=3
To determine if −2−3<x<2−3 is the solution to the inequality,test if the chosen value x=0 satisfies the initial inequality
x<−2+3 is the solution−2+3<x<−2−3 is not a solution−2−3<x<2−3 is the solutionx4=1x5=3
To determine if 2−3<x<2+3 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
x<−2+3 is the solution−2+3<x<−2−3 is not a solution−2−3<x<2−3 is the solution2−3<x<2+3 is not a solutionx5=3
To determine if x>2+3 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
x<−2+3 is the solution−2+3<x<−2−3 is not a solution−2−3<x<2−3 is the solution2−3<x<2+3 is not a solutionx>2+3 is the solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
x≤−2+3 is the solution−2−3≤x≤2−3 is the solutionx≥2+3 is the solution
The final solution of the original inequality is x∈(−∞,−2+3]∪[−2−3,2−3]∪[2+3,+∞)
x∈(−∞,−2+3]∪[−2−3,2−3]∪[2+3,+∞)
⎩⎨⎧x∈[−2+3,−2−3]∪[2−3,2+3]x=0⎩⎨⎧x∈(−∞,−2+3]∪[−2−3,2−3]∪[2+3,+∞)x2<0
Since the left-hand side is always positive or 0,and the right-hand side is always 0,the statement is false for any value of x
⎩⎨⎧x∈[−2+3,−2−3]∪[2−3,2+3]x=0⎩⎨⎧x∈(−∞,−2+3]∪[−2−3,2−3]∪[2+3,+∞)x∈/R
Find the intersection
x∈[−2+3,−2−3]∪[2−3,2+3]⎩⎨⎧x∈(−∞,−2+3]∪[−2−3,2−3]∪[2+3,+∞)x∈/R
Find the intersection
x∈[−2+3,−2−3]∪[2−3,2+3]x∈/R
Find the union
x∈[−2+3,−2−3]∪[2−3,2+3]
x∈[−2+3,−2−3]∪[2−3,2+3],x1≥04−x2≥x1,x1<0
Solve the inequality
x∈[−2+3,−2−3]∪[2−3,2+3],x>04−x2≥x1,x1<0
Since the left-hand side is always positive or 0,and the right-hand side is always negative,the statement is true for any value of x
x∈[−2+3,−2−3]∪[2−3,2+3],x>0x∈R,x1<0
Solve the inequality
x∈[−2+3,−2−3]∪[2−3,2+3],x>0x∈R,x<0
Find the intersection
2−3≤x≤2+3x∈R,x<0
Find the intersection
2−3≤x≤2+3x<0
Find the union
x∈(−∞,0)∪[2−3,2+3]
Check if the solution is in the defined range
x∈(−∞,0)∪[2−3,2+3],x∈[−2,0)∪(0,2]
Solution
x∈[−2,0)∪[2−3,2+3]
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