Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
Solve for x
x∈(−∞,0)∪(35,+∞)
Evaluate
5(x−4)x>−7x2
Multiply the terms
5x(x−4)>−7x2
Move the expression to the left side
5x(x−4)−(−7x2)>0
Subtract the terms
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Evaluate
5x(x−4)−(−7x2)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
5x(x−4)+7x2
Expand the expression
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Calculate
5x(x−4)
Apply the distributive property
5x×x−5x×4
Multiply the terms
5x2−5x×4
Multiply the numbers
5x2−20x
5x2−20x+7x2
Add the terms
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Evaluate
5x2+7x2
Collect like terms by calculating the sum or difference of their coefficients
(5+7)x2
Add the numbers
12x2
12x2−20x
12x2−20x>0
Rewrite the expression
12x2−20x=0
Factor the expression
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Evaluate
12x2−20x
Rewrite the expression
4x×3x−4x×5
Factor out 4x from the expression
4x(3x−5)
4x(3x−5)=0
When the product of factors equals 0,at least one factor is 0
4x=03x−5=0
Solve the equation for x
x=03x−5=0
Solve the equation for x
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Evaluate
3x−5=0
Move the constant to the right-hand side and change its sign
3x=0+5
Removing 0 doesn't change the value,so remove it from the expression
3x=5
Divide both sides
33x=35
Divide the numbers
x=35
x=0x=35
Determine the test intervals using the critical values
x<00<x<35x>35
Choose a value form each interval
x1=−1x2=1x3=3
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
5(−1)(−1−4)>−7(−1)2
Simplify
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Evaluate
5(−1)(−1−4)
Subtract the numbers
5(−1)(−5)
Any expression multiplied by 1 remains the same
5×5
Multiply the numbers
25
25>−7(−1)2
Simplify
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Evaluate
−7(−1)2
Evaluate the power
−7×1
Any expression multiplied by 1 remains the same
−7
25>−7
Check the inequality
true
x<0 is the solutionx2=1x3=3
To determine if 0<x<35 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
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Evaluate
5×1×(1−4)>−7×12
Simplify
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Evaluate
5×1×(1−4)
Subtract the numbers
5×1×(−3)
Rewrite the expression
5(−3)
Multiplying or dividing an odd number of negative terms equals a negative
−5×3
Multiply the numbers
−15
−15>−7×12
Simplify
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Evaluate
−7×12
1 raised to any power equals to 1
−7×1
Any expression multiplied by 1 remains the same
−7
−15>−7
Check the inequality
false
x<0 is the solution0<x<35 is not a solutionx3=3
To determine if x>35 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
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Evaluate
5×3(3−4)>−7×32
Simplify
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Evaluate
5×3(3−4)
Subtract the numbers
5×3(−1)
Any expression multiplied by 1 remains the same
−5×3
Multiply the terms
−15
−15>−7×32
Multiply the terms
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Evaluate
−7×32
Evaluate the power
−7×9
Multiply the numbers
−63
−15>−63
Check the inequality
true
x<0 is the solution0<x<35 is not a solutionx>35 is the solution
Solution
x∈(−∞,0)∪(35,+∞)
Show Solution
