Asked by Daniel
Can anyone please help with this inequality,(x+20)(x-10)(x+11)>0
The solution set is {x| }
The solution set is {x| }
Answers
Answered by
MathMate
The distinct roots of the equation
f(x) = (x+20)(x-10)(x+11) = 0
are x=-20, x=10 and x=-11
Therefore the graph changes sign three times.
When x-> -∞, f(x)<0
When x-> ∞, f(x)>0
Therefore
f(x) > 0 when
x=(10,∞) and
x=(-20,-11)
Graph the function to verify the solution.
f(x) = (x+20)(x-10)(x+11) = 0
are x=-20, x=10 and x=-11
Therefore the graph changes sign three times.
When x-> -∞, f(x)<0
When x-> ∞, f(x)>0
Therefore
f(x) > 0 when
x=(10,∞) and
x=(-20,-11)
Graph the function to verify the solution.
Answered by
Daniel
OK what you say make sense but I am unsure of the solution sets of {x| }
Answered by
MathMate
x=(10,∞) and
x=(-20,-11)
represent the two intervals on which x>0.
The solution set is the union of the two intervals.
If you need help with the interval notation, you can try:
http://zonalandeducation.com/mmts/miscellaneousMath/intervalNotation/intervalNotation.html
x=(-20,-11)
represent the two intervals on which x>0.
The solution set is the union of the two intervals.
If you need help with the interval notation, you can try:
http://zonalandeducation.com/mmts/miscellaneousMath/intervalNotation/intervalNotation.html