Asked by kadir
solve -x(4+x)>0 by using sign chart
Answers
Answered by
mathhelper
-x(4+x)>0 , then
x(4+x) < 0
Here is how I do these:
"critical" values are x=0 , x=-4
put those on a number line, giving us 3 segments on the line
1. below -4
2. between -4 and 0
3. above 0
I then pick a number in each reason and test it in x(4+x) < 0
e.g. x < -4 , say x = -5
-5(4-5) < 0 , false, so that doesn't work, so x < 0 is out
pick x = -1
-1(4-1) < 0, true, so that works, so x is between -4 and 0
pick x = 3
3(4+3) < 0 is false, so x cannot be greater than 0
-4 < x < 0
simple way:
consider y = -x(4+x)
this is a parabola opening downwards, with x-intercepts of -4 and 0
so the part above the x-axis is positive, that is
for -x(4+x) > 0 , we have - 4 < x < 0
x(4+x) < 0
Here is how I do these:
"critical" values are x=0 , x=-4
put those on a number line, giving us 3 segments on the line
1. below -4
2. between -4 and 0
3. above 0
I then pick a number in each reason and test it in x(4+x) < 0
e.g. x < -4 , say x = -5
-5(4-5) < 0 , false, so that doesn't work, so x < 0 is out
pick x = -1
-1(4-1) < 0, true, so that works, so x is between -4 and 0
pick x = 3
3(4+3) < 0 is false, so x cannot be greater than 0
-4 < x < 0
simple way:
consider y = -x(4+x)
this is a parabola opening downwards, with x-intercepts of -4 and 0
so the part above the x-axis is positive, that is
for -x(4+x) > 0 , we have - 4 < x < 0
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