How do you solve #x(2x+7) >=0#?

2 Answers
Mar 4, 2018

Answer:

#x<=(-7)/2 or x>=0#

Explanation:

#x(2x+7)>=0#

Distribute

#(x)(2x)+(x)(7)>=0#

#2x^2 + 7x >=0#

Now we need to find the critical points of the inequality!

#2x^2 + 7x =0#

We need to factorize this again

#x(2x+7)=0

Set factors equal to #0#

#x=0 or 2x+7=0#

#x=0 or 2x=0-7#

#x=0 or 2x=-7#

#x=0 or x=(-7)/2#

Now we need to check the intervals between the critical points!

#x<=(-7)/2# (This works in original inequality)

#(-7)/2<=x<=0# (This doesn't work)

#x>=0# (This works)

Thus,

The answer is:

#x<=(-7)/2 or x>=0#

Mar 4, 2018

Answer:

Use the Sign Chart Method

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Explanation:

Factor (make sure it is equal to zero) and Solve; this is already factored and equal to zero:
Pretend the inequality sign is an equal sign:

#x=0,-7/2#

Plot these on a number line:

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The two points create #3# sections. Plug in any point into the inequality from each section and check the sign:

#-4(2*-4+7)>=0->"negative times negative is positive, this works"#

#-2(2*-2+7)>=0->"negative times positive is negative, does not work"#

#2(2*2+7)>=0->"positive times positive is positive, this works"#

Write in interval notation:

#(-∞,-7/2]U[0,∞)#

(Use brackets as the answer can be equal to)