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Mathematics 21 Online
OpenStudy (anonymous):

solve for x: x^2 + 5x is > or = 6

myininaya (myininaya):

x^2+5x-6>=0 (x+6)(x-1)>=0 when is this LHS zero?

myininaya (myininaya):

when x=___ and x=___?

OpenStudy (anonymous):

What is LHS..?

OpenStudy (anonymous):

You would use the (x + 6)(x - 1) >= 0 When x <=-6 or x >= 1

OpenStudy (anonymous):

from the (X+6) you would then solve for x for both giving you the positive 1 and negative 6 correct?

OpenStudy (anonymous):

That would give you when it = 0, but when would it be > 0... see my earlier response

OpenStudy (anonymous):

Then where are you getting the positive and negative...... your doing x+6=0 or x-1+0 ... right

myininaya (myininaya):

(-inf,-6]u[1,inf)

OpenStudy (anonymous):

What.... does that say

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