Radicals and rational eqns: How are k=-2,-3 the soln's to (K+2)(k+3)=0?

Illvoices said:
(K+2)(k+3)=0

How are the possible solutions k=-2 and k=-3
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K is a place holder for a number. Since k is just a number (not known yet but still a number), then k+2 is just a number. Likewise k+3 is just a number We know that when we multiply these two numbers (k=2 and k+3) we get 0.

Well if you multiply two numbers and get 0, then one or both of the numbers must be 0.

So when is k+2 =0 since -2 + 2 = 0 we have k=-2

When is k+3 = 0? Since -3 + 3 = 0, it follows that k= -3

What were you getting when you plugged in k= -2 into the equation? What did you get when you plugged in k = -3 into the equation?
 
They are in Khan i left them out b/c I thought they were unnecessary given the title description above
 
Illvoices said:
They are in Khan i left them out b/c I thought they were unnecessary given the title description above
Click to expand...
So there are radicals? So is the radical the 2nd root or the 3rd root or is it the 1/3 root, all of the above or something else? Is the unknown radical on the k or k+2 or 2 or k+3, ...? Since this is rational expression where is the division sign. Is it between .....? How are we suppose to know? Why not just simply write the exact problem?
 
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