How do you solve #sqrt(m+2)<sqrt(3m+4)#?

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Answer 1 · 2017-08-14T01:47:37

#m > -1#.

If we square both sides, we can get:

#(sqrt(m + 2))^2 < (sqrt(3m + 4))^2 #

#m + 2 < 3m + 4#

#-2m < 2#

We now divide both sides by #-2#, not forgetting to switch the direction of the inequality symbol.

#m > -1#

We now test our answer to see if it is true. Let #m = 1#.

#sqrt(3) < sqrt(7) color(green)(√)#

This is obviously true.

Hopefully this helps!