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

1 Answer
Aug 14, 2017

Answer:

#m > -1#.

Explanation:

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!