How do you solve #sqrt(2n+3)=n# and check your solution?

Question

4211 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-25T20:46:56

#n=3, n=-1#

Square both sides:

#sqrt(2n+3)^2=n^2#

Recall that #sqrt(a)^2=a#. Thus,

#2n+3=n^2#

Get this into standard quadratic form, meaning we move everything to one side. We want #n^2# to stay positive, so we'll move everything to the right.

#n^2-2n-3=0#

Solve this for #n.# Easiest way is by factoring. Two numbers that add to give #-2# and multiply to give #-3# are #-3, 1.#

#(n-3)(n+1)=0#

So,

#n=3, n=-1#

Let's plug these into the square root to ensure we don't get a negative root, as negative roots will not give us a solution we desire.

#sqrt(2(3)+3)>0#, #n=3# is fine.

#sqrt(2(-1)+3)>0, # #n=-1# is fine.