How do you solve #5=13-sqrt(25-2x)# and identify any restrictions?

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Answer 1 · 2017-04-22T21:42:19

Please see the explanation.

The restriction that I would add is #x<= 25/2#

#5=13-sqrt(25-2x);x<= 25/2#

Subtract 13 from both sides:

#-8=-sqrt(25-2x);x<= 25/2#

Square both sides:

#64 = 25 - 2x;x<= 25/2#

We can drop the restriction on the next step, because the root is clearly not going to violate it.

Add 2x-64 to both sides:

#2x = 25-64#

#2x = -39#

#x = -39/2#

Check:

#5=13-sqrt(25-2(-39/2))#

#5 = 13 - sqrt(64)#

#5 = 5#

This checks.