How do you solve #5+(sqrt(x+2))=8+(sqrt(x-7))#?

1 Answer
Jul 6, 2016

Answer:

#x=7.#

Explanation:

#5+sqrt(x+2)=8+sqrt(x-7)#
#:.sqrt(x+2)-sqrt(x-7)=3#
#:.{sqrt(x+2)-sqrt(x-7)}^2=3^2#
#:.x+2-2*sqrt(x+2)*sqrt(x-7)+x-7=9#
#:.2x-14=2*sqrt(x+2)*sqrt(x-7)#
#:.x-7=sqrt(x+2)*sqrt(x-7)..........................(I)#
#:.(sqrt(x-7))^2=sqrt(x+2)*sqrt(x-7).................(II)#

And, here is a Word of Caution : We can not cancel out (x-7) directly from both sides without checking whether (x-7)=0.

The reason behind this is, in case, #(x-7)=0,# cancelling it directly from both sides means that we are dividing both sides by 0, which is not possible!

Accordingly, we have to consider two cases :#(1): (x-7)=0#, & #:(2): (x-7)!=0.#

#Case :(1): (x-7)=0.#

In this case, we see that #(II)# is satisfied, and hence, #x=7# is a soln.

#Case :(2):(x-7)!=0.#

In this case, now we can divide both sides by #(x-7),# {bcz., it is not zero] and get,

#sqrt(x-7)=sqrt(x+2),# squaring which, we have an impossibility # -7=2.#

Hence, #x=7# is the only soln.

Another way to solve it, is, to square #(I)#, to get,
#(x-7)^2-(x+2)(x-7)=0#
#:. (x-7){(x-7)-(x+2)}=0#
#:. (x-7)(-9)=0#
#:. x=7.#

How's that? Enjoyable Maths.!