5+sqrt(x+2)=8+sqrt(x-7)5+√x+2=8+√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.!