SOLUTION: I am 45 years old and I am trying to help my exchange student with the problem: {{{sqrt(2*t + 3) + 2 = sqrt(t -2)}}}

Question 20987: I am 45 years old and I am trying to help my exchange student with the problem:

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!

.....IF WE ARE DEALING WITH REAL NUMBERS,
this prolem has no solution..reason is as follows ...we have square root of (t-2) on the r.h.s.for it to be real or meaning full,t should be equal to 2 or more than 2,since then only t-2 can have a square root.but if t>=2,then on the l.h.s.we have 2t+3 which will be definitely more than t-2,on the r.h.s.so
sqrt.of (2t+3)on l.h.s.will be more than sq.rt.of (t-2) on the r.h.s.then obviously we cant get the r.h.s. by adding 2 to the l.h.s.
IF YOU ARE DOING COMPLEX NUMBERS PLEASE TELL ME...TO GIVE YOU THE METHOD OF ATTACK FOR THAT
OR LET ME MODIFY THE PROBLEM TO GIVE YOU THE METHOD OF SOLUTION
.....
.....
square both sides,we get
2t+3+t-2+2*sq.rt.{(2t+3)*(t-2)}=4
3t+1-4=2sq.rt(2t^2-4t+3t-6)
3t-3=2sq.rt(2t^2-t-6).....square both sides again
(3t-3)^2=4*(2t^2-t-6)
9t^2+9-2*3*3*t=8t^2-4t-24
9t^2+9-18t-8t^2+4t+24=0
t^2-14t+33=0
t^2-11t-3t+33=0
t(t-11)-3(t-11)=0
(t-3)(t-11)=0
hence t=3 or 11....since we squared the given eqn.twice it is a good practice to back check the answers to ensure that we have not introduced any additional roots by squaring.
so putting t=3 in original eqn. we get
.....or...3-2=1....correct
so putting t=11 in original eqn. we get
.....or.....5-2=3...correct
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