How do you solve #sqrtx- sqrt(x-5)=1#?

1 Answer
R V SUBBARAO
Oct 13, 2015

#x# = 9

Explanation:

#sqrt (x) -sqrt(x-5) = 1# equation 1

multiply both sides with #sqrt (x) +sqrt(x-5)#

(#sqrt (x) -sqrt(x-5)) (sqrt (x) +sqrt(x-5))# = 1#(sqrt (x) +sqrt(x-5))#

L H S is in the form of (a+b)(a-b) =# a^2 -b^2#

#(sqrt(x))^2 - (sqrt(x-5))^2 # = #sqrt (x) +sqrt (x-5)#
#x - (x-5) =sqrt (x) +sqrt(x-5)#
#x-x+5 = (sqrt (x) +sqrt(x-5))#
#5 = sqrt (x) +sqrt(x-5)# equation 2

Solve equation 1 and 2 for #x#

#sqrt (x) -sqrt(x-5) = 1# equation 1

# sqrt (x) +sqrt(x-5) = 5# equation 2

Sum of equation 1 & 2

2#sqrt(x) = 6#
#sqrt(x) = 6/2# = 3
Squaring on both side to get #x#

#x = 9#

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