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

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Answer 1 · 2017-03-24T02:16:07

#x = 4 # is applicable for +ve value of both sides
#x = 1# is applicable for -ve value of both sides

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

square both sides
#((sqrt x) - 1 )^2= (sqrt (5-x))^2#

#x - 2 sqrt x + 1 = 5-x#

#x + x - 2 sqrt x + 1 -5= 0#

#2 x - 2 sqrt x - 4= 0#

Let say #sqrt x = y#, then #x =(sqrt x)^2 = y^2#

#2 y^2 -2 y - 4 = 0#
# 2(y^2 -y -2) = 0#
#2(y - 2)(y + 1) = 0#
#y = 2, y = -1#

When #y = 2#, #x = 2^2 = 4#
When #y = -1#, #x = (-1)^2 = 1#