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

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Answer 1 · 2016-04-18T12:22:20

#x={9}#

square both sides.
x-5=(x-7)².
x-5=x²-14x+49.
0=x²-15x+54.

#Delta=sqrt((-15)^2-4*1*54)#
#Delta=sqrt(225-216)#
#Delta=sqrt(9)#
#Delta=±3#

#x_1=(15-3)/2#

#x_1=6#

#x_2=(15+3)/2#

#x_2=9#
#x={9}#

Answer 2 · 2016-04-18T12:22:55

#x = {9}#

#sqrt(x-5)=x-7#
#(sqrt(x-5))^2=(x-7)^2#

#x-5=(x-7)^2#

#x-5=x^2-14x+49#

#x^2-14x-x+49+5=0#

#x^2-15x+54=0#

#(x-6)(x-9)=0#

#if (x-6)=0" "then" "x=6#

#if (x-9)=0" "then" "x=9#

Watch out for extraneous solutions! 6 does not work in the original equation, however, 9 does. Thus, #{x =9}#