How do you solve #sqrt(x+3)= x-3#?

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Answer 1 · 2016-05-23T10:36:34

#x=+1" and "x=+6#

Square both sides

#x+3=(x-3)^2" "->" "x+3=x^2-6x+9#

Subtract #x# from both sides

#3=x^2-7x+9#

Subtract 3 from both sides

#0=x^2-7x+6#

Notice that #(-1)xx(-6) = +6" and "-1-6=-7#

Factorising

#=>0=(x-1)(x-6)#

#x=+1" and "x=+6#