How do you solve #sqrt(x + 6) = x# and find any extraneous solutions?

1 Answer
Jim G.
Oct 9, 2017

#x=3,x=-2" is extraneous"#

Explanation:

#color(blue)"squaring both sides"#

#(sqrt(x+6))^2=x^2#

#rArrx+6=x^2#

#"rearrange in standard form "ax^2+bx+c=0#

#rArrx^2-x-6=0#

#"the factors of - 6 which sum to - 1 are - 3 and + 2"#

#rArr(x-3)(x+2)=0#

#"equate each factor to zero and solve for x"#

#x-3=0rArrx=3#

#x+2=0rArrx=-2#

#color(blue)"As a check"#

#"substitute each of the possible solutions into the"#
#"original equation to test their validity"#

#x=3tosqrt(3+6)=sqrt9=3=xrArr" valid solution"#

#x=-2tosqrt(-2+6)=sqrt4=2!=xrArr"extraneous"#

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