How do you solve #x-6 = (x-4)^(1/2)# and find any extraneous solutions?

2 Answers
Sep 5, 2016

Answer:

#x=8 or 5#

If #x<4# then #x-4# is negative #=># solution is in the complex number set of values.

Explanation:

Square both sides giving:

#(x-6)^2=x-4#

#x^2-12x+36=x-4#

#x^2-13x+40=0#

completing the square

#0=(x-13/2)^2+40 -169/4#

#0=(x-13/2)^2+160/4-169/4#

#0=(x-13/2)^2-9/4#

#x=13/2+-sqrt(9/4)#

#x=8 or 5#

Sep 5, 2016

Answer:

#x=5 or x =8#

Explanation:

Remember than the index of #1/2# is the same as the square root.
We cannot find the square root of a negative number, so

#x-4 >=0 rArr x >= 4#
#color(red)("remember to check this later")#

To get rid of the square root, you need to square both sides. A common error is to square each of the terms of the left, inside of treating them as a binomial.

#(x-6)^2 = ((x-4)^(1/2))^2#

#x^2 -12x +36 = x-4 " "larr # a quadratic, so make = 0.

#x^2 -13x +40 = 0 " "larr # check for factors.

'Find factors of 40 which add (because of +40) to give 13.'
The signs will be the same, (because of +40) they are both minus (because of -13).

#"8 and 5 will work ": 8xx5=40 and 8+5=13#

#(x-5)(x-8)=0#

#"letting each factor equal 0 gives " x=5 or x =8#

#color(red)("Both answers are valid")#