Is there a real number solution for #7 + \sqrt(x+12) = 6#? Explain.

1 Answer
Binayaka C.
Jan 30, 2018

Solution : #x= -11 ; sqrt(1)=+-1 #

Explanation:

#7+ sqrt(x+12)= 6 or sqrt(x+12)= 6-7 # or

#sqrt(x+12)= -1# . Squaring both sides we get,

#(x+12)= (-1)^2 or x+12 =1 : x=1-12 = -11#

Check : #7+ sqrt(-11+12)= 6 #

#7+ sqrt(1)= 6# here #sqrt(1)=+-1#

Solution : #x= -11 ; sqrt(1)=+-1 # [Ans]

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