How do you solve #sqrt(x^2+5)=3-x# and find any extraneous solutions?

2 Answers
Aug 28, 2016

Answer:

#x=2/3#

Explanation:

Given:

#sqrt(x^2+5) = 3-x#

Square both sides (noting that this may introduce spurious solutions) to get:

#x^2+5 = 9-6x+x^2#

Subtract #x^2# from both sides to get:

#5 = 9-6x#

Subtract #9# from both sides to get:

#-4 = -6x#

Divide both sides by #-6# to get:

#2/3 = x#

This is a valid solution of the original equation:

#sqrt((2/3)^2+5) = sqrt(4/9+5) = sqrt(49/9) = 7/3 = (9-2)/3 = 3-2/3#

Aug 28, 2016

Answer:

#x=2/3#

Explanation:

#color(blue)(sqrt(x^2+5)=3-x#

#"Square both sides to cancel out the radical sign-"# #sqrt#

#rarr(sqrt(x^2+5))^2=(3-x)^2#

#rarrx^2+5=(3-x)^2#

#"Use the formula "# #color(orange)((a-b)^2=a^2-2ab+b^2#

#rarrx^2+5=3^2-2(3)(x)+x^2#

#rarrx^2+5=9-6x+x^2#

#"Cancel"# #x^2# #"both sides"#

#rarrcancel(x^2)+5=9-6x+cancel(x^2#

#rarr5=9-6x#

#"Rewrite the equation"#

#rarr6x=9-5#

#rarr6x=4#

#color(green)(rArrx=4/6=2/3#