How do you solve the equation: #x=sqrt(3x+40)#?

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Answer 1 · 2018-05-28T13:19:31

#x=8#

#x = sqrt{3x + 40}#

Over reals the radical sign refers to the principal square root, so squaring both sides may introduce extraneous solutions.

#x^2 = 3x+40#

#x^2 - 3x -40 = 0#

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

#x=8 or x=-5#

The principal square root won't ever be #-5# so we reject that as extraneous.

#x=8#

Check:

# sqrt{3(8)+40}=sqrt{64}=8 quad sqrt#