How do you solve #sqrt(12x+13)=2x+1 # and find any extraneous solutions?

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Answer 1 · 2018-05-06T11:55:50

#color(indigo)("Extraneous solutions are : " x = 3, -1#

$sqrt(12x + 13) = 2x + 1#

#"Squaring both sides," 12x + 13 = (2x + 1)^2#

#12x + 13 = 4x^2 + 4x + 1#

#4x^2 - 8x - 12 = 0#

#4x^2 + 4x - 12x - 12 = 0#

#4x (x + 1) - 12 (x + 1) = 0#

# (4x -12) * (x + 1) = 0#

# x = 12/4 = 3, -1#