how do you solve (6x^2+27x+4)^0.5-2=x

Question

solve and identify all solutions as either real, imaginary, or extraneous

Answer 1

√(6x²+27x+4)=x+2, squaring:

6x²+27x+4=x²+4x+4,

5x²+23x=0=x(5x+23), so x=0 or -23/5, but we need to check for extraneous solutions.

When x=0: √4-2=0 OK.

When x=-23/5 we have √(169/25)=13/5.

This is the positive square root and 13/5-2=⅗≠-23/5 so this is an extraneous solution.

(The negative square root does satisfy the original equation: -13/5-2=-23/5 but we cannot use this because the original equation explicitly specifies the positive square root.)