Which of the following is an extraneous solution of StartRoot negative 3 x minus 2 EndRoot = x + 2? x = –6 x = –1 x = 1 x = 6

The extraneous solution is x=-1 and x=-6Step-by-step explanation:Given the question as;[tex]\sqrt{-3x-2} =x+2[/tex]Eliminate the root in the left hand side[tex](\sqrt{-3x-2}) =x+2\\\\\\(\sqrt{-3x-2}  )^2=(x+2)^2\\\\-3x-2=x^2+4x+4\\\\0=x^2+4x+3x+4+2\\\\\\0=x^2+7x+6[/tex]solve the quadratic equation by factorizationFind multiples of 6 that add up to 7-------- 1*6Rewrite the equation as;x²+7x+6=0x²+6x+x+6=0x(x+6) +1 (x+6) =0(x+1)(x+6) =0x+1=0x=-1 orx+6=0x=-6The roots are, x=-1, and x=-6Learn MoreQuadratic equations : : extraneous, solution#LearnwithBrainly