Answer:
x=-6 is an extraneous solution
Step-by-step explanation:
we have
[tex]\sqrt{-3x-2} =x+2[/tex]
squared both sides
[tex]-3x-2=(x+2)^2[/tex]
[tex]-3x-2=x^2+4x+4[/tex]
[tex]x^2+7x+6=0[/tex]
solve the quadratic equation by graphing
The solution is x=-6 and x=-1
see the attached figure
Verify each solution
substitute the value of x in the original expression
For x=-6
[tex]\sqrt{-3(-6)-2} =-6+2[/tex]
[tex]\sqrt{16} =-4[/tex]
[tex]4=-4[/tex] ----> is not true
so
Is an extraneous solution
For x=-1
[tex]\sqrt{-3(-1)-2} =-1+2[/tex]
[tex]\sqrt{1} =1[/tex]
[tex]1=1[/tex] ----> is true
so
Is the solution