Respuesta :

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

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