Respuesta :

W0lf93
4*sqrt(x-3)= 12. On squarng both the sides, we get, 16.(x-3) =144. or (x-3)= 144/16 = 9 or we can say that x = 3+9 = 12. Thus, x = 12 and it is not an extraneous solution. It would have been extraneous if x would have been 3.

Answer:

Given the equation: [tex]-4\sqrt{x-3}=12[/tex]          .....[1]

Solve for x;

Divide both sides by -4 in [1];  we get;

[tex]\sqrt{x-3} = -3[/tex]

Squaring both the sides we get;

[tex]x-3 = (-3)^2[/tex]

or

[tex]x -3 = 9[/tex]

Add 3 both sides we get;

[tex]x -3+3= 9+3[/tex]

Simplify:

x= 12

Extraneous solution states that it is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.

Substitute  x= 12 in [1]

[tex]-4\sqrt{12-3}=12[/tex]  

[tex]-4\sqrt{9}=12[/tex]  

[tex]-4\cdot 3=12[/tex]  

[tex]-12 =12[/tex]   False.

Therefore, the value of x is 12 and it is an extraneous solution.

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