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Given the equation 2 Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

x = 6, solution is not extraneous

x = 6, solution is extraneous

x = 11, solution is not extraneous

x = 11, solution is extraneous

Respuesta :

W0lf93
2 square root (x - 5) = 2. 2^2 (x -5) = 2^2. 4(x - 5) = 4. 4x - 20 = 4. 4x = 24. x = 6. This solution is not extraneous, because extraneous solutions emerge from solving the problem but are not actually valid solutions for the initial problem. With rounding, this solution is valid for the initial problem.

Answer:

x = 6, solution is not extraneous

Step-by-step explanation:

Given the equation 2 Square root of x minus 5 = 2

[tex]2\sqrt{x-5} =2[/tex]. solve for x

Divide by 2 on both sides

[tex]\sqrt{x-5} =1[/tex]

Take square on both sides to remove square root

x-5= 1

add 5 on both sides

x= 6

Now we plug in x=6 in our original equation

[tex]2\sqrt{6-5} =2[/tex]

2 = 2 is true

So x= 6 satisfies our given equation

X=6 is not an extraneous solution

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