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

x = 6, solution is not extraneous

Step-by-step explanation:

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

. solve for x

Divide by 2 on both sides

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

2 = 2 is true

So x= 6 satisfies our given equation

X=6 is not an extraneous solution

step-by-step explanation:

it's is b

45

step-by-step explanation: