Solve. Check for extraneous solution.
6x= √20+6x
Are you kidding?
6 x = sqrt ( 20 ) + 6 x
6 x - 6 x = sqrt ( 20 )
0 = sqrt ( 20 ) ???????????
the way you typed it ....
0 = √20
---> silly statement, thus no solution
if you meant:
6x = √(20+6x)
then square each side
36x^2 = 20+6x
36x^2 - 6x - 20 = 0
(6x - 5)(6x + 4) = 0
x = 5/6 or x = -2/3
since we squared, all answers must be verified
if x = 5/6
LS = 6(5/6) = 5
RS = √(20 + 5) = √25 = 5 , good
if x = - 2/3
LS = 6(-2/3) = -4
Since the √(anything) by definition is positive, this is not valid
so x = 5/6
To solve the equation 6x = √20 + 6x and check for extraneous solutions, we can follow these steps:
Step 1: Move all terms involving x to one side of the equation.
Subtract 6x from both sides:
6x - 6x = √20 + 6x - 6x
This simplifies to:
0 = √20
Which means the equation becomes:
0 = √20
Step 2: Solve for √20:
√20 is the square root of 20, so we need to find the value of √20.
√20 is approximately 4.47 (rounded to two decimal places).
Step 3: Check for extraneous solutions:
Now, we substitute the value of √20 back into the original equation and see if it holds true:
6x = √20 + 6x
6x = 4.47 + 6x
Subtract 6x from both sides:
6x - 6x = 4.47 + 6x - 6x
0 = 4.47
Since 0 is not equal to 4.47, this means that there are no solutions that satisfy the original equation.
There is no extraneous solution to check in this case because the equation itself does not have a valid solution.