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No solution.
We will find and check the solution of the given equation.
To solve an equation with a variable expression inside a radical, we need to make sure that the radical expression is isolated. Since the radical is isolated, we can raise both sides of the equation to a power equal to the index of the radical. In this case, we will raise both sides of the equation to the second power. Let's do it!
LHS^2=RHS^2
( sqrt(a) )^2 = a
(- a)^2=a^2
Calculate power
.LHS /16.=.RHS /16.
The solution of our equation is y= 4. Now, let's check whether our solution is extraneous.
To check our solution, we will substitute 4 for y into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
We obtained a false statement, so y=4 is an extraneous solution to the equation. Therefore, there is no solution to the equation.