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To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!
LHS-2=RHS-2
LHS^2=RHS^2
( sqrt(a) )^2 = a
LHS-x=RHS-x
Subtract terms
Rearrange equation
We now have a quadratic equation, and we need to find its roots. To do it, let's identify the values of a, b, and c. x^2 -5x +4 = 0 ⇕ 1x^2+( -5)x+ 4=0
Substitute values
Using the Quadratic Formula, we found that the solutions of the given equation are x= 5± 32.
| x=5± 3/2 | |
|---|---|
| x_1=5+3/2 | x_2=5-3/2 |
| x_1=8/2 | x_2=2/2 |
| x_1= 4 | x_2= 1 |
Therefore, the solutions are x_1= 4 and x_2= 1. Let's check them to see if we have any extraneous solutions.
We will check x_1=4 and x_2=1 one at a time.
Let's substitute x=4 into the original equation.
We got a true statement. Therefore, x=4 is a solution of the original equation.
Now, let's substitute x=1.
In this case, we got a false statement, so x=1 is an extraneous solution. Therefore, x=4 is the only solution of the original equation.