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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
( sqrt(a) )^2 = a
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. s^2-2s-8=0 ⇔ 1s^2+( - 2)s+( -8)=0
Substitute values
- (- a)=a
(- a)^2=a^2
Identity Property of Multiplication
- a(- b)=a* b
Add terms
Calculate root
Using the Quadratic Formula, we found that the solutions of the given equation are s= 2± 6 2.
| s=2± 6/2 | |
|---|---|
| s_1=2+6/2 | s_2=2-6/2 |
| s_1=8/2 | s_2=-4/2 |
| s_1= 4 | s_2= -2 |
Therefore, the solutions are s_1=4 and s_2=-2. Let's check them to see if we have any extraneous solutions.
We will check s_1=4 and s_2=-2 one at a time.
Let's substitute s=4 into the original equation.
In this case we got a true statement. Therefore, s=4 is a solution of the original equation.
Now, let's substitute s= -2.
We got a false statement, so s=-2 is an extraneous solution. Therefore, s=4 is the only solution of the original equation.