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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!
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 -2x -3 = 0 ⇕ 1x^2+( - 2)x+( -3)=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 x= 2± 4 2.
| x=2± 4/2 | |
|---|---|
| x_1=2+4/2 | x_2=2-4/2 |
| x_1=6/2 | x_2=-2/2 |
| x_1= 3 | x_2= -1 |
Therefore, the solutions are x_1=3 and x_2=-1. Let's check them to see if we have any extraneous solutions.
We will check x_1=3 and x_2=-1 one at a time.
Let's substitute x=3 into the original equation.
In this case we got a true statement. Therefore, x=3 is a solution of the original equation.
Now, let's substitute x= -1.
We got a false statement, so x=-1 is an extraneous solution. Therefore, x=3 is the only solution of the original equation.