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Raise both sides of the equation to a power equal to the index of the radical.
x=2
To solve equations with a variable expression inside a radical, 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. x^2-x-2=0 ⇔ ( 1)x^2+( - 1)x+( - 2)=0
Substitute values
- (- a)=a
Calculate power
Identity Property of Multiplication
- a(- b)=a* b
Add terms
Calculate root
This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.
| x=1± 3/2 | |
|---|---|
| x_1=1+3/2 | x_2=1-3/2 |
| x_1=4/2 | x_2=- 2/2 |
| x_1=2 | x_2=- 1 |
Therefore, the solutions are x_1= 2 and x_2= - 1. Let's check them to see if we have any extraneous solutions.
We will check x_1=2 and x_2=- 1 one at a time.
Let's substitute x= 2 into the original equation.
We got a true statement. Therefore, x=2 is the 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, we obtain that x=2 is the only solution of the original equation.