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Rearrange the radical equation so that the radical expression is isolated. Then raise both sides of the equation to the second power.
5
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-x=RHS-x
LHS^2=RHS^2
( sqrt(a) )^2 = a
(a-b)^2=a^2-2ab+b^2
Rearrange equation
Commutative Property of Addition
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 -15x +50 = 0 ⇔ 1x^2+( - 15)x+ 50=0
Substitute values
Using the Quadratic Formula, we found that the solutions of the given equation are x= 15± 5 2. Therefore, the solutions are x_1=10 and x_2=5. Let's check them to see if we have any extraneous solutions.
We will check x_1=10 and x_2=5 one at a time.
Let's substitute x=10 into the original equation.
We got a false statement, so x=10 is an extraneous solution.
Now, let's substitute x=5.
In this case we got a true statement. Therefore, x=5 is the only solution of the original equation.