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 -16x +39 = 0 ⇔ 1x^2+( - 16)x+ 39=0
We can see that a= 1, b= - 16, and c= 39. Let's substitute these values into the Quadratic Formula.
Using the Quadratic Formula, we found that the solutions of the given equation are x= 16± 10 2. Therefore, the solutions are x_1= 13 and x_2=3.
Let's check them to see if we have any extraneous solutions.
Checking the Solutions
We will check x_1=13 and x_2=3 one at a time.
x_1=13
Let's substitute x= 13 into the original equation.
