Math 112  9.  Equations with Radicals

 

Since  Ãa can be written as a1/2 these are just a special form of the "quadratic types" of equations.  Since we must raise both sides to the power 2 to eliminate the radical, all solutions must be checked  to see if they are valid. 

 

The steps to solve are:

 

1.  Isolate a radical term.

2. Square both sides of the equation to eliminate the radical.

3.  Repeat steps 1 & 2 until all radical terms are eliminated.

4.  Solve the problem.

5.  Check all solutions in the original problem.

 

Example1:

     Since a radical term is already isolated, we can square both sides and get

()2 = ((x + 2)1/2)2 =  x+2 = 16, which yields x = 14

 

checking this solution in the original problem we have , which is true, so x = 14 is a solution.

 

Example 2:

 

         Since a radical term is already isolated, and keeping in mind that the square of 5-y = 25 -10y + y2 , when we square both sides we get

 

()2 = ((4y +1)1/2)2 = 4y + 1 = (5 - y)2  = 25 -10y + y2

 

writing this in standard form we get

 

y2 - 14y + 24 = 0, which factors as (y-2)( y-12) = 0, and has solutions y = 2, y = 12

 

checking y = 2 we have , so y = 2 is a solution

 

checking y = 12 we have , so 12 is not a solution.

 

Example 3:

 

              This time we must first isolate a radical term, which we will do by adding 2Ãt to both sides, which gives us

 

              now we will square both sides and get

 

((5t + 4)1/2)2 = 5t +4 = (2Ãt +1)2

 

Now we have eliminated the radical on the left, but 2Ãt is a radical term on the right.  Before we can isloate this term we must first square the binomial (2Ãt + 1) very carefully.  Doing this we get

5t +4 = 4t + 4Ãt + 1               Now we can islolate the radical term 4Ãt and get

 

t + 3 = 4Ãt                  Now, squaring both sides to eliminate the radical we get,

 

t2 + 6t + 9 = 16 t         and, writing this in standard form so we can solve it gives

 

t2 - 10t + 9 = 0            which factors as (t - 9)( t- 1) = 0 , and has solutions t = 9 and t = 1

 

Checking t = 9 in the original problem we have

 

, so t = 9 is a solution

 

Checking t = 1 in the original problem we have

 

 , so t = 1 is also a solution.

 

Problems:

 

1.                has solution x = 8

 

2.     has solution y = 6

 

3.              has no solution