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1.4 Solving Radical Equations  More detail

Example 1.  Solve   

To get rid of the square root, we need to square each

side of the equation as a unit:                                            

The left side is NOT going to be fun to square because it will have to be multiplied out, or FOILED.

An easier way would be to isolate the radical first

before squaring the equation.                                                       

Now square both SIDES of the equation.                                     

This is much easier to do but be sure to multiply out, or FOIL the expression on the right side of the equation.  You cannot just “distribute”  the exponent to each term since the expression involves subtraction.

                                                                                              (3.1)

Equation (3.1) is now a quadratic equation.

Set the equation equal to zero:                                                                  

Factor:                                                                                         

Solve:                                                                                       

There is one more step that must be completed when solving radical equations.  We The MUST check all solutions in the ORIGINAL equation.   In the squaring process, extraneous solutions may have occurred.

CHECK:   

For x = 6:   

            Does  ?

            Does   

            Does   

            Does  3 + 3  =  6  ?  YES!  So 6 is a solution.

For x = 2:           

           Does  

           Does   

           Does    

           Does  1 + 3  =  2  ?  NO!  So 2 is an extraneous solution.

If we look at the graphs of , we see that they intersect at the point with , but NOT at the point where :

Example 2.  Solve    

Isolate the radical                                                                                                                                               

We could quit right now. The symbol  represents the primary or positive square root of the number represented by x.  The number, -2, is NOT a positive square root, therefore there is no solution to this equation.  

If we overlooked this fact and continued to solve this equation, however, we discover that there is no solution when we check the answer we found.  Continuing,

we square both sides of the equation:     

CHECK:          

For  x = 4                                 

               Does   +  4  =  2  ?

              Does    2 +  4  =  2  ?    NO!  So, 4 is an extraneous solution

If we look at the graphs of , we see that they do not intersect:

Example 3.  Solve     

Isolate the radical                                        

Since this is a cube root,

we cube both sides to eliminate the radical:

Set the equation equal to zero:                                   

Factor                                   

Set each factor containing a variable equal to zero and solve for each equation.

With an odd indexed radical equation, checking the solutions is optional.  

The graphical solution does support the solution we obtained algebraically:

.

When solving equations containing two radicals, it is easier to solve if you first separate the radicals so that there is only one radical on each side of the equation.

Example 4.                Solve                                                           

The radicals are already separated

so we just square both sides:                                              

                                              We use the rules of exponents here,    

                                                   taking everything inside the ( ) to the power.

Solve for x:                                                                                     

REMEMBER to check the solution back in the original equation:

CHECK:    

                  Does     ?

                              NO!  You CANNOT have a negative radicand under an even indexed

                             radical, so  -2  is an extraneous solution.

Notice that the graphical representation supports our algebraic solution.  The graphs of the two sides of the equation DO NOT intersect.

Example 5.             Solve      

Move one of the radicals over to the other side: 

Square both sides

                                                             There are two terms here so be sure to FOIL this

There is still a radical left in the equation,

so we need to isolate this radical.                                

Square both sides:                                                                                                                                     

                                                                             (3.2)

Equation (3.2) is a quadratic equation so we set it equal to zero.

                                                                                       (3.3)

Divide each term in equation (3.3) by 4 to make the factoring easier:

Now factor:                         

Set each factor equal to zero and then solve.   

Because we squared both sides, extraneous solutions are possible.  We need to check each solution back in the ORIGINAL equation to see if it works.

CHECK:                                 

For x = 14   

NO!  So 14 is an extraneous solution.

For  

YES!  So -1 is a solution.

Notice that the graphical representation supports our algebraic solution.  The graphs of the two sides of the equation intersect at .

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