Lesson Advanced problems on solving equations containing radicals

Advanced problems on solving equations containing radicals

Problem 1

 +  +  = 1.      (1)


The domain, where all included functions are defined, is the segment  [0,1].


Two obvious solutions to the given equation in this domain are x= 0  and  x= 1.


Below I will show that the given equation HAS NO other solutions.



Indeed, let  0 < x < 1.

Then         is defined and is positive number   > 0.

Similarly,   is defined and is positive number   > 0.



    For any two real positive numbers "a" and "b" the following inequality is valid

        a + b > .    (2)


    To prove it, square both sides. You will get

        a^2 + 2ab + b^2 > a^2 + b^2,

    which is valid for all positive "a" and "b".



Now apply the inequality (2) for  a=   and  b= .  You will get

     +  >  =  =  = 1.


Thus,  the sum   +   at  0 < x < 1  is just greater than 1.


With the added positive addend  ,  the sum   +  +   is just even more than 1.  


Therefore,  the sum   +  +   can not be equal to 1  at  0 < x < 1.



Thus, it  PROVED  that the given equation has no solutions inside the segment [0,1].  

So, the endpoints  x= 0  and  x= 1 are the only solutions.

Problem 2

 +  +  = 17      (1)

 +  = 17 -  


Square both sides


(10-x) +  + (3+x) = 17 -  + 4*(30 +7x - x^2)


Notice that  (10-x)*(3+x) = 30 + 7x - x^2, and continue transform preceding equations


(10-x) +  + (3+x) = 289 -  + 4*(30 +7x - x^2)

13 +  = 289 -  + 4*(30+7x-x^2)

0 = 276 -  + 4*(30+7x-x^2)      (2)


Introduce new variable t = .


Then equation (2) takes the form


4t^2 - 70t + 276 = 0.


Solve it using the quadratic formula


 =  =  = .


Case 1.  t =  = -6.


         Then t =  = -6 implies (after squaring both sides)

              30 + 7x - x^2 = 36

              x^2 - 7x + 6 = 0

              (x-1)*(x-6) = 0

              The roots are  x= 1  and  x= 6.

              You can easily check that both these roots satisfy the original equation.



Case 2.  t =  = -11.5.


         Then t =  = -11.5 implies (after squaring both sides)

              30 + 7x - x^2 = 132.25

              x^2 - 7x + 102.25 = 0

              Discriminant d = b^2 - 4ac = 7^2 - 4*102.25 is negative,

              Hence, this case does not produce real solutions.



The solution is completed.


The  ANSWER  is:  the original equation has two solutions  x= 1  and  x= 6.

Problem 3

Your starting equation is

     +  = 2.    (1)


Let  a = ;  b = .


Raise both sides of equation (1) to degree 3.

Use

     =  + 3ab*(a+b),


     =   +  = 14,

    a + b = 2   <<<---===  given by equation,

    ab = .


You will get

     = 8.


Simplify and find x

     = ,

     = -1.


Raise both sides of the last equation to degree 3

    49 - x = -1,

    49 + 1 = x,

     x     = 50.


ANSWER.  x = 50.


CHECK.    +  = 2.414213562 + (-0.414213562) = 2.0000000