SOLUTION: √x+√(1-x)+√(x(1-x))=1

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Question 1152368: √x+√(1-x)+√(x(1-x))=1
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!


sqrt%28x%29%2Bsqrt%281-x%29%2Bsqrt%28x%281-x%29%29=1

sqrt%28x%29%2Bsqrt%281-x%29=1-sqrt%28x%281-x%29%29.............square both sides

%28sqrt%28x%29%2Bsqrt%281-x%29%29%5E2=%281-sqrt%28x%281-x%29%29%29%5E2



x%2B2sqrt%28x%281-x%29%29%2B1-x=1-2sqrt%28x%281-x%29%29%2Bx-x%5E2

2sqrt%28x%281-x%29%29%2B1=1-2sqrt%28x%281-x%29%29%2Bx-x%5E2

x%5E2-x%2B1-1=-2sqrt%28x%281-x%29%29-2sqrt%28x%281-x%29%29

x%5E2-x=-4sqrt%28x-x%5E2%29%29.........square both sides

%28x%5E2-x%29%5E2=%28-4sqrt%28x-x%5E2%29%29%5E2+

x%5E4+-+2+x%5E3+%2B+x%5E2=16+%28x+-+x%5E2%29

x%5E4+-+2+x%5E3+%2B+x%5E2=16+x+-+16x%5E2

x%5E4+-+2+x%5E3+%2B+x%5E2-16+x+%2B+16x%5E2=0

x%5E4+-+2+x%5E3+%2B+17x%5E2-16+x+=0

x%28x%5E3+-+2+x%5E2+%2B+17x-16+%29+=0

x%28x+-+1%29+%28x%5E2+-+x+%2B+16%29+=+0

real solutions:
x=0
x=1
use the discriminant b%5E2-4ac to determine if solutions of +x%5E2+-+x+%2B+16+=+0+ are real or complex

b%5E2-4ac=%28-1%29%5E2-4%2A1%2A16=1-64=-63
since b%5E2-4ac%3C0+=> we have complex solutions and we do+not need them
​


Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

sqrt%28x%29 + sqrt%281-x%29 + sqrt%28x%2A%281-x%29%29 = 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        sqrt%28x%29 is defined and is positive number  sqrt%28x%29 > 0.

Similarly,  sqrt%281-x%29 is defined and is positive number  sqrt%281-x%29 > 0.



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

        a + b > sqrt%28a%5E2+%2B+b%5E2%29.    (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= sqrt%28x%29  and  b= sqrt%281-x%29.  You will get

    sqrt%28x%29 + sqrt%281-x%29 > sqrt%28%28sqrt%28x%29%29%5E2+%2B+%28sqrt%281-x%29%29%5E2%29 = sqrt%28x+%2B+1-x%29 = sqrt%281%29 = 1.


Thus,  the sum  sqrt%28x%29 + sqrt%281-x%29  at  0 < x < 1  is just greater than 1.


With the added positive addend  sqrt%28x%2A%281-x%29%29,  the sum  sqrt%28x%29 + sqrt%281-x%29 + sqrt%28x%2A%281-x%29%29  is just even more than 1.  


Therefore,  the sum  sqrt%28x%29 + sqrt%281-x%29 + sqrt%28x%2A%281-x%29%29  can not be equal to 1  at  0 < x < 1.



Thus, it is  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.

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Solved.