Question 1152368



{{{sqrt(x)+sqrt(1-x)+sqrt(x(1-x))=1}}}


{{{sqrt(x)+sqrt(1-x)=1-sqrt(x(1-x))}}}.............square both sides


{{{(sqrt(x)+sqrt(1-x))^2=(1-sqrt(x(1-x)))^2}}}


{{{(sqrt(x))^2+2sqrt(x)*sqrt(1-x)+(sqrt(1-x))^2=1-2sqrt(x(1-x))+(sqrt(x(1-x)) )^2}}}


{{{x+2sqrt(x(1-x))+1-x=1-2sqrt(x(1-x))+x-x^2}}}


{{{2sqrt(x(1-x))+1=1-2sqrt(x(1-x))+x-x^2}}}


{{{x^2-x+1-1=-2sqrt(x(1-x))-2sqrt(x(1-x))}}}


{{{x^2-x=-4sqrt(x-x^2))}}}.........square both sides


{{{(x^2-x)^2=(-4sqrt(x-x^2))^2 }}}


{{{x^4 - 2 x^3 + x^2=16 (x - x^2)}}}


{{{x^4 - 2 x^3 + x^2=16 x - 16x^2}}}


{{{x^4 - 2 x^3 + x^2-16 x + 16x^2=0}}}


{{{x^4 - 2 x^3 + 17x^2-16 x =0}}}


{{{x(x^3 - 2 x^2 + 17x-16 ) =0}}}


{{{x(x - 1) (x^2 - x + 16) = 0}}}


real solutions:

{{{x=0}}}
{{{x=1}}}

use the discriminant {{{b^2-4ac}}} to determine if solutions of {{{ x^2 - x + 16 = 0 }}} are real or complex
 
{{{b^2-4ac=(-1)^2-4*1*16=1-64=-63}}}

since {{{b^2-4ac<0 }}}=> we have {{{complex}}} solutions and we {{{do}}}{{{ not}}} need them   
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