Given that x is a real number, determine all possible solutions of:
√(x-1) + √(3-x) = x² - 4x + 6

√(x-1)+√(3-x)=x^2-4x+6
x-1+2√(x-1)(3-x)+3-x=(x^2-4x+6)^2
2√(x-1)(3-x)=(x^2-4x+6)^2-2
4(x-1)(3-x)={(x^2-4x+6)^2-2}^2
-4(x^2-4x+3)={(x^2-4x+6)^2-2}^2
Let x^2-4x=y.Therefore the expression is equal to
-4y-12={(y+6)^2-2}^2
-4y-12=y^4+24y^3+212y^2+816y+1156
y^4+24y^3+212y^2+820y+1168=0
(y-4)(y^3+20y^2+132y+292)=0
y-4=0
y=4
x^2-4x=4
x^2-4x-4=0
(x-2)^2=0
x-2=0
x=2

Posted by Danish Ahmed Khan on 2012-10-26 05:04:12