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A Real Number problem (Posted on 20060816) 

Given that x is a real number, determine all possible solutions of:
√(x1) + √(3x) = x²  4x + 6

Submitted by K Sengupta

Rating: 4.0000 (2 votes)


Solution:

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Clearly, 1<= x< =3, since x is a real number. Otherwise, one of the two expressions in the LHS would be imaginary.
Let, y = sqrt(x1) + sqrt(3x)
or, y^{2} = 2 + 2*sqrt {(x1)(3x)}
= 2 + 2*sqrt {1  (x2)^2} < = 2+2*1 = 4
or, y < =2(#)
Now, x^{2} 4x + 6 = (x2)^{2}+2 >=2(##)
Now, (#)and (##) is feasible only when :
x^{2} 4x + 6 = 2
or, x = 2.
Consequently, x =2 is the only feasible solution to the given problem. 
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