# Write the number of real roots of the equation.

given (x-1)2+(x-2)2+(x-3)2=0

x2 + 1 - 2x + X2 + 4 - 4x + X2 + 9 - 6x = 0

3X2 - 12x+ 14 = 0

Comparing it with aX2 + bx+ c = 0 and substituting them in b2 – 4ac, we get

= (-12)2-4(3)(14)

= 144 – 168

= - 24 < 0 .

Hence the given equation do not have real roots. It has imaginary roots.

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