the no. of real roots of eqn (x-1)POWER 2 + (X-2) POWER 2 +(X-3) POWER 2= 0 IS:

Answer 1

lose all the nonsense text. You have

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

Now, you know that (x-1)^2 is a parabola, which is tangent to the x-axis at x=1, because of the double root.

Similarly, the other two terms are never negative. So, since we are adding up three positive values for each x, the graph will never cross the x-axis, meaning there are no real roots.

Or, try expanding things to get

3x^2-12x+14=0

Since the discriminant is negative, there are no real roots.