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

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.

To determine the number of real roots of the equation (x - 1)² + (x - 2)² + (x - 3)² = 0, we can use the concept of quadratic equations and some algebraic manipulation.

First, expand the equation:
(x² - 2x + 1) + (x² - 4x + 4) + (x² - 6x + 9) = 0
Combine like terms:
3x² - 12x + 14 = 0

To find the number of real roots, we can use the discriminant, which is the expression under the square root sign in the quadratic formula.

The discriminant (D) is given by the formula: D = b² - 4ac, where a, b, and c are the coefficients of the quadratic equation ax² + bx + c = 0.

In our equation, the coefficients are:
a = 3
b = -12
c = 14

Now, substitute these values into the discriminant formula:
D = (-12)² - 4(3)(14)
D = 144 - 168
D = -24

Since the discriminant is negative (D < 0), the quadratic equation has no real roots. Therefore, the given equation (x - 1)² + (x - 2)² + (x - 3)² = 0 has no real roots.