Question

The number of real roots of equation (x-1)^2 + (x-2)^2 + (x-3)^2 = 0 is?

A. 2       B. 1

C. 0       D. 3

Answer 1

Note:

(1) b^2 - 4ac > 0  ----- Positive      ---- 2 (Number of real roots)

(2) b^2 - 4ac < 0  ---- 0                 ---- 1(Number of real roots)

(3) b^2 - 4ac > 0  ----- Negative  ------ 0(Number of real roots).


Now,

 Given Equation is (x - 1)^2 + (x - 2)^2 + (x - 3)^2 = 0

= > x^2 + 1 - 2x + x^2 + 4 - 4x + x^2 + 9 - 6x = 0

= > 3x^2 - 12x + 14 = 0 .

Given that the equation has real roots.

= > b^2 - 4ac 

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

= > 144 - 168

= > -24

The discriminant is -ve, Therefore it has Zero real roots.



Hope this helps!