Question

the no of real roots ofequation(x-1)^2+(x-2)^2+(x-3)^2+0 is....

Answer 1

Step-by-step explanation:

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

x^2 + 1 - 2x + x^2 + 4 - 4x + x^2 + 9 - 6x

3x^2 - 12x + 14

Here,a=3, b=-12 & c=14

D= b^2-4ac

= (-12)^2 - (4.3.14)

= 144 - 168

= -24

Since, D<0

Therefore, It has no real roots