Question

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

Answer 1

I exapanded the backets and got the answer , but i am sure there is a better way , so i have asked this poblem here

Answer 2

its quadratic equation at the end right ?
hmm so u got 2 roots ?

Answer 3

yep

Answer 4

yeah equation is missing

Answer 5

No real roots

Answer 6

Which equation is missing?

Answer 7

I don't see an equation above, are we to assume that expression = 0 ?

Answer 8

yes thats what i assumed

Answer 9

should have asked really...

Answer 10

a square can never be negative, so if the equation has a real solution, then it should evaluate the value of expression to 0. hmm still thinking...

Answer 11

(x-1)^2 + (x-2)^2 + (x-3)^2 >=0
lets assume it zero then the only way x=1 &x=2&x=3 (which is a contradiction )
thus no real
else expand and see hmmm

Answer 12

nice :)

Answer 13

I didn't get ikram , sorry it is equal to zero , i forgot to put
@ikram002p

Answer 14

@ganeshie8 @ikram002p

Answer 15

what i mean ur equation is :-
(x-1)^2 + (x-2)^2 + (x-3)^2=0
now left side is always positive ok ? (sum of squars )
and each term is also positive
thus the only way to get zero is each term =0
xD
hope this is help ='(

Answer 16

SO,
(x-1)^2 + (x-49)^2 + (x+5665)^2 +(x-66)^2 =0
can never have real roots??

Answer 17

with the same logic

Answer 18

humm yes

Answer 19

nice , if you are sure ^_^

Answer 20

positive sure :P

Answer 21

the logic is good , thank you