Question

Can a quadratic really have 3 solutions!
I will type up the equation in the equation editor.

Answer 1

Quad means 4. :O

Answer 2

\[\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}=1\]
and a<b<c
has solutions:
x=a, x=b, x=c
how?

Answer 3

yet this is a quadratic equation

Answer 4

@saifoo.khan - quadratic equations usually have just 0, 1 or 2 real solutions. so how does this one have 3 real solutions?

Answer 5

BTW: this is a fallacy - can you spot what is wrong?

Answer 6

two factor should cancel out?

Answer 7

what do you mean @imranmeah91?

Answer 8

nothing concrete yet, but one factor is negative of other so they cancel?

Answer 9

if you put x=a into the above, you will see that you get 1 = 1
same with x=b and x=c

Answer 10

hmm i think if we open this up LHS =RHS

Answer 11

what i mean to say is the equation on LHS isn't a quad. but it's equal to 1 when we open it up, i am not sure i must do the algebra to prove it

Answer 12

yes @Ishaan94 you are correct - it lloks like a quad but all terms in x cancel out leaving 1 = 1. well done!

Answer 13

how about

(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a)=1

Answer 14

?

Answer 15

this doesn't simplify to 1

Answer 16

@Ishaan94 has got the answer - basically this equation is completely independent of 'x'. any value for x will satisfy the equation. if you do all the hard work and add these 3 fractions you will see that the left hand side simplifies to just "1".

Answer 17

no , it's a new question

Answer 18

so whats the new question asking?