Prove: If a, b, c real, uneql, then (c-a)x^2-2(a-b)x+(b-c)=0

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leeyiu

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Question: If a,b and c are real numbers and not all equal, prove that the quadratic equation (c-a)x^2 -2(a-b)x+(b-c)= 0 has unequal real roots.

I can only get to the step 4a^2+4b^2+4C^2-4ab-4ac-4bc
can anyone help to continue on? please
 
Re: Prove quadratic equation/discriminant

Hello Leeyiu:

Are you familiar with the Discriminant? It's a value; here's its definition:

b^2 - 4*a*c

This value tells us information about the roots of a quadratic polynomial. Go to the Great God Google (or use a textbook), and check it out.

Let us know what you discover, if you need more help with this exercise.

Cheers,

~ Mark :)
 
Re: Prove quadratic equation/discriminant

I know what's that, but i cant go on with 4a^2+4b^2+4C^2-4ab-4ac-4bc,even if i factorize it, I cant get the result to prove the discriminant is larger than zero
can u tell me if I have made any mistakes
 
leeyiu said:
can u tell me if I have made any mistakes
Click to expand...
Since we can't see your work, I'm afraid there is no way to do this. Sorry! :oops:

Hint: Look at the simplified form of (a - b)[sup:9wxfybn7]2[/sup:9wxfybn7] + (b - c)[sup:9wxfybn7]2[/sup:9wxfybn7] + (c - a)[sup:9wxfybn7]2[/sup:9wxfybn7].... :D

Eliz.
 
alright I'll show you my work
(c-a)x^2-2(a-b)x+(b-c)=0
discriminant=
4(a-b)^2 -4(c-a)(b-c)=
4a^2 -8ab +4b^2 + 4c^2 -4cb-4ca-4ab =
4a^2 +4b^2 +4c^2 - 4cb -4ca -4ab
that is the farthest point that i can reach,
I know if the discriminant>0 means that the the equation has unequeal reall roots
but how do you prove it is >0 from this point
Can you five my some more hints?
 
Re: Prove quadratic equation/discriminant

leeyiu said:
I know what's that ...
Click to expand...

You kept this fact a secret in your original post; thank you for revealing this secret.

4a^2+4b^2+4C^2-4ab-4ac-4bc ... can u tell me if I have made any mistakes
Click to expand...

I think that the answer to your question is "no", despite the fact that you've not asked if you made any mistakes. If you are trying to ask whether or not the expression that you typed is the Discriminant in your exercise, then my answer to that question is "yes" because it is.

Do you understand what Elizabeth suggested? Are you willing to reveal any more secrets?

~ Mark :|
 
Re: Prove quadratic equation/discriminant

leeyiu said:
4a^2+4b^2+4C^2-4ab-4ac-4bc
Click to expand...

Pull out the 4's and you're left with a^2 + b^2 + c^2 - (ab + ac +bc)

Squares are larger than rectangles...kapish?
 
leeyiu said:
how do you know that (-ab-bc-ca) is a positive numbers?
Click to expand...


Hello Leeyiu:

This knowledge is not necessary.

Consider this expression's absolute value, instead.

Cheers,

~ Mark :)

 
leeyiu said:
alright I'll show you my work
(c-a)x^2-2(a-b)x+(b-c)=0
discriminant=
4(a-b)^2 -4(c-a)(b-c)=
4a^2 -8ab +4b^2 + 4c^2 -4cb-4ca-4ab =
4a^2 +4b^2 +4c^2 - 4cb -4ca -4ab

= 2[(a^2 + b^2 - 2ab) + (b^2 + c^2 - 2bc) + (c^2 + a^2 - 2ac)]

= prove this is a positive number and you are done
Click to expand...
 
Sub, isn't it enough/easier to prove that a^2 + b^2 + c^2 - (ab + ac +bc) is positive ?
 
I think - Sum squares is positive - is - conceptually easier than - squares are larger than rectangles.
 
Je suis d'accord; I meant "squares are larger than rectangles" as a hint.
 
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