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 August 3rd, 2013, 07:57 PM #1 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Binomial as a,b and c in a quadratic equation. the question : Show the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are always real and they can not be equal unless a=b=c. a= (x-a)(x-b) , b=(x-b)(x-c) , c = (x-c)(x-a). (x-a)(x-b) = x^2-bx-ax-+ab. ......... I ended up with this $D= x^4-cx^3+bx^3-cx^3+cx^2+bx^3-b^2x^2+b^2cx-4x^2+4bx+4ax-4ab$ I tried to solve this with quadratic formula but I could not do it. Each binomial turned out be a quadratic equation.. i missed some method to factor out this equation.. please help.
August 3rd, 2013, 09:03 PM   #2
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Re: Binomial as a,b and c in a quadratic equation.

Quote:
 Originally Posted by Dean1884 $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
x = {a + b + c +-SQRT[a^2 + b^2 + c^2 - (ab + ac + bc)]} / 3

August 3rd, 2013, 09:05 PM   #3
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Re: Binomial as a,b and c in a quadratic equation.

Quote:
 Originally Posted by Dean1884 the question : Show the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are always real and they can not be equal unless a=b=c. a= (x-a)(x-b) , b=(x-b)(x-c) , c = (x-c)(x-a). (x-a)(x-b) = x^2-bx-ax-+ab. ......... I ended up with this $D= x^4-cx^3+bx^3-cx^3+cx^2+bx^3-b^2x^2+b^2cx-4x^2+4bx+4ax-4ab$ I tried to solve this with quadratic formula but I could not do it. Each binomial turned out be a quadratic equation.. i missed some method to factor out this equation.. please help.
$a=(x-a)(x-b) ; b=(x-b)(x-c) ; c =(x-c)(x-a)$ are the conditions imposed?
Thank You!

 August 3rd, 2013, 09:19 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 One needs to know that a, b and c are real. (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 3x² - 2(a + b + c)x + ab + bc + ca = 0. Discriminant = 4(a + b + c)² - 12(ab + bc + ca) = 2((a - b)² + (b - c)² + (c - a)²), etc.
August 4th, 2013, 11:47 AM   #5
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Re: Binomial as a,b and c in a quadratic equation.

Quote:
 Originally Posted by Dacu $a=(x-a)(x-b) ; b=(x-b)(x-c) ; c =(x-c)(x-a)$ are the conditions imposed? Thank You!
no , I guessed I had to solve like this.

August 4th, 2013, 11:49 AM   #6
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Re:

Quote:
 Originally Posted by skipjack One needs to know that a, b and c are real. (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 3x² - 2(a + b + c)x + ab + bc + ca = 0. Discriminant = 4(a + b + c)² - 12(ab + bc + ca) = 2((a - b)² + (b - c)² + (c - a)²), etc.
I see.. so I have to factor the equation first before calculating the Discriminant.

 August 4th, 2013, 12:14 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 You have to put the original quadratic expression into a form that allows you to find its discriminant easily. The discriminant can then be arranged to make it easy to see that it cannot be negative, and under what circumstances it can be zero.
August 8th, 2013, 07:45 PM   #8
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Re:

Quote:
 Originally Posted by skipjack One needs to know that a, b and c are real. (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 3x² - 2(a + b + c)x + ab + bc + ca = 0. Discriminant = 4(a + b + c)² - 12(ab + bc + ca) = 2((a - b)² + (b - c)² + (c - a)²), etc.
bro , $4(a+b+c)^2 - 12(ab+bc+ac)$
$4(a^2+b^2+c^2+2ab+2ab+2bc)- 12 (ab+bc+ac)$
$4(a^2+b^2+c^2-ab-bc-ac)$

I am getting the above equation .. where am I mistaking ?

 August 8th, 2013, 07:53 PM #9 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: Binomial as a,b and c in a quadratic equation. $a^2+b^2+c^2-ab-ac-bc$ can be written as $a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}(2a^2+2b^2+2c^2-2ab-2ac-2bc)=\frac{1}{2}(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2)$ $=\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$
 August 9th, 2013, 03:20 AM #10 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Re: Binomial as a,b and c in a quadratic equation. thanks

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