My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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 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
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.
Dean1884 is offline  
 
August 3rd, 2013, 09:03 PM   #2
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,597
Thanks: 1038

Re: Binomial as a,b and c in a quadratic equation.

Quote:
Originally Posted by Dean1884
x = {a + b + c +-SQRT[a^2 + b^2 + c^2 - (ab + ac + bc)]} / 3
Denis is offline  
August 3rd, 2013, 09:05 PM   #3
Senior Member
 
Joined: Apr 2013

Posts: 425
Thanks: 24

Re: Binomial as a,b and c in a quadratic equation.

Quote:
Originally Posted by Dean1884
the question : Show the roots of the equation 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
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.
are the conditions imposed?
Thank You!
Dacu is offline  
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) = 3x - 2(a + b + c)x + ab + bc + ca = 0.

Discriminant = 4(a + b + c) - 12(ab + bc + ca) = 2((a - b) + (b - c) + (c - a)), etc.
skipjack is offline  
August 4th, 2013, 11:47 AM   #5
Member
 
Joined: Jul 2013

Posts: 58
Thanks: 0

Re: Binomial as a,b and c in a quadratic equation.

Quote:
Originally Posted by Dacu
are the conditions imposed?
Thank You!
no , I guessed I had to solve like this.
Dean1884 is offline  
August 4th, 2013, 11:49 AM   #6
Member
 
Joined: Jul 2013

Posts: 58
Thanks: 0

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) = 3x - 2(a + b + c)x + ab + bc + ca = 0.

Discriminant = 4(a + b + c) - 12(ab + bc + ca) = 2((a - b) + (b - c) + (c - a)), etc.
I see.. so I have to factor the equation first before calculating the Discriminant.
Dean1884 is offline  
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.
skipjack is offline  
August 8th, 2013, 07:45 PM   #8
Member
 
Joined: Jul 2013

Posts: 58
Thanks: 0

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) = 3x - 2(a + b + c)x + ab + bc + ca = 0.

Discriminant = 4(a + b + c) - 12(ab + bc + ca) = 2((a - b) + (b - c) + (c - a)), etc.
bro ,



I am getting the above equation .. where am I mistaking ?
Dean1884 is offline  
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.

can be written as



crom is offline  
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
Dean1884 is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
binomial, equation, quadratic



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
simplify equation to get a quadratic equation mich89 Algebra 3 January 9th, 2013 01:22 PM
Complex Exponential Equation -- Binomial ^ Large Fraction EkajArmstro Probability and Statistics 1 November 9th, 2010 04:53 PM
Quadratic equation RougeRay Abstract Algebra 7 December 27th, 2009 05:18 AM
Another quadratic equation grainne Elementary Math 7 December 18th, 2009 12:52 PM
quadratic equation Caithsith Elementary Math 0 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.