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Working with Quadratic Roots

Date: 08/15/2002 at 10:22:53
From: Mandeep Singh
Subject: Quadratic equations


Could you help me with this question:

Given that a and b are the roots of the quadratic equation

    x^ - 5x + 3 = 0

form, without solving the equation, the quadratic equation whose 
roots are 

   a    b
   -    -
   b    a

Please could you show me step by step how to do this?

Date: 08/15/2002 at 12:24:35
From: Doctor Peterson
Subject: Re: Quadratic equations

Hi, Mandeep.

Since a and b are the roots, you know that

    x^2 - 5x + 3 = (x - a)(x - b)

An equation with roots a/b and b/a will be

    (x - a/b)(x - b/a) = 0

Try expanding this into a quadratic in standard form, and compare 
with the original, using what you know about the sum and product of 
its roots.

If you need more help, please write back and show me what you found.

- Doctor Peterson, The Math Forum

Date: 08/16/2002 at 18:15:08
From: Doctor Greenie
Subject: Re: Quadratic equations

Hi, Mandeep -

I find this kind of problem to be kind of fun, once you have done a 
couple and see the general procedure.

If a and b are the roots of a quadratic equation, then the equation is

  (x-a)(x-b) = 0

So with the given equation

  x^2-5x+3 = 0

we have

  (x-a)(x-b) = x^2 - (a+b)x + ab = x^2 - 5x + 3 = 0

Equating coefficients of the linear terms, we see that

  -(a+b) = -5  (or a+b = -(-5) = 5)

And equating the constant terms, we see that

  ab = 3

What we have shown is that if we have a quadratic equation with 
leading coefficient 1 (i.e., the coefficient of the x^2 term is 1), 
then the sum of the two roots is the opposite of the coefficient of 
the linear term, and the product of the two roots is the constant 

This is equivalent to properties with which you may already be 

  Given a quadratic equation px^2+qx+r = 0

  (1) the product of the roots is r/p
  (2) the sum of the roots is -q/p

In your problem, we are supposed to find a quadratic equation whose 
roots are a/b and b/a. We know this quadratic equation will be of the 

  x^2 + Ax + B = 0

where the coefficient A is the opposite of the sum of the two roots 
and the constant term B is the product of the two roots.

The sum of the two given roots (a/b and b/a) is (a^2+b^2)/ab; the 
product of these two roots is 1.

Thus we know that the equation we are looking for will be

   x^2 - ((a^2+b^2)/ab)x + 1 = 0

We will be done with the problem if we can evaluate the expression


We can do this because

  a^2+b^2   a^2+2ab+b^2 - 2ab   (a+b)^2 - 2ab   5^2 - 2(3)   19
  ------- = ----------------- = ------------- = ---------- = --
     ab            ab                 ab             3        3

So the equation we are looking for is

   x^2 - (19/3)x + 1 = 0

A little messy algebra - solving the original equation to find the 
roots a and b, then calculating a/b and b/a, then forming the equation 
with roots a/b and b/a - will show this to be the correct result.

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
Associated Topics:
High School Basic Algebra

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