A close look at quadratic equations

1.    Introduction

A rather common topic in maths: how to best solve quadratic equations. Not really rocket science, but there are a few tricks to do it fast, systematically and minimize chances of errors.

CHOOSE THE SIMPLE METHODS FIRST TO SAVE TIME, DEPENDING ON THE PROBLEM AT HAND !

2.    Sequential approach

Let us take the usual formula for a quadratic equation: f(x) = ax2 + bx + c = 0.

1- First, if you can recognize one of the remarkable identities, for instance α2 - β2 = (α – β) (α + β), then you are immediately done.  For the α2 - β2 situation, there are two roots corresponding to α = β and α = -β.

 

Example:  f(x) = 4x2 – 4x + 1  

This is the same pattern as the identity α2 - 2 α β + β2 = (α – β)2

Therefore:  f(x) = (2x – 1)2 with a double root x = 1/2

 

2- When case 1 is not applicable, maybe it is possible to find an obvious root of the equation by direct observation. This often happens, for instance roots such as 1, 2, -1 ,… Let us call this obvious root x1 . Since x1 is a root then you know that f(x) needs to be of the form (x – x1)(ax +d) and the problem is almost solved. The constant terms in  ax2 + bx + c   and     (x – x1)(ax +d) have to match, therefore c = - d x1  and d = -c / x1. As a consequence the second root is x2 = -d/a = -c/(ax1)

 

Example:  g(x) = 2x2 +5x - 7 = 0   

One obvious solution x = 1

g(x) = (x – 1) (2x + constant); the constant can be easily seen to be equal to 7.

Therefore:  g(x) = (x – 1) (2x + 7) with roots 1 and -7/2

 

3- If the quick first and second methods are not applicable, the traditional “quadratic formula” method including the calculation of the discriminant b2 – 4ac can be applied; it works all the time, however at the expense of the required calculations.

 

An interesting and effective alternative to the quadratic formula is to transform f(x) into its so-called “canonical form” a(x- α)2 + β . The method is also called “solving by completing the square”:

 

  • For the x-terms to match between ax2 + bx + c and a(x- α)2 + β = a(x2 -2αx+α2) + β, you need b = - 2 αa, therefore   α = -b/2a  .

 

  • Since ax2 is replaced by a(x- α)2 in the expression ax2 + bx +c, in order to get the constant terms correct, we need to subtract aα2:

f(x) = ax2 + bx + c = a(x- α)2 - aα2 +c ; this gives   β = - aα2 + c .  

 

  • Conclusion: the canonical form of ax2 + bx + c is a(x- α)2 + β, with α = -b/2a and β = - aα2 +c

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  • The roots can be derived directly from the canonical form by solving a(x- α)2 + β = 0, which has real roots when β/a ≤ 0.

 

Example:  h(x) = x2 – 4x -1   

h(x) can be re-written as h(x) = (x2 – 4x + 4) - 4 -1   = (x – 2)2 – 5

The roots are such that (x – 2)2 = 5     i.e.    x - 2 = ±√5

therefore the two roots are: 2 + √5 ; 2 - √5

 

As a bonus, the canonical form gives directly the axis of symmetry (x = α) of the parabolic curve corresponding to f(x) = ax2 + bx + c as well as the coordinates of the extremum (α, β) of the parabol.