**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) = ax^{2} + bx + c = 0.

1- First, if you can recognize one of the remarkable identities, for instance α^{2} - β^{2} = (α – β) (α + β), then you are immediately done. For the α^{2} - β^{2} situation, there are two roots corresponding to α = β and α = -β.

Example: f(x) = 4x^{2} – 4x + 1

This is the same pattern as the identity α^{2} - 2 α β + β^{2} = (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 x_{1 }. Since x_{1} is a root then you know that f(x) needs to be of the form (x – x_{1})(ax +d) and the problem is almost solved. The constant terms in ax^{2} + bx + c and (x – x_{1})(ax +d) have to match, therefore c = - d x_{1} and d = -c / x_{1}. As a consequence the second root is x_{2} = -d/a = -c/(ax_{1})

Example: g(x) = 2x^{2} +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 b^{2} – 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 ax
^{2}+ bx + c and a(x- α)^{2}+ β = a(x^{2}-2αx+α^{2}) + β, you need b = - 2 αa, therefore α = -b/2a .

- Since ax
^{2 }is replaced^{ }by a(x- α)^{2}in the expression ax^{2}+ bx +c, in order to get the constant terms correct, we need to subtract aα^{2}:

f(x) = ax^{2} + bx + c = a(x- α)^{2} - aα^{2 }+c ; this gives β = - aα^{2 }+ c .

- Conclusion: the canonical form of ax
^{2}+ bx + c is a(x- α)^{2}+ β, with α = -b/2a and β = - aα^{2 }+c

.

- 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) = x^{2} – 4x -1

h(x) can be re-written as h(x) = (x^{2} – 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) = ax^{2} + bx + c as well as the coordinates of the extremum (α, β) of the parabol.

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