Quadratic Equation

General Solution of a Quadratic Equation
Nature of roots of a quadratic equation
Symmetric Functions
Solved Examples

Introduction




A quadratic equation is a second-order polynomial equation in a single variable x
         ax2 + bx + c = 0
with a ≠ 0 where the letters a , b and c are the coefficients and the letter x is the variable or unknown . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real , or both complex .


General Solution of a Quadratic Equation

The general solution of an equation is as follow :
The ± means you need to do a plus and a minus , and so there are normally two solutions.
(b2 - 4ac) is called the discriminant , because it can "discriminate" between the possible types of answer :

if it is positive, you will get two solutions
if it is zero, you get just one solution,
and if it is negative you get two solutions that include imaginary numbers.


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Nature of roots of a quadratic equation

The nature of the roots of a quadratic equation depends on b2 - 4ac:

(1) If b2 - 4ac > 0, then the equation has two real and distinct (or different) roots.
(2) If b2 - 4ac = 0, then the equation has two real and equal roots .
(3) If b2 - 4ac < 0, then the equation has no real roots, i.e. the roots are complex .



Sum of the roots of a quadratic equation

In all the above three cases, sum of the two roots = - b/a

Product of the roots of a quadratic equation

In all the three cases, product of the two roots = c/a


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Symmetric Functions

Suppose α and β are the roots of a quadratic equation, then x = α and x = β

x – α = 0, and x – β = 0
(x – α) (x – β) = 0
x2 – (α + β)x + αβ = 0
x2 – (Sum of the roots)x + Product of the roots = 0
x2 – Sx + P = 0

Here, S = Sum of the roots, and P = Product of the roots. Thus, equation whose roots are α and β , is x2 – Sx + P = 0 , which is known as quadratic equation.


Any expression f(a,b) involving two numbers a and b is said to be symmetric if it remains unchanged when a and b are interchanged . [i.e. if f(a,b) = f(b,a)]
Some of the symmetric functions are :








We can express symmetric functions in terms of (α + β) and αβ .
Some examples are:


(1)

(2)        α3 + β3 = (α + β)3 - 3αβ(α + β)

(3)


(4)


(5)
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Solved Examples

Example:  Solve 5x2 + 6x + 1 = 0

Solution : a = 5, b = 6, c = 1

x = [ -b ± √(b2 - 4ac) ] / 2a

x = [ -6 ± √(62- 4×5×1) ] / (2×5)

x = [ -6 ± √(36 - 20) ]/10
x = [ -6 ± √(16) ]/10
x = ( -6 ±  4 )/10
x = -0.2 and -1

The solution to the given quadratic equation is x = -0.2 and -1


Example
: Solve x2 - 6x + 9 = 0.

Solution : a = 1, b = -6, c = 9

x = [ -b ± √(b2 - 4ac) ] / 2a

x = [ -(-6) ± √(62- 4×1×9) ] / (2×5)

x = [ +6 ± √(36 - 36) ]/2
x = [ +6 ± √0 ]/2
x = 3

The solution to the given quadratic equation is x = 3

Example : Solve 2x2 - 3x + 2 = 0.

Solution : a = 2, b = -3, c = 2

x = [ -b ± √(b2 - 4ac) ] / 2a

x = [ -(-3) ± √(32- 4×2×2) ] / (2×2)

x = [ +3 ± √(9 - 16) ]/4
x = [ +3 ± √-7 ]/4

√-7 is an imaginary number, thus there is no real roots for this equation.

Example : If α and β are the roots of the equation x2 – 7x + 10 = 0; find α3 + β3 .

Solution : a = 1, b = – 7, and c = 10

Sum of the roots = α + β = -b/a = 7/1 = 7
Product of the roots = αβ = c/a = 10/1 = 10

α3 + β3 = (α + β)3 - 3αβ(α + β)

= (7)3 – 3(10) (7)
= 343 – 210 = 133
α3 + β3 = 133


Example Find the value of k, if the roots of the equation 3x2 + 6x + k = 0, are reciprocals of each other.

Solution : a = 3, b = 6, c = k ;
roots are α and 1/α

Sum of roots = α + 1/α = -b/a = -6/3 = -2 .......... (1)
Product of roots = α(1/α) = c/a = k/3 ......... (2)

From (2), k/3 = α(1/α) = 1
Therefore, k = 3
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