Quadratic Equation

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.

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

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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