History of quadratic equation

It is believed that the Babylonians were the first one to solve and use quadratic equations around 1500 B.C. before that there is no evidence which shows that Egyptians mathematicians used or solved quadratic equations.

Babylonians came across the problems based upon area of rectangles eg. Let us suppose that the area of given rectangle is 55 sq. units and one side of rectangle is 6 units greater than other. What are the sides of rectangle. In modern mathematics we will convert the given problem into algebraic expression or quadratic equation and solve. But Babylonians used a unique method to solve such problems, let us try to understand their solution.

In around 500BC Pythagoras proved that sum of squares of two adjacent sides of rectangle is equals to the square of diagonal but rejected the idea of irrational number.

Later in 300 BC Euclid proved the existence of irrational numbers in his book elements

Indian mathematicians were responsible for the next advancement which took place in quadratic equations and its solutions, mainly Brahmagupta did lot of work on quadratic equations:

He gave the general formula for the equation

He was the first one to realise that there are two possible solutions in case of a quadratic equation and to consider negative solutions also.

In 820 AD near Baghdad, the Islamic mathematician Mohammad bin Al-Khwarismi also derived the quadratic formula, but he rejected negative solutions. Mohammad’s formula was brought to Europe.

In 1545, the algebraist Girolamo Cardano combined Al-Khwarismi’s solutions with those of Euclid. He also took into account complex, or imaginary numbers.

Finally, in 1637, René Descartes published a modern volume of geometry, which contained the quadratic formula in the form used today.

Definition of a quadratic equation

Any equation of the form  ax^2+bx+c=0

is known as quadratic equation in one variable.

Solution of quadratic equation

The value(s) of x which satisfies  ax^2+bx+c=0 is(are) known as solutions of quadratic equation.

Solution of quadratic equation/roots of quadratic equation/ zeroes of polynomial ax^2+bx+c=0 are exactly same.

Graphs of quadratic polynomial ax^2+bx+c represents an upward parabola if a is positive and downward parabola if a is negative.

Methods to solve the quadratic equation

1. Factorising by splitting the middle term
2. Completing the square
3. Using quadratic formula

Solving quadratic equations using factorisation

2x^2-5x+3=0

Split the middle term -5x in two parts such that the product of those parts is

–5x =  –3x – 2x

So our equation becomes

2x^2-3x-2x+3=0

Factorising

(2x-3)(x-1)=0

x = 2/3 and x=1

Solving quadratic equations using completing the square method

2x^2-5x+3=0

Step – I: Make the co-efficient of  x^2 = 1, by dividing the whole equation by the co-efficient of x

x^2 -5x/2+3/2=0

Step – II: Shift constant term on RHS

x^2 -5x/2=-3/2

Step – III: Add  both sides of the equation by (1/2 co-eff of x)^2

x^2 -5x/2+(5/2)^2=-3/2+(5/2)^2

Step – IV: Complete the square on LHS and simplify RHS

(x-5/4)^2 = 1/16

Note: If RHS is negative in this step the equation will have no real roots

Step – V: Taking square root both sides of the equation and solving:

(x-5/4) = +-(1/4)

x = 3/2 or x = 1

Applying the same method of completing the square to the general quadratic equation

ax^2+bx+c=0

Step – I: Make the co-efficient of  x^2= 1, by dividing the whole equation by the co-efficient of x i.e. a

x^2+bx/a+c/a=0

Step – II: Shift constant term on RHS

x^2+bx/a=-c/a

Step – III: Add  both sides of the equation

x^2+bx/a+(b/2a)^2=-c/a+(b/2a)^2

Step – IV: Complete the square on LHS and simplify RHS

(x+b/2a)^2 = (b^2-4ac)/4a^2

Step – V: Taking square root both sides of the equation and solving:

(x+b/2a) = {(b^2-4ac)^(1/2)}/2a

x = [-b+-{(b^2-4ac)^(1/2)]/2a

the quantity  b^2-4ac is represented by D, known as discriminant of the equation.

Nature of roots:

Case-I :  D = b^2-4ac  is negative

In such case x will not be a real number & the equation will have no real roots

Case-II : D = b^2-4ac is zero

Substituting the same in last step we will get

Means equation will have real and equal roots

Case-III : D = b^2-4ac is a positive real number

In such case equation will have real and distinct roots.

Relation between roots and quadratic co-efficients

Sum of roots = -b/a

product of roots = c/a