# Quadratic Equation - 1

Quadratic equation is nothing but polynomial equation with degree 2.The graph of quadratic equation i.e. represents a parabola.

It is often claimed that Babylonians (about 400 BC) were first to solve quadratic equation. However this is not true, Savasorda in 1145 came with complete solution of quadratic equation about 900 years from then we have many interesting things to calculate like intervals of values of x for which quadratic expression is of

one sign and many more things.

So let us start solving “QUADRATIC EQUATION”

**INTRODUCTION:**

An equation of form = 0 where a, b, c e C is called quadratic equation.

- If a = 0 then one root is and other root is
- If a=b=0 then both roots are .

In general roots of equation are given by

Where is known as Discriminant.

__Nature of roots:__

1) The quadratic equation has real and equal roots if D=0.

2) The quadratic equation has real and distinct roots if D>0

3) The quadratic equation has complex roots with non-zero imaginary parts if D<0.

4) p+iq is a root of quadratic equation if p-iq is a root of equation.

More:

a) In general if a polynomial equation with all real coefficients has a root p+iq then p-iq will also be root of equation.

b) So, any polynomial equation with all real coefficients will have non-real complex roots in conjugate pairs.

5) If a, b, c Q and p+√q (q is not a perfect square) is an irrational root of quadratic equation then p-√q is also a root.

More:

a) In general if polynomial equation with all coefficients rational has an irrational root p+√q then p-√q will also be a root of equation.

b) So polynomial equation with all rational coefficients will have irrational roots in conjugate pair.

6) If quadratic equation is satisfied by more than two distinct numbers (real or imaginary) then it becomes an identity i.e. a=b=c=0

**Illustration 1:**

If a+b+c=0 then find the nature of the root of the equation ?

Solution:

So the roots are real and equal.

**Relation between roots and coefficients:**

= 0 a ¹ 0, a, b, cÃŽC, If a,b are roots then a+b = , ab =

Factorization is = a(x-a) (x-b), If a and b are given then equation is

x2-(a+b) x+ab=0.

**Illustration 2:**

If one root of equation is square of other then find the relation between p and q.

Solution: Let one root be a, so other root is a2

So, a+a2 = -p and a(a2) = q

Now (a+a2)3 = (-p)3

__Common roots__

If and

1) have a common root, then and common root is given by,

How?

Let a be a common root then

And

Solving we get

Eliminating a we get

**Illustration 3:**

Find the condition on a, b, c and d such that equations and have a common root.

Solution: Let a be a common root

Now from (2) and (3) we find the condition i.e.

**Lagrange’s Identity:**

If a1, a2, a3, b1, b2, b3ÃŽR then,

**Cauchy Swartz inequality:**

**Illustration 4:**

If a1, a2, ----------an ÃŽR+ then show that

Solution:

Since a1, a2, ----------an ÃŽR+ so we can take root of each of the ai’s

Now using the Cauchy Swartz inequality we get,

**Equation of Higher Degree:**

Consider the equation;

Also,