Topics: Polynomial, Quadratic equation, Real number Pages: 21 (1457 words) Published: October 28, 2014
Equations

MODULE - I
Algebra

2
Notes

Recall that an algebraic equation of the second degree is written in general form as ax 2 + bx + c = 0, a ≠ 0
It is called a quadratic equation in x. The coefficient ‘a’ is the first or leading coefficient, ‘b’ is the second or middle coefficient and ‘c’ is the constant term (or third coefficient). For example, 7x² + 2x + 5 = 0,

5
1
x² + x + 1 = 0,
2
2

1
= 0, 2 x² + 7x = 0, are all quadratic equations.
2
In this lesson we will discuss how to solve quadratic equations with real and complex coefficients and establish relation between roots and coefficients. We will also find cube roots of unity and use these in solving problems.

3x² − x = 0, x² +

OBJECTIVES
After studying this lesson, you will be able to:
• solve a quadratic equation with real coefficients by factorization and by using quadratic formula;
• find relationship between roots and coefficients;
• form a quadratic equation when roots are given; and
• find cube roots of unity.

EXPECTED BACKGROUND KNOWLEDGE
• Real numbers
• Quadratic Equations with real coefficients.

MATHEMATICS

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MODULE - I
Algebra

2.1 ROOTS OF A QUADRATIC EQUATION
The value which when substituted for the variable in an equation, satisfies it, is called a root (or solution) of the equation.
If  be one of the roots of the quadratic equation

Notes

then,

ax 2 + bx + c = 0, a ≠ 0

... (i)

a 2 + b + c = 0
In other words, x −  is a factor of the quadratic equation (i)

In particular, consider a quadratic equation x2 + x − 6 = 0

...(ii)

If we substitute x = 2 in (ii), we get
L.H.S = 22 + 2 – 6 = 0

L.H.S = R.H.S.

Again put x = − 3 in (ii), we get
L.H.S. = ( − 3)2 –3 –6 = 0

L.H.S = R.H.S.

Again put x = − 1 in (ii) ,we get
L.H.S = ( − 1)2 + ( − 1) – 6 = –6 ≠ 0 = R.H.S.
∴ x = 2 and x = − 3 are the only values of x which satisfy the quadratic equation (ii)

There are no other values which satisfy (ii)
∴ x = 2, x = − 3 are the only two roots of the quadratic equation (ii)

Note:

If  ,  be two roots of the quadratic equation
ax 2 + bx + c = 0, a ≠ 0
...(A)
then (x −  ) and (x −  ) will be the factors of (A). The given quadratic equation can be written in terms of these factors as (x −  ) (x −  ) = 0

2.2 SOLVING QUADRATIC EQUATION BY FACTORIZATION
Recall that you have learnt how to factorize quadratic polynomial of the form

p ( x ) = ax 2 + bx + c, a ≠ 0, by splitting the middle term and taking the common factors. Same method can be applied while solving quadratic equation by factorization. p
If x − q and x − sr are two factors of the quadratic equation p
ax2 + bx + c = 0 ,a ≠ 0 then ( x − q )( x − sr ) = 0
p
either x = q or, x = sr

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MATHEMATICS

Equations

MODULE - I
Algebra

p
The roots of the quadratic equation ax + bx + c = 0 are q , sr 2

Example 2.1 Using factorization method, solve the following quadratic equation : 6x 2 + 5x − 6 = 0
Solution:

Notes

6x + 5x − 6 = 0
2

... (i)

Splitting the middle term, we have
6x 2 + 9x – 4x − 6 = 0
or, 3x (2x + 3) –2 (2x + 3) = 0
or, (2x + 3)(3x – 2) = 0
∴ Either 2x + 3 = 0 ⇒ x = −

or,

3x – 2 = 0 ⇒ x =

3
2

2
3

∴ Two roots of the given quadratic equation are −

3 2
,
2 3

Example 2.2 Using factorization method, solve the following quadratic equation: 2
3 2 x + 7x − 3 2 = 0

Solution:

Splitting the middle term, we have
2
3 2 x + 9x – 2x − 3 2 = 0

or,

3x ( 2 x + 3) −

or,

( 2 x + 3)(3x − 2 ) = 0

Either

or,

3x − 2 = 0

2 ( 2 x + 3) = 0

2x+3=0

3

⇒ x= − 2

⇒ x=

2
3

∴ Two roots of the given quadratic equation are −

MATHEMATICS

3
2
,
2 3

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MODULE - I
Algebra

Example 2.3 Using factorization method, solve the following...