Qudratic eqn practice for cbse11

8/13/2019 Qudratic eqn practice for cbse11

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Quadratic EquationsQuadratic Equations

MATHEMATICS

Notes

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2

QUADRATIC EQUATIONS

Recall that an algebraic equation of the second degree is written in general form as2 0, 0ax bx c a+ + =

It is called a quadratic equation inx. The coefficient a is the first or leading coefficient, bis the second or middle coefficient and c is the constant term (or third coefficient).

For example, 7x+ 2x+ 5 = 0,2

5x+

2

1x+ 1 = 0,

3xx= 0, x+2

1= 0, 2x+ 7x= 0, are all quadratic equations.

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.

OBJECTIVES

After studying this lesson, you will be able to:

solve a quadratic equation with real coefficients by factorization and by using quadraticformula;

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.


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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.

Ifbe one of the roots of the quadratic equationthen, 2 0, 0ax bx c a+ + = ... (i)

2 0a b c + + =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)23 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 andx=3 are the only values ofxwhich 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 quadraticequation 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

( ) 2 , 0,p x ax bx c a= + + by splitting the middle term and taking the common factors.

Same method can be applied while solving quadratic equation by factorization.

If xpq and x rs are two factors of the quadratic equation

ax2+ bx + c = 0 ,a 0 then (xpq )(x rs ) = 0

either x =pq or, x =

r

s


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The roots of the quadratic equation ax2+ bx + c = 0 arepq ,

r

s

Example 2.1 Using factorization method, solve the following quadratic equation :

6x 2+ 5x6 = 0

Solution: The given quadratic equation is

6x 2+ 5x6 = 0 ... (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 = 23

or, 3x 2 = 0 x =3

2

Two roots of the given quadratic equation are2

3,

3

2

Example 2.2 Using factorization method,solve the following quadratic equation:

23 x2

+ 7x 23 = 0Solution: Splitting the middle term, we have

23 x2 + 9x 2x 23 = 0

or, 3x ( 2 x + 3) 2 ( 2 x + 3) = 0

or, ( 2 x + 3)(3x 2 ) = 0

Either 2 x + 3 = 0 x =2

3

or, 3x 2 = 0 x =3

2

Two roots of the given quadratic equation are2

3,

3

2


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Example 2.3 Using factorization method, solve the following quadratic equation:

(a + b)2x2+ 6 (a2 b2) x + 9 (a b)2= 0


(a + b)2

x2

+ 6 (a2

b2

) x + 9 (a b)2

= 0Splitting the middle term, we have

(a + b)2x2+ 3(a2 b2) x + 3(a2 b2) x + 9 (a b)2 = 0

or, (a + b)x {(a + b) x + 3 (ab) } + 3 (a b) {(a +b) x + 3 (a b) } = 0

or, {(a + b) x + 3 (ab) } {(a + b) x + 3 (a b) } = 0

either (a + b) x + 3 (a b) =0 x =ba

)ba(3

+

=ba

)ab(3

+

or, (a + b) x + 3 (a b) =0 x =ba

)ba(3

+ =

ba

)ab(3

+

The equal roots of the given quadratic equation areba

)ab(3

+

,ba

)ab(3

+

Alternative Method

The given quadratic equation is

(a + b)2x2+ 6(a2 b2) x + 9(a b)2= 0

This can be rewritten as

{(a + b) x}2+ 2 .(a + b)x . 3 (a b) + {3(a b)}2= 0

or, { (a + b)x + 3(a b) }2= 0

or, x =ba

)ba(3

+

=ba

)ab(3

+

The quadratic equation have equal rootsba

)ab(3

+

,ba

)ab(3

+

CHECK YOUR PROGRESS 2.1

1. Solve each of the following quadratic equations by factorization method:

(i) 3 x2 + 10x + 8 3 = 0 (ii) x

2 2ax + a2 b = 0

(iii) x2+

ab

c

c

abx 1 = 0 (iv) x2 4 2 x + 6 = 0


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2.3 SOLVING QUADRATIC EQUATION BY QUADRATIC

FORMULA

Recall the solution of a standard quadratic equation

a

x2

+ bx+ c = 0, a 0 by the Method of Completing SquaresRoots of the above quadratic equation are given by

x1=

a2

ac4bb 2 +

and x2=

a2

ac4bb 2

=a2

Db+, =

a2

Db

where D = b2 4ac is called the discriminant of the quadratic equation.

For a quadratic equation 2 0, 0ax bx c a+ + = if

(i) D>0, the equation will have two real and unequal roots

(ii) D=0, the equation will have two real and equal roots and both roots are

equal to b

2a

(iii) D0.

The equation will have two real and unequal roots


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Verification:By quadratic formula, we have

x =2

419

The two roots are2

419+,

2

419which are real and unequal.

(ii) The given quadratic equation is 9 y2 6 2 y + 2 = 0

Here, D = b2 4 ac

= ( 6 2 )2 4.9.2

= 72 72 = 0

The equation will have two real and equal roots.


y =9.2

026 =

3

2

The two equal roots are3

2,

3

2.

(iii) The given quadratic equation is 2 t2 3t + 3 2 = 0

Here, D = ( 3)2 4. 2 .3 2

=15< 0

The equation will have two conjugate complex roots.


t =22

153

3 15,

2 2

i= where 1i =

Two conjugate complex roots are22

i153+,

22

i153


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Example 2.5 Prove that the quadratic equation x2+ py 1 = 0 has real and distinct roots

for all real values of p.

Solution: Here, D = p2+ 4 which is always positive for all real values of p.

The quadratic equation will have real and distinct roots for all real values of p.

Example 2.6 For what value of k the quadratice equation

(4k+ 1) x2+ (k + 1) x + 1 = 0 will have equal roots ?


(4k + 1)x2+ (k + 1) x + 1 = 0

Here, D = (k + 1)2 4.(4k + 1).1

For equal roots, D = 0

(k + 1)2 4 (4k + 1) = 0

k2

14k 3 = 0

k =2

1219614 +

or k =2

20814

= 7 2 13 or 7 + 2 13 , 7 2 13

which are the required values of k.

Example 2.7 Prove that the roots of the equation

x2(a2+ b2) + 2x (ac+ bd) + (c2+ d2) = 0 are imaginary. But if ad = bc,

roots are real and equal.

Solution: The given equation is x2 (a2+ b2) + 2x (ac + bd) + (c2+ d2) = 0

Discriminant = 4 (ac + bd)2 4 (a2+ b2) (c2+ d2)

= 8 abcd 4(a2d2+ b2c2)

= 4 ( 2 abcd + a2d2+ b2c2)

= 4 (ad bc) 2


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1. Solve each of the following quadratic equation by quadratic formula:

(i) 2x2 3 x + 3 = 0 (ii) 2 2 1 0x x + =

(iii) 24 5 3 0x x + = (iv) 23 2 5 0x x+ + =

2. For what values of k will the equation

y2 2 (1 + 2k) y + 3 + 2k = 0 have equal roots ?

3. Show that the roots of the equation

(x a) (x b) + (x b) (x c) + (x c) (x a) = 0 are always real and they can not be

equal unless a = b = c.

2.4 RELATION BETWEEN ROOTS AND COEFFICIENTS OF

A QUADRATIC EQUATION

You have learnt that, the roots of a quadratic equation ax2+ bx + c = 0, a 0

area2

ac4bb 2 +

anda2

ac4bb 2

Let = a2

ac4bb 2 +

...(i) and = a2

ac4bb 2

... (ii)

Adding (i) and (ii), we have

+ =a2

b2=

a

b

Sum of the roots= 2coefficient of x b

coefficient of x a= ... (iii)

= 2

22

a4

ac)4(bb +

= 2a4

ac4

=a

c


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Product of the roots= 2constant term

coefficient of

c

x a= ...(iv)

(iii) and (iv) are the required relationships between roots and coefficients of a given quadratic

equation. These relationships helps to find out a quadratic equation when two roots aregiven.

Example 2.8 If,, are the roots of the equation 3x2 5x + 9 = 0 find the value of:

(a) 2+ 2 (b) 21

+ 2

1

Solution: (a) It is given that , are the roots of the quadratic equation 3x2 5x +9 = 0.

5

3 + = ... (i)

and9

33

= = ... (ii)

Now, ( )22 2

2 + = +

=

2

3

5

2.3 [By (i) and (ii)]

=9

29

The required value is9

29

(b) Now, 21

+ 2

1

= 22

22

+

=9

929

[By (i) and (ii)]

=81

29


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Example 2.9 If , are the roots of the equation 3y2+ 4y + 1 = 0, form a quadratic

equation whose roots are 2, 2

Solution: It is given that, are two roots of the quadratic equation 3y2+ 4y + 1 = 0.

Sum of the roots

i.e., + = 2yoftcoefficienyoftcoefficien

=3

4... (i)

Product of the roots i.e., = 2yoftcoefficien

termttancons

=3

1... (ii)

Now, 2+ 2 = (+ )2 2

=

2

3

4

2.

3

1[ By (i) and (ii)]

=9

16

3

2

=9

10

and 2 2 = ( )2=9

1[By (i) ]

The required quadratic equation is y2 (2+ 2)y+2 2 = 0

or, y2 9

10y +

9

1= 0

or, 9y2 10y + 1 = 0

Example 2.10 If one root of the equation 2 0, 0ax bx c a+ + = be the square ofthe other, prove that b3+ ac2+ a2c = 3abc

Solution:Let , 2be two roots of the equation ax2+ bx + c = 0.

+2=a

b... (i)


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and .2=a

c

i.e., 3=a

c. ... (ii)

From (i) we have

(+ 1) = a

b

or, { }3

1)( + =3

a

b

= 3

3

a

b

or, 3(3+ 32+ 3+1) = 33

a

b

or,a

c

+

+ 1

a

b3

a

c= 3

3

a

b... [ By (i) and (ii)]

or, 2

2c

a 2

bc3

a+

a

c= 3

3

a

b

or, ac2 3abc + a2c =b3

or, b3

+ ac2

+ a2

c = 3abcwhich is the required result.

Example 2.11 Find the condition that the roots of the equation ax2+ bx + c = 0 are in the

ratio m : n

Solution: Let mand nbe the roots of the equation ax2+bx + c = 0

Now, m+ n= a

b... (i)

and mn2 =a

c... (ii)

From (i) we have, (m + n ) = a

b

or, 2(m + n)2= 2

2

a

b


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or,a

c(m + n)2= mn

2

2

a

b[ By (ii)]

or, ac (m+n)2= mn b2

which is the required condition


1. If , are the roots of the equation ay2+ by + c = 0 then find the value of :

(i) 21

+ 2

1

(ii) 4

1

+ 4

1

2. If

,

are the roots of the equation 5x2 6x + 3 = 0, form a quadratic equation

whose roots are:

(i) 2, 2 (ii) 3 , 3

3. If the roots of the equation ay2+ by + c = 0 be in the ratio 3:4, prove that

12b2= 49 ac

4. Find the condition that one root of the quadratic equation px2 qx + p = 0

may be 1 more than the other.

2.5 SOLUTION OF A QUADRATIC EQUATION WHEN D < 0

Let us consider the following quadratic equation:

(a) Solve for t :

t2+ 3t + 4 = 0

t =2

1693 =

2

73

Here, D= 7 < 0

The roots are2

73 +and

2

73

or,2

i73 +,

2

i73

Thus, the roots are complex and conjugate.


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(b) Solve for y :

3y2+ 5 y 2 = 0

y = )3(2)2).(3(455

or y =6

195

Here, D = 19 < 0

The roots are5 19 5 19

,6 6

i i +

Here, also roots are complex and conjugate. From the above examples , we can make thefollowing conclusions:

(i) D < 0 in both the cases

(ii) Roots are complex and conjugate to each other.

Is it always true that complex roots occur in conjugate pairs ?

Let us form a quadratic equation whose roots are

2 + 3i and 4 5i

The equation will be {x (2 + 3i)} {x (4 5i)} = 0

or, x2 (2 + 3i)x (4 5i)x + (2 + 3i) (4 5i) = 0

or, x2+ (6 + 2i)x + 23 + 2i = 0

which is an equation with complex coefficients.

Note :If the quadratic equation has two complex roots, which are not conjugate

of each other, the quadratic equation is an equation with complex coefficients.

2.6 CUBE ROOTS OF UNITY

Let us consider an equation of degree 3 or more. Any equation of degree 3 can be expressed

as a product of a linear and quadratic equation.

The simplest situation that comes for our consideration is

x 1 = 0 ...(i)

x1 = (x1) (x +x+ 1) = 0

either x1 = 0 or, x +x+ 1 = 0


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or, x = 1 or, x =2

i31

Roots are 1, 2i3

2

1

+ and 2i3

2

1

These are called cube roots of unity.

Do you see any relationship between two non-real roots of unity obtained above ?

Let us try to find the relationship between them

Let w =2

i3

2

1+

Squaring both sides, we have

w =

2

2

i3

2

1

+

= )i32i31(4

1 2 +

= )i3231(4

1

=4

i322

=( )

4

i312 +=

( )2

i31+

w =2

i3

2

1 = other complex root.

We denote cube roots of unity as 1, w, w


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Roots are1,2

i3

2

1 and

2

i3

2

1+ which can also be written as1 ,w andw2

Therefore, cube roots of1 are1 ,w andw2

In general, roots of any cubic equation of the form x 3= a3 would be a , aw and aw2

Example 2.12 If 1, w and w are cube roots of unity, prove that

(a) 1 + w2+ w7= 0

(b) (1 w + w2) (1 + w w2) = 4

(c) (1 + w)3 (1 + w2)3= 0

(d) (1 w + w2

)3

=8 and (1 + w w2

)3

= 8Solution:

(a) 1 + w2+ w7= 0

L.H.S = 1 + w2+ (w3)2. w

= 1 + w2+ w [ since w3=1]

= 0 [ since 1+ w + w2= 0]

= R.H.S

L.H.S = R.H.S

(b) (1w + w2) (1 + w w2) = 4

L.H.S = (1w + w2) (1 + w w2)

= (1 + w2 w) (1 + w w2)

(since 1 + w2=w and 1 + w =w2)

= ( ww) (w2w2)= ( 2w) ( 2w2)

= 4w3

= 4.1 = 4 =R.H.S

L.H.S = R.H.S


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(c) (1 + w)3 (1 + w2)3= 0

L.H.S = (1 + w)3 (1 + w2)3

= ( w2)3 (w)3 ( 1 + w =w2

= w6

+ w3

and 1 + w2

=w)= (w3)2+ 1

= (1)2+ 1

= 0 =R.H.S

L.H.S = R.H.S

(d) (1 w + w2)3 =8 and (1 + w w2)3=8

Case I :L.H.S = (1w + w2)3

= (1 + w2

w)3

= ( w w)3

= ( 2w)3

= 8w3

= 8 = R.H.S

L.H.S = R.H.S

Case II : L.H.S = (1 + ww)

= ( w2

w2

)3

= ( 2w2)3

= 8w6

= 8(w3)2

= 8 = R.H.S

L.H.S = R.H.S


1. Solve each of the following cubic equations:

(i) x3= 27 (ii) x3=27

(iii) x3 = 64 (iv) x3=64


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2. If 1, w, w2are cube roots of unity, show that

(i) (1 + w) (1 + w2) (1 + w4) (1 + w8) = 1

(ii) (1 w) (1 w2

) (1 w

4

) (1 w

5

) = 9

(iii) (1 + w)4+ (1 + w2)4=1

(iv) (1 + w3)3= 8

(v) (1 w + w2)6= (1 + w w2)6= 64

(vi) (1 + w)16+ w = (1 + w2)16+ w2=1

LET US SUM UP

Roots of the quadratic equation ax2+ bx + c = 0 are complex and conjugate of eachother, when D < 0.

If, be the roots of the quadratic equation

ax2+ bx + c = 0 then + =b

aand =

a

c

Cube roots of unity are 1, w, w2

where wi

= +1

2

3

2and w

i2 = 1

2

3

2

Sum of cube roots of unity is zero i.e., 1+ w + w2=0

Product of cube roots of unity is 1 i.e., w3= 1

Complex roots w and w2are conjugate to each other.

In general roots of any cubic equation of the form

x3= a3 would be a , aw and aw2


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TERMINAL EXERCISE

SUPPORTIVE WEB SITES

http://www.wikipedia.org

http://mathworld.wolfram.com


2(a2 + b2)x2 + 2(a + b)x + 1=0 are imaginary,when a b


bx2 + ( b c)x = c + a b are always real if those of

ax2 + b( 2x + 1) = 0 are imaginary.

3. If , be the roots of the equation 2x2 6x + 5 = 0 , find the equation whose

roots are:

(i)

,

(ii) +

1, +

1(iii) 2+2, 2

1

+ 2

1

4. If 1, w and w2are cube roots of unity , prove that

(a) (2 w) (2 w2) (2 w10) (2 w11) = 49

(b) ( x y) (xw y) ) (xw2 y) = x3 y3

5. If x = a + b , y = aw + bw2and z = aw2+ bw , then prove that

(a) x2+ y2+ z2= 6 ab (b) x y z = a3+ b3
http://mathworld.wolfram.com/http://mathworld.wolfram.com/http://www.wikipedia.org/


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ANSWERS


1. (i)

2 34

3, (ii) a b a + b,

(iii)ab

c,

c

ab (iv) 2,23


1. (i)4

i153 (ii)2

i1

(iii)8

i435(iv)

6

i582

2. 1,2

1


1. (i) 2

2

c

2acb (ii)

4

2222

c

c2a2ac)b(

2. (i) 25x2 6x + 9 = 0 (ii) 625x2 90x + 81 = 0

4. q2 5p2= 0


1. (i) 3, 3w, 3w2 (ii) 3,3w ,3w2

(iii) 4, 4w, 4w2 (iv)

4,

4w ,

4w2

TERMINAL EXERCISE

3. (i) 5x2 8x + 5 = 0 (ii) 10x2 42x + 49 = 0 (iii) 25x2 116x + 64 = 0


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Qudratic eqn practice for cbse11

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