Quad Removal Equa

8/10/2019 Quad Removal Equa

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After learning this chapter you will be able to :

* Identify a quadratic equation.

* Distinguish between a pure and an adfected quadratic equation.

* Solve simple problems on pure and adfected quadratic equations.

* Identify the standard form of a quadratic equation.

* Solve the quadratic equations by factorization.

* Solve the quadratic equations by using formula.

* Find the value of the discriminant and know the nature of the roots.

* Frame the quadratic equation for the given roots.

* Solve the quadratic equation graphically.

QUADRATIC EQUATIONS

You are familiar with the properties and area of a square and a rectangle.

Consider a square of side x

units and its perimeter 16 units, then the lengthof the side is

Perimeter = 4 (length of the side)

16 = 4x

or

4x = 16 .............. (1)

In equation (1) the degree of the variable is one.

Can you recall the name of such an equation?

Such an equation is a linear equation.

4x = 16

x =4

16

x = 4

An equation involving a

variable of degree one only

is a linear equation.

A Linear equation has

only one root.

x

x

xx

5 QUADRATIC EQUATIONS


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An equation involving a

variable of degree 2 is

quadratic equation.

a

a

aa

m m

(m+2)

(m+2)

Consider a square of side a units and its area 25 square units.

Area of the square = (side)2

25 = a2

or

a2 = 25 .............. (2)

In equation (2) the degree of the variable is two

What do you call such an equation?

Such an equation is a quadratic equation.

a2 = 25

a = 5

a = + 5 or a = 5

Consider a rectangle of sides m and (m + 2) units and its area is 8 sq units.

Area of a rectangle = (length) (breadth)

8 = (m) (m + 2)

8 = m2 + 2m

or

m2 + 2m = 8 .......................... (3)

Compare the equation (2) and (3)

In equation (2) a2 = 25, variable occurs only in second degree.

In equation (3) m2 + 2m = 8, variable occurs in second degree as well as in first degree.

Quadratic equation involving a variable only in second degree is a

Pure Quadratic Equation.

Example :

(1) x2 = 9 (2) 2a2 = 18

If the terms in the RHS are transposed to LHS then,

(1) x2 9 = 0 (2) 2a2 18 = 0

An equation that can be expressed in the form ax2 + c = 0, where a and

c are real numbers and a 0 is a pure quadratic equation.

A Quadratic equation has

only two roots.


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Quadratic equation involving a variable in second degree as well as in first

degree is an Adfected Quadratic Equation

Example :

(1) x2 + 3x = 10 (2) 3a2 a = 2

If the terms in the RHS are transposed to LHS then,

(1) x2 + 3x 10 = 0 (2) 3a2 a 2 = 0

ax2 + bx + c = 0 is the standard form of a quadratic equation where a, b

and c and variables and a

0.

1. Solving Pure Quadratic equation

Example 1 : Solve the equation 3x2 27=0

Solution : 3x2 27 = 0

3x2 = 27

x =3

27

x2 = 9

x = 9x = +3 or x = 3

Example 2 : Solve the equation 4y2 9 = 0

Solution : 4y2 9 = 0

4y2 = 9

y2 =4

9

y =4

9

y =2

3

y =2

3+ or y =

2

3


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Example 3 : Solve the equation 99 = 4r2 1

Solution : 99 = 4r2 1

4r2 1 = 99

4r2 = 99 + 1

4r2 = 100

r2 =4

100 = 25

r = 25

r = 5

r = +5 or r = 5

Example 4 : Solve the equation (m + 8)2 5 = 31

Solution : (m + 8)2 5 = 31

(m + 8)2 = 31 + 5

(m + 8)2 = 36

(m + 8)2 = 36

(m + 8) = 36 m = 8 6

m = 8 + 6 o r m = 8 6

m = 2 or m = 14

Example 5 : If A =2r ; Solve for r.

Solution : A = 2r

2r = A

r2 =A

r =

A


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l2

Example 6 :If l2 = r2 + h2. Solve for h and find the value of h if l = 15 and r = 9.

Solution : l2 = r2 + h2

or

r2 + h2 = l2

h2 = l2 r2

h = 22 rl

h = 22 915 (substituting l = 15, r = 9)

h = 81225

h = 144 h = 12

h = +12 or h = 12

Example 7 : If B =4

a.32

Solve for a and find the value of a if B = 16 3 .

Solution : B =4

a.32

a2 =3

B4

a =3

B4 (Substituting B = 16 3 )

a =3

3164

//

a = 64 , a = 8

a = + 8 or a = 8

Exercise : 5.1

A. Classify the following equations into pure and adfected quadratic equation.

1) x2 + 2 = 6 2) a2 + 3 = 2a 3) p (p 3) = 1

4) 2m2 = 72 5) k 2 k = 0 6) 7y =y

35


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If mn = 0, then either

m =0 o r n =0

B. Solve the equations

1) 5x2 = 125 2) m2 1 = 143 3) 4a =a

81

4)2

2x

4

3 =

4

17 5) (2m 5)2 = 81 6)

18

)4(2x

=9

2

C.

1) If A = 2 2r Solve for r and find the value of r if A = 77 and =7

22

2) If V = hr2 Solve for r and find the value of r if V = 176 and h = 14

3) If r2 = l2 + d2 Solve for d and find the value of d if r = 5 and l = 4.

4) If c2 = a2 + b2 Solve for b. If a = 8 and c = 17 and find the value of b.

5) If K = 1/2mv2 Solve for v and find the value of v if K = 100 and m = 2

6) If v2 = u2 + 2as. Solve for v. If u = 0, a = 2 and s = 100, find the value

of v.

2. Solving the adfected quadratic equation by factorization :

Example 1 : Solve the quadratic equation a2

3a + 2 = 0Solution : a2 3a + 2 = 0

i. Resolve the expression a2 2a 1a + 2 = 0

ii. Factorize a(a 2) 1 (a 2) = 0

iii. Taking the common factor (a 2) (a 1) = 0

iv. Equate each factor to zero a 2 = 0 or a 1 = 0

v. The roots are a = 2 or a = 1

Example 2 : Solve the quadratic equation m2 m = 6

Solution : m2 m = 6

m2 m 6 = 0m2 3m + 2m 6 = 0

m(m 3) +2 (m 3) = 0

( m 3 ) ( m + 2 ) = 0

Either (m 3) = 0 or (m + 2) = 0

m = +3 or m = 2


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x2

1x 1x

Example 3 : Solve the quadratic equation 2x2 3x + 1 = 0

Solution : 2x2 3x + 1 = 0

2x2 2x 1x + 1 = 0

2x (x 1 ) 1 (x 1) = 0

(x 1) (2x 1) = 0Either (x 1) = 0 or (2x 1) = 0

x = 1 or x = 21

Example 4 : Solve the quadratic equation 4k (3k 1) = 5.

Solution : 4k (3k 1) = 5

12k2 4k 5 = 012k2 1 0 k + 6 k 5 = 0

2k (6k 5) + 1(6k 5) = 0

(6k 5) (2k + 1) = 0Either (6k 5) = 0 or (2k + 1) = 0

k = 6

5or k =

2

1

Exercise : 5.2

A. Find the roots of the following equations1) x(x 3) = 0 2) a (a + 5) = 0 3) m2 4m = 0

4) 3k 2 + 6k = 0 5) (y + 6) (y + 9) = 0 6) (b 3) (b 5) = 0

7) (2n + 1) (3n 2) = 0 8) (5z 2) (7z + 3) = 0

B. Solve the quadratic equations

1) x2 + 15x + 50 = 0 2) a2 5a + 6 = 0 3) y2 = y + 2

4) 6 p2 = p 5) 30 = b2 b 6) 2x2 + 5x 12 = 0

7) 6y2 + y 15 = 0 8) 6a2 + a = 5 9) 13m = 6(m2 + 1)

10) 0.2t2 0.04t = 0.03

Consider the equation x2 + 3x + 1 = 0

It cannot be factorised by splitting the middle term.

How do you solve such an equation ?

It can be solved by using Formula.


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3. Solving quadratic equation by formula method

General form of a quadratic equation ax2 + bx + c = 0

Divide by a 0a

c

a

bx

a

ax2

=++

Transpose the constant term to R.H.S. a

c

a

bx

x

2

=+

Add

2

a2

b

to both the sides

22

2

a2

b

a

c

a2

b

a

bxx

+=

++

2

22

a4

b

a

c

a2

bx +=

+

2

22

a4

bac4

a2

bx +=

+

Simplify 2

22

a4

ac4b

a2

bx

=

+

Taking square root 2

2

a4

ac4b

a2

bx

=+

a2

ac4b

a2

bx

2 =+

a2

ac4b

a2

bx

2

=

Roots are a2ac4bb

x

2 =

a2

ac4bbx

2 += or

a2

ac4bbx

2 =

Roots of the equation ax2 + bx + c = 0 are x =a2

ac4bb 2

Note : The roots of the equation ax2 + bx + c = 0 can also be found using

Sridharas method.

x x

xx

xx

x

x

x

x

x x

x

x

x


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Example 1 : Solve the equation x2 7x + 12 = 0

consider x2 7x + 12 = 0

This is in the form ax2 + bx + c = 0

the coefficients are a = 1, b = 7 & c = 12

The roots are given by x = a2

ac4bb 2

Substituting the values a = 1, b = 7 and c = 12

1x2

)12)(1(4)7()7(x

2 =

2

48497x

=

Simplify x =2

17

x =2

17 +or x =

2

17

x =2

8or x =

2

6

Roots are x = 4 or x = 3

Example 2 : Solve the equation 2p2 p = 15

Consider 2p2 p = 15

2p2 p 15 = 0


The coefficients are a = 2, b = 1 and c = 15

The roots are given bya2

ac4bbx

2 =


p =)2(2

)15)(2(4)1()1(2

x

x

x

x

x x

x

x x


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x

p =4

12011 ++

p =4

1211

p =4

111

p =4

111+or p =

4

111

p =4

12or p =

4

10

p = 3 or p =2

5

Example 3 : Solve the equation 2k2 2k 5 = 0

Consider 2k 2 2k 5 = 0


The coefficients are a = 2, b = 2 and c = 5

The roots are given by x =a2

ac4bb 2


k =

)2(2

)5)(2(4)2()2( 2

k =4

4042 + =

4

442

k =4

1122 =

4

1112

The roots are k = 2111+

or k = 2

111


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Example 4 : Solve the equation m2 2m = 2

Consider m2 2m = 2

m2 2m 2 = 0


Comparing the coefficients a = 1, b = 2 and c = 2

The roots are given by x =a2

ac4bb2

m =)1(2

)2)(1(4)2()2( 2

m =2

842 ++

m =2

122

m =2

312

m = 31+ or m = 31Exercise : 5.3

Solve the following equations by using formula

1) a2 2a 4 = 0 2) x2 8x + 1 = 0 3) m2 2m + 2 = 0

4) k2 6k = 1 5) 2y2 + 6y = 3 6) 8r2 = r + 2 7) p = 5 2p2

8) 2z2 + 7z + 4 = 0 9) 3b2 + 2b = 2 10) a2 = 4a + 6

4. Equations reducible to the form ax2 + bx + c = 0

Example 1 : Solve the equation (x + 6) (x + 2 ) =x

Solution :(x

+ 6) (x

+ 2 ) =x

x2 + 6x + 2x + 1 2 =x

x2 + 8x + 1 2 x = 0

x2 + 7x + 12 = 0

x2 + 4x + 3x + 12 = 0

x(x + 4) + 3 (x + 4) = 0

(x + 4) (x + 3) = 0

Either (x + 4) = 0 or (x + 3 ) = 0

x = 4 or

x = 3

x


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Example 2 : Solve the equation (a 3)2 + (a + 1)2 = 16

Solution : (a 3)2 + ( a + 1 )2 = 16

Using (a + b)2 = a2 + 2 a b + b2

(a b)2 = a 2 a b + b2

[(a)2 2(a)(3) + 32] + [a2 + 2 (a) (1) + 12] = 16

a2

6a + 9 + a2

+ 2a + 1 16 = 02a2 4a 6 = 0

a2 2a 3 = 0

a2 3a + 1a 3 = 0

a ( a 3 ) + 1 ( a 3 ) = 0

(a 3) (a + 1) = 0

Either (a 3) = 0 or (a + 1) = 0

a = 3 or a = 1

Example 3 : Solve 5(p 2)2 + 6 = 1 3 ( p 2 )

Solution : 5(p 2)2 + 6 = 1 3 ( p 2 )

Let p 2 = b

then 5b2 + 6 = 1 3 b

5b2 1 3 b + 6 = 0

5b2 1 0 b 3 b + 6 = 0

5b (b 2) 3 (b 2) = 0

(b 2) (5b 3) = 0

Either (b 2) = 0 or (5b 3) = 0

b = 2 or b = 5

3

p 2 = 2 or p 2 = 5

3(

b = p 2)

p = 2 + 2 or p = 5

3 +

1

2

p = 4 or p =5

13


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Example 4 : Solve the equation1k

1k

5k2

2k3

+

=++

Consider1k

1k

5k2

2k3

+

=++

Cross multiplying (3k + 2) (k 1) = (2k + 5) (k + 1)

3k2

+ 2k 3k 2 = 2k 2

+ 5k + 2k + 53k2 1 k 2 2 k 2 7k 5 = 0

3k2 1 k 2 2 k 2 7k 5 = 0

On simplification k2 8k 7 = 0

This is in form of ax2 + bx + c = 0

The co-efficients are a = 1, b = 8, c = 7

The roots of the equationa2

ac4)b(bx

2 +=

k =1x2

)7)(1(4)8()8(2

k =2

28648 ++

k = 2

928

k =2

2328 =

( )2

2342

k = 234

Example 5 : Solve the equation 1

y2

3

4

y=

Consider 1y2

3

4

y=

Taking L.C.M. 1y4

6y2=

By cross multiplication y2 6 = 4y

y2 4y 6 = 0

x


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this is in the form ax2 + bx + c = 0

comparing coefficients a = 1, b = 4, c = 6

y =)1(2

)6)(1(4)4()4( 2

the roots of the equation are =2

24164 +

y =2

404

y =2

1024 =

( )2

1022

y = 102+ or y = 102

Example 6 : Solve1m2

4

3m

1

2m

4

+=

+

+

1m2

4

)3m)(2m(

)2m(1)3m(4

+=

++++

1m2

4

6m3m2m

2m12m42 +=+++

+

1m2

4

6m5m

10m32 +

=++

+

On Cross multiplying, 4m2 + 20m + 24 = 6m2 + 20m + 3m + 10

4m2 + 20m + 24 6m2 2 3 m 1 0 = 0

2m2 3m + 14 = 0

This is in the Standard form 2m2 + 3m 14 = 0

2m2 + 7m 4m 14 = 0

m(2m + 7) 2 (2m + 7) = 0

(2m + 7) (m 2) = 0

Either (2m + 7) = 0 or (m 2) = 0

m =

2

7or m = 2


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Example 2 : Sum of a number and its reciprocal is 55

1. Find the number.

Solution : Let the number be = y

Reciprocal of the number =y

1

(Number) + (its reciprocal) = 55

1

y +y

1 =

5

26

y

1y2 + =

5

26

5(y2 + 1) = 26y

5y2 + 5 = 2 6 y

5y2 2 6 y + 5 = 0

5y2 2 5 y 1 y + 5 = 0

5y (y 5) 1 (y 5) = 0

(y 5) (5y 1) = 0

Either (y 5) = 0 or (5y 1) = 0

y = 5 or y = 5

1

The required number is 5 or 51

Example 3 :The base of a triangle is 4 cms longer than its altitude. If the area of

the traingle is 48 sq cms. Find the base and altitude. Solution : Let the altitude = x cms.

Base of the triangle = (x + 4) cms.

Area of the triangle = 2

1 (base) (height)

48 = 2

1 (x + 4)x


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[ ] [ ]

xx

xx

5

60300602 +

+ = 1

1

1

5

3002

=+ xx

x2 + 5x = 300

x2 + 5x 300 = 0

x2 + 20x 15x 3 0 0 = 0

x(x + 2 0 ) 1 5 (x + 2 0 ) = 0

(x + 20) (x 1 5 ) = 0

Either (x + 20) = 0 or (x 1 5 ) = 0

x

= 20 orx

= 15 Number of books = x = 15

Cost of each book =x

60 =

15

60 = Rs. 4

Example 5 : The speed of a motor boat in still water is 15 km/hr. If it goes down

the stream 30 kms and again returns to the starting point in total time

of 4 hrs and 30 minutes, find the speed of the stream.

Solution : Speed in Still water is = 15 km/hr

Total distance travelled = 30 km

Let the speed of the stream = x km/hr

Speed up the stream = (15 x) km/hr

Speed down the stream = (15 + x) km/hr

Total time taken = 4hrs and 30 minutes

Time taken to row down the stream =x+15

30

Time taken to row up the stream =x15

30

Time taken to row + time taken to row = 4 hours 30 minutes

Down the stream up the stream

x+15

30+

x15

30=

2

14

x = 20 cannot be

considered because

number of books is always

positive


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2

9

)15()15(

)15(30)15(30=

+++xx

xx

2

9

225

30450304502

=

++x

xx

2

9

225

9002=

x

2225

9

)2(900x=

200 = 225 x2

x2 = 225 200

x2 = 25

x = 5

x = + 5 or x = 5

Speed of the stream = x = 5km/hr

Exercise : 5.5

1) The sum of a number and twice its square is 105. Find the number.

2) Product of two consecutive integers is 182. Find the integers.

3) The sum of the squares of three consecutive natural numbers is 194. Find the integers.

4) The length of rectangular field is 3 times its breadth. If the area of the field is

147 sq mts. Find the length of the field.

5) Hypotenuse of a right-angled triangle is 20 mts. If the difference between the lengths

of other two sides is 4 mts. Find the measures of the sides.

6) An Aero-plane takes 1 hr. less for a journey of 1200 km. If its speed is increasedby 60 km/hr from its initial speed find the initial speed of the plane.

7) Some students planned a picnic. The budget for the food was Rs. 480. As eight

of them failed to join the party the cost of the food for each member increased

by Rs. 10. How many students participated in the picnic?

8) Sailor Raju covered a distance of 8 km in 1 hr 40 minutes down stream and returns

to the starting point. If the speed of the stream is 2 km/hr, find the speed of the

boat in still water.


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9) A dealer sells an article for Rs. 24 and gains as much percent as the cost price

of the article. Find the Cost price of the article.

10) Sowmya takes 6 days less than the number of days taken by Bhagya to complete

a piece of work. If both Sowmya and Bhagya together can complete the same

work in 4 days. In how many days will Bhagya complete the work?

6. Nature of the roots of a quadratic equation.1) Consider the equation x2 2x + 1 = 0


The coefficients are a = 1, b = 2, c = 1

x =a2

ac4bb 2

x =1x2

1x1.4)2()2( 2 +

x =2

442

x =2

02+

x =2

02+ or x =2

02

x = 1 or x = 1 roots are equal

2) Consider the equation x2 2x 3 = 0


the coefficients are a = 1, b = 2, c = 3

x =a2

ac4bb 2

x =1x2

16)2(

x =2

42+


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x =2

42+or x =

2

42

x =2

6or x =

2

2

x = 3 or x = 1 roots are distinct

3) Consider the equation x2 2x + 3 = 0



x =a2

ac4bb2

x =1x2

)3)(1(4)2()2(22

x =2

1242

x =2

82

x =2

222

x =( )

2

212 = 21

x = 21 + or 21 roots are imaginary

From the above examples it is clear that,

1) Nature of the roots of quadratic equation depends upon the value of (b2 4ac)

2) The Expression (b2 4ac) is denoted by (delta) which determines the natureof the roots.

3) In the equation ax2 + bx + c = 0 the expression (b2 4ac) is called the discriminant.


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Discriminant (b2 4ac) Nature of the roots

= 0 Roots are real and equal > 0 (Positive) Roots are real and distinct < 0 (negative) Roots are imaginary

Example 1 : Determine the nature of the roots of the equation 2x2 5x 1 = 0.

Consider the equation 2x2 5x 1 = 0

This is in form of ax2 + bx + c = 0

The co-efficient are a = 2, b = 5, c = 1

= b2 4ac = (5)2 4(2) (1) = 25 + 8 = 33

> 0Roots are real and distinct

Example 2 : Determine the nature of the roots of the equation 4x2 4x + 1 = 0

Consider the equation 4x2 4x + 1 = 0

This is in the form of ax2 + bx + c = 0

The co-efficient are a = 4, b = 4, c = 1

= b2 4ac

= (4)2 4 (4) (1)

= 16 16 = 0 Roots are real and equal

Example 3 : For what values of m roots of the equation x2 + mx + 4 = 0 are

(i) equal (ii) distinct

Consider the equation x2 + mx + 4 = 0


the co-efficients are a = 1, b = m, c = 4

= b2 4ac = m2 4(1) (4) = m2 16

1) If roots are equal = 0 m2 16 = 0

m2 = 16

m = 16 m = 4


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2) If roots are distinct > 0 m2 16 > 0 m2 > 16

m2 > 16

m > 4

Example 4 : Determine the value of k for which the equation kx2 + 6x + 1 = 0 has

equal roots.

Consider the equation kx2 + 6x + 1 = 0


the co-efficients are a = k, b = 6, c = 1

= b2 4ac

since the roots are equal, b2

4ac = 0 (

= 0)(6)2 4(k)(1) = 0

3 6 4 k = 0

4k = 36

k =4

36 = 9

k = 9

Example 5 : Find the value of p for which the equation x2 ( p + 2 )x + 4 = 0 hasequal roots.

Consider the equation x2 ( p + 2 )x + 4 = 0


Coefficients are a = 1, b = (p + 2), c = 4

since the roots are equal = 0b2 4 a c = 0

[(p + 2)]2 4(1)(4) = 0

(p + 2)2 16 = 0

p + 2 = 16p + 2 = 4p + 2 = + 4 or p + 2 = 4

p = 4 2 or p = 4 2

p = 2 or p = 6


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If m and n are the roots of the

quadratic equation

ax2 + bx + c = 0

Sum of the rootsa

b=

Product of rootsa

c+=

Exercise : 5.6

A. Discuss the nature of roots of the following equations

1) y2 7y + 2 = 0 2) x2 2x + 3 = 0 3) 2n2 + 5n 1 = 0

4) a2 + 4a + 4 = 0 5) x2 + 3x 4 = 0 6) 3d2 2d + 1 = 0

B. For what positive values of m roots of the following equations are

1) equal 2) distinct 3) imaginary

1) a2 ma + 1 = 0 2) x2 mx + 9 = 0

3) r2 (m + 1) r + 4 = 0 4) mk 2 3k + 1 = 0

C. Find the value of p for which the quadratic equations have equal roots.

1) x2 px + 9 = 0 2) 2a2 + 3a + p = 0 3) pk 2 1 2 k + 9 = 0

4) 2y2 py + 1 = 0 5) (p + 1) n2 + 2 ( p + 3 ) n + ( p + 8 ) = 0

6) (3p + 1)c2

+ 2 (p + 1) c + p = 0

7. Relationship between the roots and co-efficient of the terms of the quadratic

equation.

If m and n are the roots of the quadratic equation ax2 + bx + c = 0 then

m =a2

ac4bb 2 +, n =

a2

ac4bb 2

m + n =a2

ac4bb 2 + +a2

ac4bb 2

m + n =a2

ac4bbac4bb 22 +

m + n =a2

b2

m + n =a

b-

mn =

+a2

ac4bb2

a2

ac4bb2

mn =

( )2

222

a4

ac4b)b(


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mn =( )

2

22

a4

ac4bb

mn = 2

22

a4

ac4bb +

mn = 2a4ac4 = ac mn = ac

Example 1 : Find the sum and product of the roots of equation x2 + 2x + 1 = 0

x2 + 2x + 1 = 0



Let the roots be m and n

i) Sum of the roots m + n =a

b =

1

2

m + n = 2

ii) Product of the roots mn =a

c =

1

1

mn = 1

Example 2 : Find the sum and product of the roots of equation 3x2 + 5 = 0

3x2 + 0x + 5 = 0



Let the roots are p and q

i) Sum of the roots p + q =a

b =

3

0

p + q = 0

ii) Product of the roots pq =a

c =

3

5 pq =

3

5


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Example 3 : Find the sum and product of the roots of equation 2m2 8m=0

2m2 8m + 0 =0


Let the roots be and

i) Sum of the roots a

b

=+ 2

)8(

= = 4

ii) Product of the rootsa

c=

2

0= = 0

Example 4 : Find the sum and product of the roots of equation x2 (p+q)x + pq = 0

x2 ( p + q )x + pq = 0

The coefficients are a = 1, b = (p + q), c = pq


b

m + n =( )[ ]1

qp +

m + n = (p + q)

ii) Product of the roots mn = a

c = 1

pq

mn = pq

Exercise : 5.7

Find the sum and product of the roots of the quadratic equation :

1) x2 + 5x + 8 = 0 2) 3a2 10a 5 = 0 3) 8m2 m = 2

4) 6k 2 3 = 0 5) pr2 = r 5 6) x2 + (ab)x + ( a + b ) = 0

8. To form an equation for the given roots

Let m and n are the roots of the equation

x = m or x = ni.e., x m = 0 , x n = 0

(x m) (x n) = 0

x2 mx nx + mn = 0

x2

( m + n )x + mn = 0


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If m and n are the roots then the Standard form of the equation is

x2 (Sum of the roots)x + Product of the roots = 0

x2 (m + n) x + mn = 0

Example 1 : Form the quadratic equation whose roots are 2 and 3

Let m and n are the rootsm = 2, n = 3Sum of the roots = m + n = 2 + 3

m + n = 5Product of the roots = mn

= (2) (3)

mn = 6Standard form x2 ( m + n )x + mn = 0

x2 (5)x + ( 6 ) = 0

x2 5x + 6 = 0

Example 2 : Form the quadratic equation whose roots are5

2 and

2

5

Let m and n are the roots

m =5

2 and n =

2

5

Sum of the roots = m + n =5

2 +

2

5 =

10

254+

m + n =10

29

Product of the roots = mn =2

5x

5

2 mn = 1

Standard form x2 (m + n)x + mn = 0

x2 10

29x + 1 = 0

10x2 29x + 10 = 0


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Example 3 : Form the quadratic equation whose roots are 3 + 2 5 and 3 2 5


m = 3 + 2 5 a n d n = 3 2 5Sum of the roots = m + n

= 3 + 2 5 + 3 2 5

m + n = 6Product of the roots = mn

= (3 + 2 5 ) (3 2 5 )

= (3)2 (2 5 )2

= 9 20

mn = 11x2 ( m + n )x + mn = 0

x2 6x 11 = 0

Example 4 : If m and n are the roots of equation x2 3x + 1 = 0 find the value

of (i) m2n + m n2 (ii)n

1

m

1+

Consider the equation x2 3x + 1 = 0


The coefficients are a = 1, b = 3, c = 1Let m and n are the roots


b =

1

)3( = 3

m + n = 3


c

mn =1

1 mn = 1

(i) m2n+mn2 = mn (m + n)

= 1(3) = 3

(ii)m

1 +

n

1=

mn

mn + =

mn

nm + =

1

3

m1

+ n

1 = 3


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Example 5 : If m and n are the roots of equation x2 3x + 4 = 0 form the

equation whose roots are m2 and n2.

Consider the equation x2 3x + 4 = 0



i) Sum of the roots = m + n =a

b =

1

)3(

m + n = 3

ii) Product of the roots = mn =a

c =

1

4

mn = 4

If the roots are m2 and n2Sum of the roots m2 + n2 =(m+n)2 2mn

= (3)2 2(4)

= 9 8

m2 + n2 = 1

Product of the roots m2n2 = (mn)2

= 42

m2n2 = 16x2 (m2 + n2)x + m2n2 = 0

x2 (1)x + (16) = 0

x2 x + 16 = 0

Example 6 : If one root of the equation x2 6x + q = 0 is twice the other, find the

value of q

Consider the equation x2 6x + q = 0


The coefficients are a = 1, b = 6, c = q

Let the m and n are the roots


b =

1

)6(

m + n = 6


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c =

1

q

mn = q

If one root is (m) then twice the root is (2m)

m = m and n = 2m

m + n = 6

m + 2m = 6

3m = 6

m =3

6 m = 2

We know that q = mn

q = m(2m)

q = 2m2

q = 2(2)2

q = 8

q = 8

Example 7 : Find the value of k so that the equation x2 2x + (k + 3) = 0 has one

root equal to zero.

Consider the equation x2

2x + (k + 3) = 0The coefficients are a = 1, b = 2, c = k + 3


Product of the roots = mn

mn =a

c

mn = 1

3k+

mn = k + 3

Since m and n are the roots, and one root is zero then

m = m and n = 0 mn = k + 3

m(0) = k + 3 0 = k + 3

k = 3


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Exercise : 5.8

A. Form the equation whose roots are

1) 3 and 5 2) 6 and 5 3) 2 and2

34)

3

2 and

2

3

5) 2 + 3 and 2 3 6) 3 + 2 5 and 3 2 5

B.

1) If m and n are the roots of the equation x2 6x + 2 = 0 find the value of

i) (m + n) mn ii)m

1 +

n

1

2) If a and b are the roots of the equation 3m2 = 6m + 5 find the value of

i)ab

ba + ii) (a + 2b) (2a + b)

3) If p and q are the roots of the equation 2a2 4a + 1 = 0 Find the value of

i) (p + q)2 + 4pq ii) p3 + q3

4) Form a quadratic equation whose roots areq

p and

p

q

5) Find the value of k so that the equationx2 + 4x + (k + 2) = 0 has one root equal

to zero.

6) Find the value of q so that the equation 2x2 3qx + 5q = 0 has one root which

is twice the other.

7) Find the value of p so that the equation 4x2 8px + 9 = 0 has roots whose

difference is 4.

8) If one root of the equationx2 + px + q = 0 is 3 times the other prove that 3p2 = 16q

Graphical method of solving a Quadratic Equation

Let us solve the equation x2 4 = 0 graphically,

x2 4 = 0

x2 = 4let y =x2 = 4

y =x2

and y = 4


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y =x2

x = 0 y = 02 y = 0

x = 1 y = 12 y = 1

x = 2 y = 22 y = 4

x = 1 y = (1)2

y = 1x = 2 y = (2)2 y = 4

x 0 1 1 2 2 3

y 0 2 2 8 8 6

(x, y) (0, 0) (1, 2) (1, 2) (2, 8) (2, 8) ( 3 ,6)

Step 1: Form table of

corresponding values

of x and y

Satisfying the equation

y =x2

Step 2: Choose the scale onx axis, 1 cm = 1 unit

y axis, 1 cm = 1 unit.

Step 3: Plot the points (0, 0);

(1, 1); (1, 1); (2, 4)

and (2, 4) on graph

sheet.

Step 4: Join the points by asmooth curve.

Step 5: Draw the straight line

y = 4 Parallel to x-axis

Step 6: From the intersecting

points of the curve and

the line y = 4, draw

perpendiculars to thex axis

Step 7: Roots of the equations are x = +2 or x = 2

The graph of a quadratic polynomial is a curve called parabola

Example 1 : Draw a graph of y = 2x2 and find the value of 3 , using the graph.

Step 1: Form the table of

corresponding values ofx and y satisfying the

equation y = 2x2

Step 2: Choose the scale on x

axis, 1 cm = 1 unit and

y axis, 1 cm = 1 unit


(1, 2) (1, 2); (2, 8) and

(2, 8) on graph sheet.


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x 0 1 1 2 2

y 0 1 1 4 4

(x, y) (0, 0) (1, 1) (1, 1) (2, 4) (2, 4)

x 0 1 1 2 2

y 2 1 3 0 4

(x, y) (0, 2) (1, 1) (1, 3) (2, 0) (2, 4)

Step 4: Join the points by a

smooth curve

Step 5: Draw the straight line

y = 6 Parallel tox-axis.



the line y = 6, draw

perpendiculars to the

x-axis.

Step 7: Value of 3 = 1.7x = 1.7 or x = + 1.7

Example 2 : Draw a graph of y = x2 and y = 2-x and hence solve the equation

x2 +x 2 = 0


corresponding values of

x and y satisfying the

equation y = x2


corresponding values ofx and y satisfying the

equation y = 2 x.


axis 1 cm = 1 unit and

y axis, 1 cm = 1 unit.


(1, 1); (1, 1); (2, 4)and (2, 4) on the graph

sheet.

Step 5: Join the points by a

smooth curve.

Step 6: Plot the points (0, 2) ;

(1, 1); (1, 3); (2, 0)

and (2, 4) on graph

sheet


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x 0 1 1 2 2

y 2 1 1 4 4

(x, y) (0, 0) (1, 1) (1, 1) (2, 4) (2, 4)

x

0 1 2 1 2y 2 3 4 1 0

(x, y) (0, 2) (1, 3) (2, 4) (1, 1) (2, 0)

Step 7: Join the points to get a line.


Curve and the line, draw

perpendiculars to the

x-axis

Step 9: Roots of the equation are x = 1 or x = 2

Example 3 : Solve the equation

Method I : x2 x 2 = 0

Split the equation

y =x2 and y = 2 +x


corresponding valuesx

and y satisfying the

equation y =x2

Step 2: Form the table ofcorresponding valuesxand y satisfying theequation y = 2 +x

Step 3: Choose the scale on

x axis, 1 cm = 1 unity axis, 1 cm = 1 unit

Step 4: Plot the points (0, 0);(1, 1); (1, 1); (2, 4)and (2, 4) on the graphsheet.

Step 5: Join the points by asmooth curve

Step 6: Plot the points (0, 2);(1, 3) (2, 4); (1, 1) and

(2, 0) on the graphsheet.

Step 7: Join the points to get astraight line

Step 8: From the intersectingpoints of Curve and theline, draw the perpendi-

culars to the x-axis.

Step 9: Roots of the equation are x = 1 or x = 2


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x 0 1 1 2 2

y 2 2 0 0 4

(x, y) (0, 2) (1, 2) (1, 0) (2, 0) (2, 4)

Method II :


corresponding values of

x and y satisfying

equation y =x2 x 2.


axis 1 cm = 1 unit and

y axis 1 cm = 1 unit.


(1 2); (1, 0); (2, 0)

and (2, 4) on the graph

sheet.

Step 4: Join the points to form

a smooth curve

Step 5: Mark the intersecting


the x axis.

Step 6: Roots of the equations are x = 1 or x = 2

Exercise : 5.9

A. 1) Draw the graph of y = x2 and find the value of 7

2) Draw the graph of y = 2x2 and find the value of 3

3) Draw the graph of y =2

1x2 and find the value of 10

B. 1) Draw the graph of y =x2 and y = 2x + 3 and hence solve the equation

x2 2x 3 = 0

2) Draw the graph of y = 2x2 and y = 3 x and hence solve the equation2x2 +x 3 = 0

3) Draw the graph of y = 2x2 and y = 3 +x and hence solve the equation

2x2 x 3 = 0

C. Solve graphically

1) x2 +x 12 = 0 2) x2 5x + 6 = 0 3) x2 + 2x 8 = 0

4) x2 +x 6 = 0 5) 2x2 3x 5 = 0 6) 2x2 + 3x 5 = 0

Quad Removal Equa

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