suppsafdsfyreedft fsdds

7/24/2019 suppsafdsfyreedft fsdds

1/17

138

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


2/17

142

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

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


3/17

143

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.


4/17

144

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


5/17

145

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


6/17

146

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(


7/17

147

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


8/17

148

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

x

2

( m + n )x+ mn = 0


9/17

149

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


10/17

150

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 = 11x

2 ( 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


11/17

151

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 = 16x

2 (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


12/17

152


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


13/17

153

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


14/17

154

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.


15/17

155

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


16/17

156

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


17/17

157

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

1x

2 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

Documents

suppsafdsfyreedft fsdds