Vedanta Optional Math Teacher's guide

11(a). The sum of the three numbers in G.P. is 38 and their product is 1728. Find the numbers.

Solution a
r
Let the three numbers b in GP be , a, ar

According to question

a +a+ar=38
r
a+ar+ar2
or r =38...............(i)

Also

a .a.ar=1728
r

or, a3=1728
or, a3=123
or, a=12
Substituting value of r is eqn(i)

or 12(1+r+r2 ) =38
r

or, 6+6r+6r2=19r

or, 6r2–13r+6=0

or, 3r(2r–3)+2(2r–3)=0

or, (2r–3)(3r–2)=0

Either

2r–3=0............(ii)

OR,

3r–2=0...........(iii)

From (ii) , r= 3
2
Fwarrho=men39(ri=i=i),238r,=,an=23u1m2baenrsdaarre=12×
3 =18
2

2 r= 2
when 3

a =122 =18, a=12 and ar=12× 2 =18
r 3

3
Hence the required numbers are 8,12 and 18 or 18, 12 and 8.

11(b). Sum of the three consecutive terms in G.P. is 28 and their product is 512. Find the numbers.

Solution a
r
Let the three consecutive terms in GP be , a, ar

Vedanta Optional Mathematics Teacher's Guide ~ 10 101

According to question

a +a+ar=28 ...............(i)
r
Also

a .a.ar=512
r

or, a3=512

or, a3=83

or, a=8

Substituting value of a is eqn(i)

8 +8+8r=28
r

or 8+8r+8r2 =28
r
or, 2+2r+2r2=7r

or, 2r2–5r+2=0

or, 2r2–4r–r+2=0

or, 2r(r–2)–1(r–2)=0

or, (r–2)(2r–1)=0

or, (2r–3)(3r–2)=0

Either

r–2=0............(ii)

OR,
2r–1=0...........(iii)

From (ii) , r=2

From (ii),r= 1
2
when r=2,numbers are

a = 3 =4, a=8 and ar=8×2=16
r 2
1
when r= 2

a = 8 =16, a=8 and ar= 8 =4
r 1 2
2

11(c). The product of three numbers in G.P. is 729 and sum of their squares is 819 ? Find
the numbers

Solution a
r
Let the three numbers is G.P. be , a and ar

According to question

a .a.ar=729 ...............(i)
r

a .a.ar=512
r

102 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, a3=729

or, a3=93

or, a=9

Also.

Substituting value of a is eqn(i)

a 2
r +a2+(ar)2=819

9 2
r +92+(9r)2=819

81 =81+81r2=819
r2

or, 9+9r2+9r4=91r2

or, 9r4 – 81r2–r2+9=0

or, 9r2(r2 –9)–1(r2–9)=0

or, (r2 –9)(9r2–1)=0

or, (r+9)(r –9)(9r2–1)=0

or, 2r2–5r+2=0

Either

r+9=0............(ii)

or, r–9=0...........(iii)

or, 9r2–1=0.........(iii)

From (i) , r = –9

From (ii), r = 9 1 2 1
3 3
From (iii), r2 = , r=±

where r =–9, numbers are

a = 9 =–1, a=9 and ar=–81
r -9
But it doesnot satisfy the second condition. So they arenot the required numbers.

Similarly when r = 9, numbers are 1,9 and 81.

They also don't satisfy second condition.

when r= 1
3
a 9 1
r = 1 =27, a=9 and ar=9. 3 =3

It sat3isfies both conditions. So the numbers are 27, 9 and 3.

when r=– 1 the required numbers are,
3
-1
a = 9 =–27, a=9 and ar=9. 3 =–3
r -1

3

Hence the required numbers are 27, 9,3 or –27,9,–3

Vedanta Optional Mathematics Teacher's Guide ~ 10 103

Questions for practice

1. Find the 10th term of a geometric sequence 2, 4, 8, 16, 32,.....................

2. If p, p–2,p+1 are in G.P., find the value of p.

3. If 1 , p, 3 and q are in G.P., find the values of p and q.
3

4. If x+6, x, x–3 are three terms of a G.P.,find the value of x.

5. If the first term and third term of a G.P. are 2 and 8–1, find the common ratio of the
progression.

6. The 6th and 10th terms of a G.P. are 64 and 1024 respectiely, find the 7th term of the G.P.

7. If 2nd term and 4th term of a G.S. are 1 and 1 , find the 7th term.
4 16
8. The product of three numbers in a G.P. is 216 and their sum is 26, find the numbers.

9. The product of three numbers in G.P. is 64 and their sum is 21 , find the numbers.

10. If the 7th term of a G.P. is 1 and its 16 times of 11th term is also 1, find the 20th term.

Geometric Means (G.M.'s)

Notes:
i) G.M. between two numbers a and b, G.M.= ab

ii) Let m1,m2,m3,........mn be n G.M.'s between a and b, then common ratio (r)= b 1
a n+1

and geometric means are given by

m1=ar
.m2=ar2
.
.

mn =arn

Some solved problems

1. Find the G.M. between 1 and 1 .
3 27

Solution 1 1
3 27
Here, a= ,b=

G.M. between a and b= ab = 1 . 1 = 1
3 27 9

2. Find the value of x,y and z if 3,x,y,z,48 are in G.P.

104 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution
Here x,y,z are in G.M.'s between 3 and 48.

Numbers of G.M.'s (n)=3,a=3, b=48

1
b n+1
Common ratio(r) = a

1
3+1
= 48
3

1
=16 4 =2

Now, first mean (m1)=x=ar=3.2=6
Second mean (m2)=y=ar2=3.22=12
Third mean (m3)=z=ar3=3.23=24

3.Insert four G.M.'s between 1 and 32. 1
n+1
b
Solution: Here a=1, b=32, Common ratio(r) = a

1

= 32 4+1
1

1

5

= 25 =2

Let required geometric means be m1, m2, m3 and m4.
m1=ar=1.2=2
m2=ar2=1.22=4
m3=ar3=1.23=8
m4=ar4=1.24=16

4. Third geometric mean between 27 and 1 is 1. Find the number of means.
27

Solution 1
27
Here, number of means(n)=?, m3 =1,a=27, b=

1

common ratio (r)= b n+1
a
11
27 n+1
27
=

1

= 1 n+1
36

6

1 n+1
3

Vedanta Optional Mathematics Teacher's Guide ~ 10 105

third mean (m3) =ar3
63

or, 1=27 1 n+1
3

1 1 18
27 3
or, = n+1

2 1 18
3
or, 1 = n+1
3

or, 3 = 18
n+1

or, 6 = n+1
⸫ n=5

5(a). There are three geometric means, between a and b. If the first mean and third mean
1 1
are 4 and 64 respectively, find the values of a and b.

Solution
Here, first term=a, last term =b

Let r be the common ratio

m1 =ar and m3 = ar3
or, 1 =ar...............(i)
4

1 = ar3...............(ii)
64
Dividing (ii) by (i), we get

1
64
ar3 = 1
ar

4

or, r2 = 1 ×4
16
1
or, r2 = 16

or, r= 1
4

put the value of r in equation (i), we get.

1 =a. 1
4 4

a=1
Again, put the value of a and r in

106 Vedanta Optional Mathematics Teacher's Guide ~ 10

b 1
1
r= n+1

1 b 1
4 1
or, = 3+1

or, 1 = b 1
4 4

Taking fourth power on both sides, we get.

1 =b
256
1
b= 256

a=1, b= 1
256

5(b). There are three G.M.'s between p and q . If the first mean and the third mean are
respectively 1 and 125, find the values of p and q.

Solution
Here, first mean(m1)=1
Third mean (m3)=125
Let r be the common ratio.

Then

1 11
q q q
r= p n+1 = p 3+1 = p 4

m1 =ar or, 1=pr..............(i)
and m3 = ar3 or 125= pr3...............(ii)
Dividing (ii) by (i), we get

pr3 = 125
pr 1

or, r2 =125

r=5 5

Put the value of r in equation(i)

1=p.5 5

p=5 1 q 1
5 4

Again, r = 1
55

q 1
or, 5 5 = 1 4

55

Taking fourth power on both sides,
15625=5 5 q

Vedanta Optional Mathematics Teacher's Guide ~ 10 107

or, q=5155625 =625 5

p=5 1 and q=625 5
5

6(a). Some G.M.'s are inserted between 4 and 128. Find the number of means where the ratio
of first mean and last mean is 1:8.

Solution
Let r be common ratio. a=4, b=128.

1
b n+1
r= a

1

= 128 n+1
4

5

= 2 n+1

But,

mmn1 = 1
8

or, ar = 1
arn 8

or, rn–1 = 23

5

But r= 2 n+1

5 n-1 =23

2 n+1

or 5(n-1) =3
n+1
or, 5n–5=3n+3

o, 2n =8

⸫ n=4

(b). There are n G.M.'s between 1 and 25. If the ratio of the first mean to the third mean
is 1:25, find the value of n. 25

Solution 1
n+1
1 b
Here, a= 25 , b=25 , common ratio(r)= a

25 1
= 1 n+1
25

625 1
4 n+1

108 Vedanta Optional Mathematics Teacher's Guide ~ 10

4 first mean 1
last mean 25
= 5 n+1

4

∴ r=5 n+1

By question, =

or, ar = 1
arn 25

or, rn–1 = 52

4

But r= 5 n+1

4 n–1
5n+1
=52

⸫ 4(n-1) =2
n+1

or, 4n–4=2n+2

or, 2n =6

∴ n=3

7(a). Find the two numbers whose A.M. and G.M. are respectively 10 and 8.

Solution
Let required two numbers be a and b.

Then, A.M. = a+b =10
2

a+b =20...........(i)

G.M. = ab =8
or, ab=64.............(ii)

we have, a–b= (a+b)2-4ab

= 400-256

= 144

a–b=12............(iii)
adding(i) and (iii),we get,
2a =32

a=16

put the value of a in equation(i), we get,
b=20–16=4

a=16,b=4

Vedanta Optional Mathematics Teacher's Guide ~ 10 109

(b). Find two numbers whose A.M. and G.M. are respectively 25 and 20.

Solution
Let a and b two required numbers

Then, A.M. = a+b =25
2

a+b =50...........(i)

G.M. = ab =20

or, ab=400.............(ii)
we have, a–b= (a+b)2-4ab

= 2500-1600

= 900

a–b=30............(i)
adding (i) and (ii),we get,
2a =80

a=40
put the value of a in equation(i), we get,
b=20–16=410

a=40, b=10

(c). Find the two numbers whose A.M. and G.M. are respectively 13 and 8.

Solution
Let a and b two required numbers

Then, A.M. = a+b =13
2
a+b =26...........(i)

G.M. = ab =12

or, ab=144.............(ii)

we have, a–b= (a+b)2-4ab

= 676-576

= 100 =10
a–b=10............(iii)
solving equations (ii) and (iii),we get,
a =18, b=8
put the value of a in equation(i), we get,
b=20–16=4

a=16,b=4

110 Vedanta Optional Mathematics Teacher's Guide ~ 10

(d). Find two numbers whose A.M. is 34 and G.M. is 16.

Solution

Let the required two numbers be a and b.

Then, A.M. = a+b =34
2

a+b =68...........(i)

Also, G.M. = ab =16

or, ab=256.............(ii)
we have, a–b= (a+b)2-4ab

= 4624-1024

= 3600

a–b=60............(iii)

adding(i) and (iii),we get,
a =64
put the value of a in (i),b=4

b=20–16=4

a=64, b=4

8(a). The ratio of two numbers is 1:16, their geometric mean is 1 . Find the numbers.
4

Solution
Let the required numbers be a and b.
Also, Let a=k, b=16k

Then, G.M. = ab = 1
4
1
or, k.16k = 4

or, 4k= 1
1 4
16
k= 1
16
Hence a=k=

b=16k=16. 1 =1
16
1
a= 16 ,b=1

(b). The ratio of two numbers is 1:81, their geometric mean is1 . Find the numbers.

Solution
Let two numbers be ak and b=81k.
Then, G.M. = ab =1

or, k.81k =1

Vedanta Optional Mathematics Teacher's Guide ~ 10 111

or, 9k=1

k= 1
9
1 1
Hence the required numbers are a= 9 and b=81. 9 =9

Questions for practice

1. Find the G.M. between 100 and 400.

2. If p+2, P+8 and 17+p are the geometric sequence, find the value of m.

3. Find the values of x,y,z from the given geometric sequence 1 ,x,y,z,2.
8
4, Insert 3 G.M.'s between 3 and 48.

5. Insert 5 G.M.'s between 35 and 2240.

6. There are some geometric means between 5 and 80. If the second mean is 20, find the
number of means. Also find the last mean.

7. Find the number of geometric means inserted between 1 and 64 in which the ratio of
first mean to the last means is 1:16.

8. There are 5 geometric means between a and b. If the second mean and last means are
63 and 1701 respectively, find the values of a and b.

Sum of Geometric Series

Notes

i) In a G.P. if the first term =a, common ratio =r, number of terms =n are given, the sum of

the first n terms is given by

Sn = a(rn-1) , when r>1
r-1

Sn = a(1-rn) , when r<1
1-r

ii) When the last term = l, common ratio =r and the first term =a, then the sum of the first

n terms is given by

Sn = lr-a , when r>1
r-1

Sn = a-lr , when r<1
1-r

iii) Three consecutive terms of a G.P. are given by
a
r , a, ar

Some solved problems

1. Find the sum of series

(a). 1+ 1 + 1 ............ upto 8 terms.
2 4

112 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution
Given series is a G.P.

First term (a)=1,common ratio (r)= 1 <1
2
number of terms (n)=8

Sum of the 8 terms is given by,

Sn = a(1–rn)
1–r

1 8
2
1 1– 1– 1
256
S8 = = = 0 255 =1 127
1 1 128 128
1– 2
2

(b). 81+27+..........+ 1
243

Solution
Here. First term (a)=81,

common ratio (r)= 27 = 1 <1
81 3
1
last term (l)= 243

Now, Sn = a-lr
1-r

81+ 1 . 1
243 3
=
1
1- 3

= 59049 × 1
729 3

= 59048
243×2

= 29524
243

3. Find the first n term of a geometric series whose last term is 1792, common ratio 2, sum 3577

Solution
Here, in a given series, a =?, r=2,

l=1792, sn=3577

By using formula,

Now, Sn = lr-a
r-1

or, 3577= 1792×2-a
2-1

or, 3577=3584–a

or, –7=–a

a=7

Vedanta Optional Mathematics Teacher's Guide ~ 10 113

4(a). How many terms of G.P. must be taken in the series 128+64+32+................ so that
the sum may be 254.

Solution
In the given G.S.

First term (a)=128,common ratio (r)= 64 = 1
128 2
Sn =254

By using formula,

Sn = a(1–rn)
1–r
n
1
128 1 – 2

or, 254 = 1
2
1–

or, 127 = 2 1– 1n
64 2

or, 127 = 1– 1n
128 2

or, 1 n 127
2 128
=1–

or, 1 n 1
2 128
=1–

or, 1 n 17
2 2
=

n=7

(b) How many terms of a G.P. must be taken in the series 64+96+144+216+......... so that
the sum may be 2059 ?

Solution a=64, r = 96 = 3
In the given G.S., 64 2

Sn =254, n= ?

By using formula,

Sn = a(rn –1)
r–1
n
64 3
or,2059 = 2 –1

3 –1
2

or, 2059 =2 3 n
64 2
–1

114 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, 2059 = 3 n
128 2
–1

or, 2059 +1= 3n
128 2

or, 2187 = 3n
128 2
3n
or, 3 7 2
2
=

n=7

(c). How many terms of G.P. must be taken in the series 3–6+12..... so that the sum may be –63 ?

Solution
Let Sn = –63
6
r=– 3 =–2

a=3

By using formula,

Sn = a(1-rn)
or, –63 1-r
3(1-(-2)n)
= 1+2

or,–63 =1–(–2)n

or,64 =(–2)n

or,(–2)6 =(–2)n

n=6

5(a). The first term of a G.P. is 5 and the sum of the first four terms is 780, find the common
ratio.

Solution Sa4 (=rn7-810)
Here, a=5, r-1
Now, Sn =

or,780 = 5(r4 -1)
r-1

or,156 = (r2 +1)(r-1)(r+1)
r-1

or, 156=r3+r2+r+1

or, r3+r2+r–155=0

or, r3–5r2+6r2–30r+31r–155=0

or, r2(r–5)+6r(r–5)+31(r–5)=0

or, (r2+6r+31)(r–5)=0

Vedanta Optional Mathematics Teacher's Guide ~ 10 115

Either r=5 or r2+6r+31=0
Here, r2+6r+31 doesnot give real solution

r=5

5(b). The first term of a G.P. is 3 and the sum of the first four terms is 120, find the common ratio.

Solution
Here, in given G.P. , a=3, S4 =120, n=4
Now,
Sn = a(rn -1)
r-1

or,120 = 3(r4 -1)
r-1
(r2 +1)(r-1)(r+1)
or,40 = r-1

or, 40=r3+r+r2+1

or, r3+r2+r–39=0

for r=3, r3+r2–39=0

By factor theorem, r–3 is a factor of

r3+r2+–39=0

Now, r3–3r2 +4r2–12r+13r–39=0

or, r2(r–3)+4r(r–3)+13(r–3)=0

or, (r–3)(r2+4r+13)=0

Either r–3=0=3 or r2+6r+31=0

Here, r2+4r+13=0 does not give real value.

r=3

6(a). The sum of the first four terms of a G.P. with common ratio 3 is 118 . Find the sum of
the first ten terms of the progression. 3

Solution 118
3
Here, S4 = , r=3

Now,

Sn = a(rn –1) , or, Sn = a(34–1)
r–1 3–1

or, 118 = a.80
3 2
118
or, 3 =40a

or, a= 118
120

= 59
60

Again,

S10 = a(r10 –1) , or, Sn = a(34–1)
r–1 3–1

116 Vedanta Optional Mathematics Teacher's Guide ~ 10

59 {310–1}
120
= 59 3–1
120 {59049–1}
=
2

= 59.59048
240
435479
= 30

S10 = 435479
30

6(b). The sum of the first three terms of a G.P. with common ratio 2 is 112. Find the sum of
the first 8 terms of the progression.

Solution
Here, S3 =112, r=2

Now, Sn = a(rn –1)
r–1

or, S3 = a(23 –1)
2–1

or, 112 = 7a
1

a= 112 =16
7

Again, S8 = 16(28 -1)
2-1

=4080

7(a). In a G.P. the sum of three consecutive terms is 21 and their product is 64. Find them.

Solution a
r
Let three terms in be , a, ar

then,

a +a+ar=64
r
or, a3 =64

a=4

Again,
oarr,.a4r.a+r=4+214r=21
or,4+4r+4r2=21r

or, 4r2–17r+4=0

or, 4r2–16r–r+4=0

or, 4r(r–4)–1(r–4)=0

Vedanta Optional Mathematics Teacher's Guide ~ 10 117

or, (r–4)(4r–1)=0

r=4, 1
4
Find r=4, the required numbers are,

a = 4 =1, a=4, ar =4.4 =16
r 4
1
when, r= 4 , the required numbers are,

a = 4 =16, a=4, ar=4. 1 =16
r 1 4

4

Hence the required numbers are 1, 4, 16 or 16,4,1

7(b). The sum of three consecutive terms in G.P. is 62 and their product is 1000, find them.

Solution a
r
Let three terms in G.P. be , a, ar

then,

a +a+ar=1000
r
a=10

Again,
oarr,.a1.r0ar+=1602+10r=62
or,10+10r+10r2=62r

or, 10r2–52r+10=0

or, 5r2–26r+5=0

or, 5r2–25r–r+5=0

or, 5r(r–5)–1(r–5)=0

or, (r–5)(5r–1)=0

r=5, 1
5
Find r=5, the required numbers are,

waarr h==e1n11500, r===2155, 0a,=,tha1=e0r1,e0aqr,u=air1r=e0d1.50n=.u515m0=b2ers are,
5

Hence the required numbers are 2, 10 and 50 or 50,10 and 2.

8(a). The sum of the first 2 terms of G.P. is 6 and that of the first four terms is 15 . Find the
sum of the first six terms. 2

Solution 15
2
Here, S2 =6 , S4=

118 Vedanta Optional Mathematics Teacher's Guide ~ 10

Now, S2 = 9
S4 15
2

a(rn –1)
r–1 12
or, a(rn –1) = 15

r–1 12
1 15
or, r2+1 =

or, 15=12r2+12

or, 3=12r2

r2= 1 or, r=± 1 1
4 2 2
Since sum is positive, we take positive value of r=

Put the value r in s2=6

By using formula,

or, a(r2 -1) =6
r-1

or, a(r2 -1) =6
r-1

a 1 –1
or, 4
=6
1
2 –1

or, a. 3 .2=6
4

a=4 a(1–rn)
Now, Sn = 1–r
16
4 1– 4 4 1– 1
64
63
or,S6 = 1– 1 = 1– 1 = 8
2 2

(b).The sum of the first 2 terms of G.P. is 8 and that of the first four terms is 80 . Find the
sum of the first 6 terms.

Solution
Here, S2 =8 and s4=80

Now,

Sn = a(rn –1) ,
r–1

or, S2= a(r2–1)
r–1

Vedanta Optional Mathematics Teacher's Guide ~ 10 119

and

S4= a(r4–1)
r–1

S2 a(r4 –1)
S4 r–1
or, = a(r2 –1)

12 r–1
15
or, =r2+1

or, 10=r2+1

or, r2=9

r=±3

Since sum are positive, taking positive square roots only.

r=3

put the value of r in

S2= a(r2–1)
r–1
a(9–1)
or, 8= 3–1

a=2

Again,

S6= a(r6 –1) ,
r–1

= 2(36–1)
3–1

=728

9(a). The sum of the first 8 terms of G.P. is 5 times the sum of the first 4 terms. Find the
common ratio.

Solution

Here, S6=5 s4

or. a(r8 –1) = 5 a(r4–1)
r–1 r–1

or, (r4+1)(r4–1)=5(r4–1)

or, r4+1=5

or, r4=4

⸫ r= 2

9(b). The sum of the first 6 terms of G.P. is 9 times the sum of the first 3 terms. Find the
common ratio.

Solution
Here, S6=9 S3

120 Vedanta Optional Mathematics Teacher's Guide ~ 10

or. a(r6 –1) = 9. a(r3–1)
r–1 r–1

or, r3+1=9

or, r3=8

r=2

10(a). The sum of the first 2 terms of G.P. icsom98moannrdattihoeasnudmthoef the last 2 terms is 1152.
There are altogether 12 terms. Find the first term.

Solution S2= 9
In given G.P. 8

or. a(r2 –1) = 9 or, a(r+1)= 9 ..................(i)
r–1 8 8
Total number of terms =12

and the sum of last two term =1152

t11+t12=1152

or, ar10+ar11=1152

or, ar10(1+r)=1152............(ii)

Dividing (ii) by (i), we get,

ar10(1+ r) = 1152
a(1+a) 918

or, r10=1024

or, r10=210

r=2

put the value of r in equation (i), we get

a(2+1)= 9
8

or,a= 9 = 3
8×3 8

r=2, a= 3
8

10(b). The sum of the first four terms of G.P. is 30 and that of the last four terms is 960.If its
first terms is 2 and the last terms is 512,find the common ratio and the number of terms.

Solution
In a given G.P.

First term (a)=3, S4 =30

Let us find common ratio r,

Sn = a(rn –1)
r–1
2(r4 –1)
or,S4 = r–1

or,30 = 2(r2 +1)(r+1)(r–1)
r–1

Vedanta Optional Mathematics Teacher's Guide ~ 10 121

or, 15=r3+r2+r+1

or, r3+r2+r–14=0 → For r = 2, 23 + 22 + 2 – 14 = 0

Hence,by factor theorem,

(r–2) is a factor of r3+r2+r–14

Now,

r3–2r2 +3r2–6r+7r–14=0
or, r2(r–2)+3r(r–2)+7(r–2)=0
or, (r–2)(r2+3r+7) =0
Either r–2=0 ⇒ r=2
or, r2+3r+7=0 It doesnot give real valueof r.
Again, last term, l=arn–1
or, 512=2.2n–1
or, 256=2n–1
or, 28=2n–1

8=n–1
or, n=9
Then sum of the last four terms is given by
ar5+ar6+ar7+ar8=960(True).
ie.
2(25+26+27+28 )
=2(32+64+128+256)
=960

Questions for practice

1. Find the sum of the series 27+9+3+.......+ 1 .
81

2. How many terms are there in the series 1+2+4+.....+256. Calculate the sum.

3. A geometric series has 8 terms whose sum of the first three terms 7 and the sum of the
sum of the last three terms is 224. Find the first term, common ratio and sum of the series.

4. In a geometric series if the fifth term is 8 times the second term and the sum of the first
four terms is 60, then find the positive common ratio and the first term of the series.

5. In a G.P. of the sixth term is 16 times the second term and the sum of the first seven terms
127
is 4 then find the positive common ratio and the first term of the series.

6. Find two numbers whose A.M. is 5 and G.M. 4.

7. Find two numbers whose A.M. is 85 and G.M. 40.

8. The sum of three numbers in A.P. 45 and if 17, 11 and 3 are subtracted from them
respectively, the resulting numbers will be in G.P. Find the numbers.

9. Find the sum of the first 8 terms of a G.P. whose 5th and 8th terms are respectively 81 and 2187.

10. The ratio of the fifth term and eighth term of a G.P. is 1:8. If its seventh term is 64, find
the sum of the first 10 ten terms.

122 Vedanta Optional Mathematics Teacher's Guide ~ 10

UNIT

four Linear Programming

Estimated teaching periods : 4

1. Objectives Objectives
S.N. Level To define inequality
(i) Knowledge(K)
To draw graph of linear equation
(ii) Understanding(U)
(iii) Application(A) To define linear programming problem.
(iv) Higher Ability(HA)
To define feasible region.
To draw graph of linear inequality

To draw graph of set of inequalities.
To draw graph of system of linear inequalities to find
feasible/convex region.

To find the vertices of convex polygon.
To find maximum and minimum values of an objective
function under given conditions graphically.

2. Teaching materials

Graph papers, pencils

3. Teaching Learning Strategies:
– Review on real number lines.
– Draw graphs of the following inequalities (as given in text book).

x≥0, x≤0, x≥4, x≤4, y≥4, y≤4

– Discuss on the graph of x>4 and x≥4.
– Define linear inequalities with examples
– Discuss how to draw graph of 3x+4y ≤ 12, with origin test and shade solution region.
– Define system of linear inequalities and their solution, on the basis of graphs of

system of linear inequalities–define feasible region and convex polygon, find the
vertices of convex polygon.
– Define linear programming graphically.
– Discuss about mathematical formulation of linear programming problems. (given
in text book).

Note :
(i) The equation of boundary line corresponding to inequalities ax+by ≤ c and ax +by ≥ c.

(ii) The inequalities x ≥ 0 and y ≥ 0 represent a feasible region on the first quadrant.

(iii) x =0 represents y–axis and y=0 represent x–axis.

(iv) Each of the boundary line is drawn on the graph paper.

(v) To find the feasible plane region formed by given equatities ax + by +≤ c and ax + by 7 c,
we use origin o(0,0) as a testing point, if the boundary line does not pass through the origin.

Vedanta Optional Mathematics Teacher's Guide ~ 10 123

(vi) Find the feasible region by shading the area which is satisfied by all the constraints.
(vii) Find the coordinates of all the vertices of feasible region / convex polygon.
(viii) Find the value of the objective function at each vertex of the feasible region / convex polygon.
(ix) Identify the value of x and y of the point for which the objective function has maximum

or minimum values.
(x) Write the conclusion at which point the objective function has maximum or minimum value.

Some solved problems

1. Draw graph of 4y–5x ≥ 20,.

Solution
Given inequality is 4y–5x ≥ 20

The corresponding boundary line of given inequality is

4y–5x ≥ 20 y

when x=0, then, y=5

when y=0, then x = –4 5
Hence the boundary line passes through the
points (–4,0) and (5,0). 4y–5x≥20

Take 0(0,0) as a testing point for 4y–5x ≥ 20 x' o 5x
, we get, 0 ≥ 20 which is false. Hence the –5
solution region is the half plane not containing
the origin.

Alternatively,

The points on the boundary line

–5x +4y = 20 can be obtained as –5
y'
x + y =1
-4 5
x y
which is in the form of a + b =1

Hence the boundary line passes through the points (–4,0) and (0,5).

2. Graph the following system of inequalities and find the vertices of convex polygon of exists.
(a) x+y+2 ≥ 0 , y ≤ 2x+4, y ≤ 4–4x .

Solution
The boundary line equations of given system of inequalities are given by
x+y+2=0..........(i)
y = 2x+4..........(ii)
y = 4–4x .........(iii)
From boundary line equation (i)

x+y+2=0

when x=0, they y=–2

124 Vedanta Optional Mathematics Teacher's Guide ~ 10

when y=0, they x=–2

The boundary line (i) passes through the points (–2,0) and (0,–2). Take 0(0,0) as a testing point

for x+y+2 ≥ 0, we get, 2≥0, which is true . Hence the solution of x+y+2 ≥ 0, is the half

plane containing the origin.

From the boundary line equation (ii), we get, y
y=2x+4, 6

For x=0, y=4 and for y=0, x=–2 the boundary 4C
line (ii) passes through the points (0,4) and (–2,0).
Take 0(0,0) as a testing point for y ≤ 2x+4 we get, x' 2 24 x
0≤4 which is true. A
Hence the solution of y≤2x+4 is the half plane –4 –2 o
containing the origin.
Again from the boundary line (iii), we get, –2
y = 4–4x
–4 B

For x=0, y=4 and for y=0, x=1. The boundary y'
line (iii) passes through the points (0,4) and (1,0).
Take 0(0,0) as a testing point for y≤4–4x. we get,
0≤4 which is true.

Hence the solution of y≤4–4x is the half plane containing the origin. All the boundary lines
are drawn on the graph the feasible region is ∆ABC whose vertices A(–2,0), B(2,–2) and C(0,4)
which is shaded in the graph.

(b) x+y≥3 and 2x–y≤4

Solution y
The boundary line equations of the corresponding
inequalities are

x+y=3...........(i)

2x–y=4..........(ii) 3
–4 –2 o
From boundary line equation (i), we get, 2x–y=4
3 x+y=3 x
x + y =1. x'
3 3

The boundary line equation (i) passes through

the points (3,0) and (0,3). Take 0(0,0) as a testing

point for x+y≥3, we get 0 ≥ 3 which is false. –4
Hence solution of x+y≥ 3 is the half plane not

containing the origin.

Again from boundary equation (ii), we get, y'

x + y =1
2 -4

The boundary line (ii), passes through the points (2,0) and (0,–4). Take 0(0,0) as a testing

Vedanta Optional Mathematics Teacher's Guide ~ 10 125

point for 2x–y≤4 , we get , 0≤4 which is true.

Hence the solution of 2x–y≤4 is the half plane containing the origin. The boundary lines are
plotted on the graph and common solution region is shaded in the graph which is not closed.
The convex polygon does not exist.

3. Maximize z=2x+3y subject to the constraints 2x+y ≤ 14, x+2y ≤ 10, x ≥ 0, y ≥ 0.

Solution
The given linear constraints are
2x+y ≤ 14 ...........(i)
x+2y ≤ 10 ............(ii)
x ≥ 0, y ≥ 0 .........(iii)
The corresponding linear equations are
2x+y = 14 ...........(i)
x+2y =10 ............(ii)
x = 0.......(iii)

From equation(i),
if x=0, then y=14

if y=0, then x=7
Hence the line (i) passes through the points (0,14) and (7,0).
From equation (ii), we get, if x=0 then y=5 and

if y=0, then x=10

Hence the boundary line (ii) passes through the points (0,5) and (10,0)

For determination of the plane region if both of the linear constraints, we get, by using origin
as testing point (0,0), put x=0, y=0, 2.0 +0 ≤14

ie. 0≤14 which is true also put the point in x+2y≤10, we get, 0≤10 which is true.

Hence the solutions of both 2x+y ≤ 14 and x+2y ≤ 10 contains the origin.

Here, x=0 represents equation of y–axis x ≥ 0 has solution of the right half plane of y–axis.
All the boundary lines are plotted in the graph. the feasible region is quadrilateral OABCD
with vertices O(0,0), A(7,0), B(6,2) and C(0,5).

S.N. vertices value of z=2x+3y Remarks
Minimum
1. O(0,0) 0
maximum
2. A(7,0) 14

3. B(6,2)18 18

4. C(0,5) 15

Hence z has the maximum value 18 at B(6,2).

126 Vedanta Optional Mathematics Teacher's Guide ~ 10

y

x=0
15
C(0,14)

10
2x+y=14

C(0,5) B(6,2)
5 x+2y=10

x' y=0 x
–5 o
5 A(7,0)10

–5

y'

4. Minimize F=2x+y subject to the constraints 2x+y ≤20, 2x+3y≤24. x ≥ 0, y ≥ 0.

Solution
The given linear constraints are 2x+y ≤20, 2x+3y≤24. x ≥ 0, y ≥ 0.
The corresponding boundary line equations of above constraints are
2x+y =20...................(i)
2x+3y=24..................(ii)
x= 0....................(iii)
y = 0....................(iv)
From equation (i), if x=0, then y=20
if y=0, then x=10
The boundary line equation(i) passes through the points (0,20) and (10,0). Taking O(0,0)
for 2x+y ≤20, we get, 0≤ 20 which is true. Hence the solution of 2x+y ≤ 0 is the half
plane containing the origin.
From equation (ii), if x=0, they y=8, if y=0, then x=10.
Hence the boundary line (ii) passes through the paints.(0,8) and (10,0)
Taking O(0,0) as a testing paint for 2x+3y≤24, we get 0≤ 24, which is true. x=0 and y=0
represents y–axis and x–axis respectively. As x ≥ 0 and y ≥ 0 a feasible region lies in the
first quadrant. Hence the solution of 2x+3y ≤ 24 is the half plane containing the origin.
All the boundary lines are plotted in the graph. The feasible region of given system of
constraints is quadrilateral OABC with vertices O(0,0), A(10,0), B(9,2) and C(0,8).

Vedanta Optional Mathematics Teacher's Guide ~ 10 127

S.N. vertices value of Objectives function F=2x+y Remarks

1. O(0,0) 0 Minimum

2. A(10,0) 20 Maximum

3. B(9,2) 20 maximum

4. C(0,8) 8

Hence the given objective function F has minimum value o at origin O(0,0).

y

x=0
20 (0,20)

15

10 2x+y=20

C(0,8) 2x+3y=24
5

B(9,2)

x' y=0 x
–5 o
5 10
A(10,0)

–5
y'

5. Find the maximum and minimum values (extreme values) of F=16x–2y+40 subject to
the constraints 3x+5y ≤ 24, 0≤ x ≤ 7, 0 ≤ y ≤ 4.

Solution
Given objective function is F=16x–2y+40
Given linear constraints are
3x+5y ≤ 24
0≤ x ≤ 7 ie, x ≥ 0, x≤7
0 ≤ y ≤ 4 ie. y ≥ 0, y≤4
The corresponding boundary line equations of above consttraints are

128 Vedanta Optional Mathematics Teacher's Guide ~ 10

3x+5y = 24,...................(i)

x=7..................(ii)

y=4....................(iii)

x=0, y = 0....................(iv)

From equation (i), if y=0, then x=8

if x=–2, then y=6

Hence the boundary line equation(i) passes through the points (8,0) and (–2,6). Taking
O(0,0) for 3x+5y ≤24, we get, 0≤ 24, which is true. So, solution of 3x+5y ≤ 24 is the half
plane containing the origin.

x=7, is a line parallel to y–axis and x ≤ 7 has the solution left if x=7.

y=4 is the boundary line parallel to x–axis and y≤4 has the solution lower half plane of
y=4.

As x ≥ 0 and y ≥ 0, a feasible region lies in the first quadrant. All the boundary lines are
plotted in the graph and the feasible region is pentagon OABC with vertices O(0,0), A(7,0),
3 4
B(7, 5 ) and C( 3 ,4) and D(0,4)

S.N. vertices value of Objectives Remarks
function F=2x+3y

1. O(0,0) 40

2. A(7,0) 152 Maximum

3. B(7, 3 ) 150.8
5

4. C( 4 ,4) 53.3
3

5. D(0,4) 32 Minimum

From above table, we conclude that the maximum value is 152 at A(7,0) and minimum
value 32 at (0,4).

8
x=7

5 C( 3 , 4) y=4
5
D(0,4) 3x+5y=24

x' B(7, 3 ) x
5

–2 o 5 A(7,0) 10

–5
6(a). In the given graph the coordinates of the points O, A, B, C, D and E are respectively

Vedanta Optional Mathematics Teacher's Guide ~ 10 129

(0,0), (2,0), (0,3). (0,5), ( 20 , 45 ), (5,0) respectively. OADB is feasible region. Find all the
19 19

inequalities and minimize p=6x+10y+20

Solution
In the graph feasible region is OADB. We have to find the equations of linear constraints.
For line AC, it cuts x–axis at A(2,0) and y–axis at c(0,5). So equation of boundary line AC is

x + y =1.
2 5

or, 5x+2y ≤ 10

Let 5x+2y ≤ 10, taking O(0,0) as the testing point we get, 0 ≤ 10 which is true. So 5x+2y ≤
10 is the inequality associated with line AC. For line BE, it cuts X–axis at E (5,0) and Y–axis
at B (0,3).So equation of boundary line is given by

x + y =1
5 3

or, 5x+3y ≤ 15

Let 5x+3y ≤ 15, taking O(0,0) as the testing point we get, 0 ≤ 15 which is true. So 5x+3y
≤ 15 is the inequality associated with boundary line BE. Feasible region is quadrilateral
OADB which is the first quatrant. Hence x ≥0 and y ≥ 0 . Now, we find the minimum value
of p =6x+10y+20.

S.N. vertices value of Objectives Remarks
function F=2x+3y

1. O(0,0) 20

2. A(2,0) 32 Maximum

3. B( 20 , 45 ) 50
19 19

4. B(0,4) 50

From above table, we conclude that P has minimum value is 20 at (0,0) .

10 y

5C

BD

E

x' oA 5 10 x

y'
130 Vedanta Optional Mathematics Teacher's Guide ~ 10

6(b) text book 8b (Page 95)

Solution
In the graph ∆ABC is the feasible region is AB is parallel to the y–axis at distance of 5 units
right of the y–axis. Hence its equation is x=5. . As feasible region is left of AB,we write x ≤
5. F. For line AC, we have A(5,0) and C(0,5), equation of AC is given by

x + y =1.
5 5

or, x+y = 5

Let x+y = 5, taking testing point O(0,0), we get, 0 ≤ 5 which is true. Hence x+y = 5 is the
linear constraint associated with line AC. Again for line OB it passes through the origin, we
have O(0,0) and B (5,10). The equation of OB is given by

y=mx

It passes through the point (5,10), So we get,

10=m.5

m=2

Hence the equation of OB is y=2x.

let 2x–y ≤0, take the testing point (1,0),

we get, 2 ≤0, which is false . Hence required linear constraints is

2x–y ≥ 0

Questions for practice

1. Find the maximum value of the objective function Z=5x+8y subject to constraints
2x +5y ≤ 15, x ≥ 2,y ≥ 1. (Ans 33 at (5,1))

2. Maximum the objective function F=x+2y subject to constraints
2x +y ≤ 14, x +2y ≤ 10 and x ≥ 0, y ≥ 0. (Ans 10 at (6,2)and (0,5))

3. Minimize F=100x+600y subject to constraints (Ans 1500 at (3,2))
x +y ≤ 10, x +3y ≤ 16 , x ≤ 3, y ≥ 2.

4. Minimize the function F(x,y)=5x+2y subject to constraints

2x +y ≤ 4, x –2y ≤ 2 , x ≥ 0, y ≥ 0 (Ans 0 at (3,2))

5. Find the extreme values of the function Z=5x+7y subject to constraints
x + 2y ≤ 20, x +y ≤ 16 , x ≥ 0, y ≥ 0.

Vedanta Optional Mathematics Teacher's Guide ~ 10 131

UNIT

five Quadratic Functions and Graphs

Estimated teaching periods : 3

1. Objectives

S.N. Level Objectives

To define a quadratic function

To define a Cubic function

(i) Knowledge(K) To define a line of symmetry of a parabola.
To write vartex of parabola with equation

y=ax2+bx+c, a ≠ 0.

To define the point of intersection of a curve and a straight line.

(ii) Understanding(U) To draw graph of a straight line

To draw graph of given equation of parabola.

(iii) Application(A) To draw graph of given Cubic function.
To find solution of a parabola and a straight line graphically.

To find the equation of parabola passing through given points.

(iv) Higher To find equation of given curve of parabola.
Ability(H.A)

2. Teaching materials
i) Graph papers, functions
ii) Chart papers with graph of parabola and cubic function.

3. Teaching learning strategies.

– Review definitions of linear function, quadratic function and cubic function with
graphs.

– Discuss graph of a straight line like 4x+3y =12.
– Discuss graph of a parabola ln the form of
y=ax2, put a=±1, ±2, ±3.

– Discuss and conclude the conclusions about nature of curves. eg. Draw graph of y=x2
+4, y=4x2–8x+3, y=(x+1)3

– Review the solution of a straight line and a quadratic equation by substitution method.
– Discuss how to solve a linear equations and quadratic equation by using graph, with

an example (y=x2+2x–8, y=–5)
– Let the student give to do some problems in each class as class work and the teacher

supervise them and give necessary feedback.

Notes :
1. General equation of parabola is y=ax2+bx+c, a≠0

2. Equations of parabola vertex

i) y=ax2, a≠0 O(0,0)

ii) y=a(x–h)2+k, a≠0, (h,k)

132 Vedanta Optional Mathematics Teacher's Guide ~ 10

iii)y=ax2 +c a≠0, (0,c)
iv) y=ax2 + bx+c, a≠0,
– b , 4ac–b2
2a 4a

v)y=ax2+bx, a≠0, – b ,–4ba2
2a

3. General Equation of cubic function is y=ax3+bx2+cx+d, a≠0

Some solved problems

1. Observe the given curve and answer the y P
following questions. 10
N
(a) Name the curve (b) Find the vertex of the curve 5 M

(c) Write the coordinates of points M and N. o 4
(d) Name the line PQ which is parallel to y-axis. y' Q
y
Solutions
(a) The name of curve is parabola.

(b) The vertex or turning point of the parabola is
A(1,0)
(c) The coordinates of M and N are (2,1) and (3,4) x' –4 x
x
respectively.

(d) The line PQ which divides the curve into
two equal parts is called the line of
symmetry. It is parallel to y–axis. Its
equation in this figure is x=1.

2(a). Draw graph of the given linear –5,8
equations 0,4
3,0
4x+3y=12
o
Solution (6,–4
Here, 4x+3y=12
y'
or, 3y=12–4x

or, y= 1 (12–4x)
3
take

x 0 3 6 –3 x'
y 4 0 –4 8

The points (0,4), (3,0),(6,–4) and (–3,8) are
plotted in a graph paper and joined them,
we get a straight line.

Vedanta Optional Mathematics Teacher's Guide ~ 10 133

(b) y = 3x y

Solution,
x 0 1 2 –1 –2
y 0 3 6 –3 –6

y=3x (2,6)
(1,3)

x' o x
(–1,3)

3. Find the graphs for the quadratic equations (–2,6) (3, 9)
y'
(a) y=x2
y
Solution
Here, y=x2 represents a parabola (–3, 9)
take
(–2, 4) y=x2
x 0 ±1 ±2 ±3
y0149 x' (1, 1) x
The points from above table are plotted in a graph o x
paper and joined them, we get a curve called
parabola. y'

(b) y=–x2 x' y
O (0,0)
Solution
Here, y=–x2 it represents a parabola with vertex (1, –1)
at the origin.

x 0 ±1 ±2 ±3
y 0 –1 –4 –9

The points from above table are plotted in a (–2,–4) y=–x2
graph paper and joined them, we get a curved as
shown in the graph along side.

y'

134 Vedanta Optional Mathematics Teacher's Guide ~ 10

(c) y=x2+4 (–3,13) y (3,13)

Solution (–2,8) (2,8)
Here, y=x2+4, comparing it with y=ax2+bx=c,we (–1,5) (1,5)
get, a=1, b=0, c=4 (0,4)

Vertex= - b , 4ac-b2
2a 4a

= - 0 , 4.1.4-02
2.1 4.1

=(0,4)

x 0 ±1 ±2 ±3 x' o x
y 4 5 8 13
The points from above table are plotted in a graph
paper and joined them, we get a curved graph
along side.

y'

(d) y = x2–4 y

Solution 0 ±2 ±3 ±4
Here, y=x2–4 –4 0 5 12
x ±1
y –3

(–2,8) y=x2–4
(–1,5)

(–2,0) (2,0)

x' o x

(0,–4)
y'
Vedanta Optional Mathematics Teacher's Guide ~ 10 135

(e) y=(x+1)2 y

Solution (–3,4) (1,4)
Here, y=(x+1)2 . It is in the form of y=a(x–h)2 +k
vertex =(h,k)=(–1,0), where , a=1, h=–1, k=0
Table

x 0 1 2 –1 –2 –3 –4 x' (–1,0) O x
y1490149 (–4,8) (4,8)
The points from above table are plotted in a graph y'
paper and joined them, we get a curved as shown y
along side.

(f) y= 1 x2
2

Solution 1
2
Here, y= x2 . It is in the form of y=ax2 , where ,

a= y= 1 , vertex =(0,0) y= 1 x2
2 2
1
y= 2 x2 (–2,2) (2,2)
x' o
Table

x 0 ±1 ±2 ±4 x

y 0 1 2 8
2
y'
The points from above table are plotted in a graph y
paper and joined them, we get a curved as
shown along side. (2,8)

4. Draw the graphs for the following equations

(a) y=x3 x' (–1,–1) (1,1) x
O
Solution
Here, y=x3. This is a equation of cubic
function
Table

x –2 –1 0 1 2
y –8 –1 0 1 –8

The points from the above table are plotted
on a graph and joined them, we get a curve as
shown alongside.

(–2,–8)

y'

136 Vedanta Optional Mathematics Teacher's Guide ~ 10

(b) y=–x3 y
Solution
Here, y=–x3............(i). It is an equation of x' y=x3 x
cubic function O
Table
x –2 –1 0 1 2
y 8 1 0 –1 –8

The points from above table are plotted
in a graph and joined them, we get a
curve as shown the graph alongside.

(c) y=(x+1)3 2 y'
8
Solution y
Here, y=(x+1)3...............(i) (1,8)
Table

x –3 –2 –1 0
y –8 –1 0 1

The points from the above table are
plotted ia graph and joined them, we get a
curve as shown in the graph alongside.

x' (–1,0) (0,1) x
(–2,–1) O

(–3,–8)

y'

Vedanta Optional Mathematics Teacher's Guide ~ 10 137

(d) y=–(x+1)3 y

Solution 1 (–3,8) 8
Here, y=–(x+1)3.=..............(i) –8 6
Table
4
x –3 –2 –1 0
y 8 1 0 –1

The points from the above table are x' 2 4 6 x
plotted ia graph and joined them, we get (–2,1) (–1,0)
a curve as shown in the graph alongside.
–4 –(02,–1) 0 2
–2

–4

–6

–8 (1,–8)
y'

5.Draw the graphs for the following quadratic functions

(a) y=4x2 – 8x+3

Solution (–1,15) y (3,15)
This equation is in the form of y=bx2 y=4x2–8x+3
–8b+c, where, a=4, b=8 and c=3.

Vertex= - b , 4ac-b2
2a 4a

=- -8 , 4.1.4-(-8)2
2.4 4.1

=(1,–1) (0,3) (2,3)
Table

x –3 –2 –1 0 1 2 3
y 69 35 15 3 –1 3 15

Plotting the above points in the graph x' O (1,–1) x
paper and joining them, we get a curved
curve as shown alongside.

y'
138 Vedanta Optional Mathematics Teacher's Guide ~ 10

(b) y=x2 +2x–8 y
(3,7)
Solution (–5,7)
Here,

y=x2 +2x–8..............(i). It is in the form y=x2+2x–8
of y=bx2 +8b+c , where, a=1, b=2 and
c=–8.

Vertex= – b , 4ac-b2 (–4,0) (2,0)
2a 4a O

= – 2 , 4.1.(–8)–22 x' x
2.1 4.1

=(–1,–9) (–3,–5) (1,–5)

x 1 2 3 0 –1 4 –2 –3 –4 –5 (–2,–8) (0,–8)
y –5 0 7 –8 –9 16 –8 –5 0 7
(–1,–9)
Plotting the points from above table and
joining them, we get a curved curve as y'
shown alongside.

(c) y=–x2 –2x+5

Solution y (3,7)
The equation is, y=–x2 –2x+5..............(i). (–1,6)
It is in the form of y=bx2 +bx+c , where,
a=–1, b=–2 and c=5. (–2,5) (0,5)

Vertex= - b , 4ac-b2
2a 4a
y=x2+2x–8
= -2 , 4(-1)(5)-(-2)2 (–3,2) (1,2)
2.1 4×(-1)

=(–1,6) x' O x
Table (–4,–3) (2,–3)
0 123
x –5 –4 –3 –2 –1 5 2 –3 –10
y –10 –3 2 5 6

Plotting the points from above table and
joining them, we get a curved curve as
shown alongside.

(–5,–10) (3,–10)
y'

Vedanta Optional Mathematics Teacher's Guide ~ 10 139

(d) y=4x2 +8x+5 (–3,17) y (1,17)
15
Solution
The eqn. is y=4x2 +8x+5..............(i). It is
in the form of y=ax2 –bx+c , where, a=4,
b=8 and c=5.

Vertex= - b , 4ac-b2 y=4x2+8x+5
2a 4a 10

= - -8 , 4×4×5-(8)2
2.4 4×(-1)

=(–1,1) (–2,5) 5
Table (0,5)

x –3 –2 –1 0 1
y 17 5 1 5 17

Plotting the points from above table and (–1,1) x
joining them, we get a curve as shown x' O
alongside.
y'

(e) y=–(x+3)2

Solution y
The eqn is y=–(x+3)2..............(i).

or, y=–[x2+6x+9]

or, y= –x2–6x–9..............(ii) x' (–3,0) O x

To find vertex of parabola, (–4,–1) (0,–1)

Vertex= – b , 4ac–b2 y=(x+3)2
2a 4a

6 4×9–(–6)2 (–5,–4) (–1,–4)
–2 –4
= ,

=(–3,0) (–6,–9) (0,–9)
eqn. (i) in table; y'

x –6 –5 –4 –3 –2 –1 0
y –9 –4 –1 0 –1 –4 –9

140 Vedanta Optional Mathematics Teacher's Guide ~ 10

6. Find the equation of parabola under the given conditions
(a). Vertex at O(0,0) and passing through (2,3).

Solution
The eqn. of the parabola whose vertex is at origin is given by:–

(i) y=ax2

The line/parabola passes through (2,3)

3=a.(2)2

or, 3 =a
4

a= 3
4

The eqn. is, y=ax2

put the value of a in the eqn.

y= 3 .x2
4
3x2
y= 4 is the required equation.

6.(b) Vertex =(2,–1), passing through (1,0).

Solution
Let, the eqn. of the parabola be,
y=a(x–h)2 +k............(i)
vertex=(h,k)=(2,–1)

the eqn. is
y=a(x–2)2 –1............(ii)
The parabola passes through (1,0)

Put value of x and y such that x=1.y=0,
0=a(1–2)2 –1
or. 0=a(–1)2 –1
or, 1=a

Put the value of a in (ii)
y=1(x–2)2 –1
or, y=x2 –4x+4–1
or, y=x2 –4x+3

y=x2 –4x+3, is the required eqn.

6.(c) Vertex =(–1,–4), passing through (2,5).

Solution
let, vertex =(–1,–4)=(h,k).
Let, the eqn. of the parabola be,

Vedanta Optional Mathematics Teacher's Guide ~ 10 141

y=a(x–h)2 +k............(i)
Put value of hand k in equation (i), we get
y=a(x+1)2 +(–4)
or, y=a(x+1)2 –4............(ii)
(2,5) lies in the parabola.
when x=2, y=5 in (ii)

Put value of x and y such that x=1.y=0,
5=a(x+1)2 –4
or. 9=a(2+1)2
or, 9=a×9

a=1
Put the value of a in (ii)
y=1(x+1)2 –4
or, y=x2 2x+1–4
or, y=x2 +2x33

y=x2 +2x–3, is the required eqn.

6.(d) passes through the points (2,4),(–1,–2) and (3,10).

Solution
let, the eqn. of parabola be :
y=ax2 +bx+c............(i)
(2,4) lies in eqn. (i)
when x=2, y=4 in (i);
4=a.(2)2 +b(2)+c
or, 4=a×4 +2b+c

4=4a +2b+c............(ii)
(–1,–2) lies in eqn. (i)
when x=–1, y=–2 in (i);
–2=(2)2 a +b(–1)+c
–2=a – b+c......(iii)
(3,10) lies in eqn. (i)
ie, when x=3, y=10 in (i);
10=a(3)2 +b(3)+c

10=9a +3b+c......(iv)
Subtract (ii) from (iii)

-2 = a - b + c
4 = 4a + 2b +c
- -- -
-6 = -4a+a-2b-b

142 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, –6=–3a–3b
or, a+b=2.............(v)
Subtract (iv) from (iii)

–2 = a – b + c
–10 =–9a +– 3b +– c
–12 =–9a+a – 3b–b

or, –12=–8a–4b
or, 3=2a+b
2a+b=3 ............. (vi)
Subtract (vi) from (v);

a+b=2
–2a +– b =–3
–a = –1

a=1
Put the value of a in (iv);
3=2(1)+b

b=1
Put the value of a and b in (iii);
–2 = 1–1+c

c=–2
Put the value of a, b and c in (i)
y=1×x2 +1×x–2
or, y=x2+x–2
which is the required equation.

6(e). passes through the points (1,4),(2,1) and (4,1).

Solution
let, the parabola represented by (a) (i)
y=ax2 +bx+c............(a)
(1,4) lies in eqn. (i)
when x=1, y=4 in (i);
4=a.(1)2 +b(1)+c
or, 4=1.a +1.b+c

4=a +b+c............(i)
(2,1) lies in eqn. (a)
when x=2, y=1
or, 1=a.(2)2 +2.b+c
or, 1=4a +2b+c......(ii)
Subtract (ii) from (i)

Vedanta Optional Mathematics Teacher's Guide ~ 10 143

4=a+b+c
–1 =–4a +– 2b –+c
3 = -3a -b

3a+b=–3.....................(iv)
(4,1)0) lies in eqn. (i)
ie, when x=4, y=1;
1=a.(4)2 +4.b+c

1=16a +4b+c......(iv)
Subtract (iv) from (iii)

1 = 4a + 2b + c
–1 =–16a +– 4b +– c
0 = -12a - 2b

or, 6a + b =0
or, 3=2a+b.............(v)
Subtract (iii) from (v);

6a + b = 0
–3a +– b =+–3
3a = 3

a=1
Put the value of a in (v);
6 × 1 +b=0

b=–6
Put the value of a , b and c in (i);
a + b+ c =4
or, 1–6+c=4
or, –5+c =4

c= 9
Put the value of a, b and c in (a)
y=1×x2 – 6×x +9
or, y=x2– 6x + 2

y=x2– 6x + 2 which is the required equation.

7. (a) From the given graphs of parabola, find their equations.

Solution
From the graph, it can be cleared that
(0,0) is the vertex of the parabola (1,1),(2,4) and (3,9) lies in the parabola.
Since the vertex is origin,
the eqn. is given by,

144 Vedanta Optional Mathematics Teacher's Guide ~ 10

y=ax2............(i)
(1,1) lies in the parabola,
when y=1, x=1,
1=a×1

a=1
The value of a also satisfies the condition
For (2,4) and (3,9) according to the equation.

The eqn. of the parabola
y=1×x2
i.e., y=x2

(b)
Solution
here, vertex =(–1,)
let, (–1,–4) = (h,k)
Also, (–2,–3),(0,–3), (–3,0), (1,0) and (–4,5) lies in the parabola.
let, the eqn. of the parabola be
y=a(x–h)2 + k...............(i)
where,
(h,k)=vertex
Then,
When h=–1, k=–4 in (i),
y=a(x+1)2 + k...............(ii)
(–2,–3) lies in the parabola,
when x=–2, y=–3 in (ii);
–3=a(–2+1)2 –4
or, 1 = a×1

a=1
put the value of a , b and k in (i);
y=1(x+1)2 –4
or, y=x2 +2x+1 – 4
or, y=x2 +2x – 3 which is the required eqn.

Vedanta Optional Mathematics Teacher's Guide ~ 10 145

Solution of linear and Quadratic Equations

Some solved problems

1. Solve the following equations by substitution method.
(a) y=x2 – 3x and y=x
Solution
Given equation are
y=x2 – 3x ............(i)
and y=x..............(ii)
put the value of y in equation (i), from (ii), we get,
x=x2 – 3x
or, x2 – 4x =0
or, x(x–4)=0
Either x=0
or, x=4

x=0,4
when x=0, then y=0
when x=4, then y=4
The required solutions are (0,0) and (4,4).

(b) y=x2 +7x+12. x–y+7 =0

Solution
Given equation are
y=x2 +7x+12 ............(i)
or, . x–y+7 =0..............(ii)
From equation (ii) y=x+7
put the value of y in equation (i), from (ii), we get,
x +7=x2 +7x+12
or, x2 +6x +5 =0
or, x2 +5x +x +5 =0
or, x(x+5) +1(x+5)=0
or, (x+5)(x+1)=0
Either x + 5 = 0, x=–5
or, x + 1=0 x=–1

x=0,4
when x=–5, then y=–5 + 7 =2
when x=–1, then y=–1 +7 =6
Hence the required solution (–5,2) and (–1,6).

146 Vedanta Optional Mathematics Teacher's Guide ~ 10

2. Solve the folowing equations graphically y
(a) x2 + 6x + 8 =0

Solution 15
Let y=x2+6x+8=0, i.e. y=0

y=x2+6x+8=0 represents an equation of
parabola, comparing at with

y=ax2+bx+c, we get, a=1, b=6 and c=8

First let us find the vertex of the parabola. 10 +6x+8
b y=x2
x–coordinate of vertex (h)=– 2a
6
=– 2.1

=–3 4ac -b2 5
y–coordinate of vertex (k) = 4a

= 4×1×8 -62
4.1

= 32 -36
4
=–1
x' –5 O5 x
Vertex (h,k)=(–3,–1) x

Table to draw curve of parabola 1 –3
x –5 –4 –3 –2 –1 0 y'

y 0 –1 0 3 8 15 y
From the above curve we observed that the (–4,16) 16
curve cuts the x–axis at points (–2,0) and
(–4,0).

x = –2,–4 14
Alternate Method 12
Here, x2 + 6x +8 =0 or, x2=–6x=8

Let y=x2 = –6x–8 10

Then, y=x2................(i) and y=– 8
6x–8.............(ii)
(–2,4) 6
The equation(i) represents a parabola and 4
(ii) represents a straight line. From equation
(i)

x ±4 ±3 ±2 ±1 0
y 16 9 4 1 0

plotting the above points on graph paper x' –6 –4 –2 2
and joining them, we get a curve of O2 4 6
parabola.
From equation (ii) y=–6x–8

x –1 –2 –3 y'
y –2 4 10

plotting there points on the same graph, we get a straight line. The straight line cuts

Vedanta Optional Mathematics Teacher's Guide ~ 10 147

the curve of parabola at points (–2,4) and (–4,16). Hence the required solution of given
equation area =–2,–4

(b) x2 – x – 2 =0

Solution
Let x2 – x – 2 =0,

i.e. y= x2 – x – 2, y=0

It represents an equation of parabola, in the term of y=ax2+bx+c, where, a=1, b=–1,c=–2

Let us find the vertex of the parabola.

Vertex (h,k)= - b , 4ac-b2
2a 4a

= - (-1) , 4.1.(-2)-(-1)2
2.1 4.1.(-2)

= 1 ,- 9
2 4

To draw a curve of parabola

x –2 –1 0 1 2 3
y 4 0 –2 –2 0 4

The points from the above table are plotted in a graph and joined them, we get a curve of
parabola.

6
5

4

3

2 y=x2 –x+2
1

x' x
–5 –4 –3 –2 –1 o
1 23 4 5

–1

–2

–3 1 ,– 9
2 4

The curve cuts x–axis at points (–1,0) and (2,0). Hence the required solutions are x=–1,2.

148 Vedanta Optional Mathematics Teacher's Guide ~ 10

3. Solve the following equations graphically.
(a) y= x2 – 5 and y=2x+3

Solution
Given equations are
y= x2 – 5 ......................(i)
and y=2x+3 ................(ii)
Equation (i) represents a parabola and (ii) represents a straight line. The point of the
intersection of the curve of parabola and straight line is the solution of given equations.

From equation (i),

Vertex (h,k)= - b , 4ac-b2
2a 4a

= - 0 , 4.1.(-5)-0
2.1 4.1

=(0,–5)

To draw the curve of parabola

x –3 –2 –1 0 1 2 3
y 4 –1 –4 –5 –4 –1 4

Plotting the points on graph and joining them we get a curve of parabola.

From equation (i) y=2x + 3

x –3 0 1
y33 5

4,11
10

5

y=x2 –5

x' x
–5 –1,–1 o
5

–5

Vedanta Optional Mathematics Teacher's Guide ~ 10 149

From the graph, we observed that the point of intersection of the parabola and the straight
lines are (–2,–1) and (4,11).
Hence the required solutions are (–2,–1) and (4,11).

(g) y=4x2+8x +5 ,x + y=3

Solution
Given equations are
4x2+8x +5 ......................(i)
and x + y=3................(ii)
In above equation, equation (i) represents a parabola and equation (ii), represents a straight
line.
From equation (i), y=4x2+8x +5

x –3 –2 –1 0 1

y 17 5 1 5 17

Vertex of parabola (h,k)

= – b , 4ac–b2 y
2a 4a

= – 8 , 4.4.5–82 20
2.4 4.4

=(–1,1) 15
Then plotting above points
from table and joining them,
we get a curve of parabola.
From equation(ii), y =3–x

x 0 –1 3 10

y340

plotting these points in same (–2,5) 5
graph, we get a straight line.
– 1 , 13
From the graph, the curve and 4 4

the straight lines intersects at x' x

points (–2,5) and – 1 , 13 . –5 o 5
4 4

Note. If the point of –5
intersection of a curve and a
straight line is not clear to read
their coordinates, we can solve
them by substitution method
in rough.

y'

150 Vedanta Optional Mathematics Teacher's Guide ~ 10