Selina ICSE Class 10 Maths Solutions Chapter 5 Quadratic Equations

Selina ICSE Solutions

Question 1. Find which of the following equations are quadratic:
(𝑖) (3π‘₯– 1)2 = 5(π‘₯ + 8)
(𝑖𝑖) 5π‘₯2 – 8π‘₯ = βˆ’3(7– 2π‘₯)
(𝑖𝑖𝑖) (π‘₯– 4)(3π‘₯ + 1) = (3π‘₯– 1)(π‘₯ + 2)
(𝑖𝑣) π‘₯2 + 5π‘₯– 5 = (π‘₯– 3)2
(𝑣) 7π‘₯3– 2π‘₯2+ 10 = (2π‘₯– 5)2
(𝑣𝑖) (π‘₯ βˆ’ 1)2 + (π‘₯ + 2)2 + 3(π‘₯ + 1) = 0
Solution:

(𝑖) (3π‘₯– 1)2 = 5(π‘₯ + 8)
We know that,
(π‘Ž βˆ’ 𝑏)2 = π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2
(3π‘₯)2 βˆ’ 2(3π‘₯)(1) + (1)2 = 5π‘₯ + 40
9π‘₯2 – 6π‘₯ + 1 = 5π‘₯ + 40
9π‘₯2 – 6π‘₯ βˆ’ 5π‘₯ = 40 βˆ’ 1
9π‘₯2 – 6π‘₯ βˆ’ 5π‘₯ = 39
9π‘₯2 – 11π‘₯ – 39 = 0
Above equation is the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0
Hence, the given equation is a quadratic equation.
(𝑖𝑖) 5π‘₯2 – 8π‘₯ = βˆ’3(7– 2π‘₯)
5π‘₯2 – 8π‘₯ = βˆ’21 + 6π‘₯
5π‘₯2 – 8π‘₯ βˆ’ 6π‘₯ = βˆ’21
5π‘₯2 – 14π‘₯ + 21 = 0
5π‘₯2– 14π‘₯ + 21 = 0
Above equation is the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0
Hence, the given equation is a quadratic equation.
(𝑖𝑖𝑖) (π‘₯ – 4)(3π‘₯ + 1) = (3π‘₯– 1)(π‘₯ + 2)
(π‘₯ – 4)(3π‘₯ + 1) = (3π‘₯– 1)(π‘₯ + 2)
3π‘₯2 + π‘₯ βˆ’ 12π‘₯ βˆ’ 4 = 3π‘₯2 + 6π‘₯ βˆ’ π‘₯ βˆ’ 2
3π‘₯2 βˆ’ 11π‘₯– 4 = 3π‘₯2 + 5π‘₯ – 2
βˆ’11π‘₯– 5π‘₯ = – 2 + 4
βˆ’16π‘₯ = 2
βˆ’16π‘₯ βˆ’ 2 = 0
16π‘₯ + 2 = 0
Above equation is not in the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0
Hence, the given equation is not a quadratic equation.
(𝑖𝑣) π‘₯2 + 5π‘₯– 5 = (π‘₯– 3)2
We know that,
(π‘Ž βˆ’ 𝑏)2 = π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2
π‘₯2 + 5π‘₯– 5 = π‘₯2 βˆ’ 2(π‘₯)(3) + 32
π‘₯2 + 5π‘₯– 5 = π‘₯2 βˆ’ 6π‘₯ + 9
5π‘₯ + 6π‘₯ = 9 + 5
11π‘₯ = 14
11π‘₯– 14 = 0
Above equation is not in the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0
Hence, the given equation is not a quadratic equation.
(𝑣) 7π‘₯3– 2π‘₯2 + 10 = (2π‘₯– 5)2
We know that,
(π‘Ž βˆ’ 𝑏)2 = π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2
7π‘₯3 – 2π‘₯2 + 10 = (2π‘₯)2 βˆ’ 2(2π‘₯)(5) + (5)2
7π‘₯3 – 2π‘₯2 + 10 = 4π‘₯2 βˆ’ 20π‘₯ + 25
7π‘₯3 – 2π‘₯2 βˆ’ 4π‘₯2 + 10 + 20π‘₯ βˆ’ 25 = 0
7π‘₯3– 6π‘₯2 + 20π‘₯– 15 = 0
Above equation is not in the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0
Hence, the given equation is not a quadratic equation.
(𝑣𝑖) (π‘₯ βˆ’ 1)2 + (π‘₯ + 2)2 + 3(π‘₯ + 1) = 0
We know that,
(π‘Ž βˆ’ 𝑏)2 = π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2
(π‘Ž + 𝑏)2 = π‘Ž2 + 2π‘Žπ‘ + 𝑏2
π‘₯2 βˆ’ 2(π‘₯)(1) + (1)2 + π‘₯2 + 2(π‘₯)(2) + 22 + 3π‘₯ + 3 = 0
π‘₯2 βˆ’ 2π‘₯ + 1 + π‘₯2 + 4π‘₯ + 4 + 3π‘₯ + 3 = 0
2π‘₯2 + 5π‘₯ + 8 = 0
Above equation is the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0
Hence, the given equation is a quadratic equation.

Question 2. (i) Is x = 5 a solution of the quadratic equation π‘₯2– 2π‘₯– 15 = 0?
Solution:

It is given that,
π‘₯2 βˆ’ 2π‘₯ – 15 = 0
π‘₯ = 5
Put the value of x in given equation
L.H.S = (5)2 – 2(5) – 15
25 – 10 – 15 = 0
L.H.S = R.H.S
Thus, π‘₯ = 5 is a solution of the quadratic equation π‘₯2 βˆ’ 2π‘₯ – 15 = 0

Question 2. (ii) Is x = -3 a solution of the quadratic equation 2π‘₯2 – 7π‘₯ + 9 = 0?
Solution:

It is given that,
2π‘₯2 – 7π‘₯ + 9 = 0
π‘₯ = βˆ’3
Put the value of x in given equation
L.H.S = 2(βˆ’3)2 – 7(βˆ’3) + 9
2(9) + 21 + 9
18 + 21 + 9 = 48
L.H.S. β‰  R.H.S
Thus, π‘₯ = βˆ’3 is not a solution of the quadratic equation 2π‘₯2 – 7π‘₯ + 9 = 0

Question 3. If √(2/3) is a solution of equation 3π‘₯2 + mx + 2 = 0, find the value of m.
Solution:

It is given that,
3π‘₯2 + mx + 2 = 0
x = √(2/3)
Put the value of x in given equation

Question 4. 2/3 and 1 are the solutions of equation π‘šπ‘₯2 + 𝑛π‘₯ + 6 = 0. Find the values of m and n.
Solution:

It is given that,
mx2 + nx + 6 = 0
Put the value of π‘₯ = 2/3 in the given equation.

4π‘š + 6𝑛 + 54 = 0……………………(i)
Put the value of π‘₯ = 1 in the given equation.
π‘š(1)2 + 𝑛(1) + 6 = 0
π‘š + 𝑛 + 6 = 0
π‘š + 𝑛 + 6 = 0……………………(ii)
Get the value of π‘š from equation (ii)
π‘š = βˆ’6 βˆ’ 𝑛……………(iii)
Put the value of π‘š in equation (i)
4(βˆ’6 βˆ’ 𝑛) + 6𝑛 + 54 = 0
βˆ’24 βˆ’ 4𝑛 + 6𝑛 + 54 = 0
2𝑛 + 30 = 0
2𝑛 = βˆ’30
𝑛 = βˆ’30/2
𝑛 = βˆ’15
Put the value of 𝑛 in equation (iii)
π‘š = βˆ’6 βˆ’ (βˆ’15)
π‘š = βˆ’6 + 15
π‘š = 9
Hence, the value of 𝑛 is -15 and π‘š is 9.

Question 5. If 3 and -3 are the solutions of equation π‘Žπ‘₯2 + 𝑏π‘₯– 9 = 0. Find the values of π‘Ž and 𝑏.
Solution:

It is given that,
π‘Žπ‘₯2 + 𝑏π‘₯– 9 = 0
Put the value of π‘₯ = 3 in the given equation.
π‘Ž(3)2 + 𝑏(3)– 9 = 0
9π‘Ž + 3𝑏– 9 = 0
3(3π‘Ž + 𝑏– 3) = 0
3π‘Ž + 𝑏– 3 = 0………………..(i)
Put the value of π‘₯ = βˆ’3 in the given equation.
π‘Ž(βˆ’3)2 + 𝑏(βˆ’3)– 9 = 0
9π‘Ž βˆ’ 3𝑏– 9 = 0
3(3π‘Ž βˆ’ 𝑏– 3) = 0
3π‘Ž βˆ’ 𝑏– 3 = 0………………..(ii)
Get the value of π‘Ž from equation (ii)
π‘Ž = 𝑏+3/3 ……………(iii)
Put the value of π‘Ž in equation (i)
3 (𝑏+3/3 ) + 𝑏– 3 = 0
𝑏 + 3 + 𝑏– 3 = 0
2𝑏 = 0
𝑏 = 0
Put the value of 𝑏 in equation (iii)
π‘Ž = 0+3/3
π‘Ž = 3/3
π‘Ž = 1
Hence, the value of π‘Ž is 1 and 𝑏 is 0.

Exercise 5B

Question 1. Without solving, comment upon the nature of roots of each of the following equations:
(𝑖) 7π‘₯2– 9π‘₯ + 2 = 0
(𝑖𝑖) 6π‘₯2– 13π‘₯ + 4 = 0
(𝑖𝑖𝑖) 25π‘₯2– 10π‘₯ + 1 = 0
(𝑖𝑣) π‘₯2 + 2√3π‘₯ – 9 = 0
(𝑣) π‘₯2– π‘Žπ‘₯– 𝑏2 = 0
(𝑣𝑖) 2π‘₯2 + 8π‘₯ + 9 = 0
Solution:

(i) It is given that,
7π‘₯2– 9π‘₯ + 2 = 0
π‘Ž = 7
𝑏 = 9
𝑐 = 2
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
D = (9)2 βˆ’ 4(7)(2)
D = 81 βˆ’ 56
D = 25
Here 𝐷 > 0 then given equation have two real and unequal roots.
(ii) It is given that,
6π‘₯2– 13π‘₯ + 4 = 0
π‘Ž = 6
𝑏 = βˆ’13
𝑐 = 4
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
D = (βˆ’13)2 βˆ’ 4(6)(4)
D = 169 βˆ’ 96
D = 73
Here 𝐷 > 0 then given equation have two real and unequal roots.
(iii) It is given that,
25π‘₯2– 10π‘₯ + 1 = 0
π‘Ž = 25
𝑏 = βˆ’10
𝑐 = 1
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
D = (βˆ’10)2 βˆ’ 4(25)(1)
D = 100 βˆ’ 100
D = 0
Here 𝐷 = 0 then given equation have two real and unequal roots.
(iv) It is given that,
π‘₯2– 2√3π‘₯ βˆ’ 9 = 0
π‘Ž = 1
𝑏 = βˆ’2√3
𝑐 = βˆ’9
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
D = (βˆ’2√3)2 βˆ’ 4(1)(βˆ’9)
D = 12 + 36
D = 48
Here 𝐷 > 0 then given equation have two real and unequal roots.
(v) It is given that,
π‘₯2– π‘Žπ‘₯ βˆ’ 𝑏2 = 0
π‘Ž = 1
𝑏 = βˆ’π‘Ž
𝑐 = βˆ’π‘2
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
D = (βˆ’π‘Ž)2 βˆ’ 4(1)(βˆ’π‘2)
D = π‘Ž2 + 4𝑏2
Here 𝐷 > 0 then given equation have two real and unequal roots.
(vi) It is given that,
2π‘₯2– 8π‘₯ βˆ’ 9 = 0
π‘Ž = 2
𝑏 = 8
𝑐 = 9
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
D = (8)2 βˆ’ 4(2)(9)
D = π‘Ž2 + 4𝑏2
Here 𝐷 > 0 then given equation have two real and unequal roots.

Question 2. Find the value of p, if the following quadratic equation has equal roots: 4π‘₯2– (𝑝– 2)π‘₯ + 1 = 0
Solution:

It is given that,
Quadratic equation has equal roots. So, D = 0
4π‘₯2– (𝑝– 2)π‘₯ + 1 = 0
π‘Ž = 4
𝑏 = βˆ’(𝑝 βˆ’ 2)
𝑐 = 1
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
0 = [βˆ’(𝑝 βˆ’ 2)]2 βˆ’ 4(4)(1)
0 = 𝑝2 βˆ’ 2(𝑝)(2) + 4 βˆ’ 16
0 = 𝑝2 βˆ’ 4𝑝 + 4 βˆ’ 16
0 = 𝑝2 βˆ’ 4𝑝 βˆ’ 12
By splitting the middle term
𝑝2 βˆ’ 6𝑝 + 2𝑝 βˆ’ 12 = 0
𝑝(𝑝 βˆ’ 6) + 2(𝑝 βˆ’ 6) = 0
(𝑝 + 2)(𝑝 βˆ’ 6) = 0
Then
𝑝 + 2 = 0             π‘ βˆ’ 6 = 0
𝑝 = βˆ’2                π‘ = 6

Question 3. Find the value of β€˜p’, if the following quadratic equations have equal roots: π‘₯2 + (𝑝 – 3)π‘₯ + 𝑝 = 0
Solution:
It is given that,
Quadratic equation has equal roots. So, D = 0
π‘₯2 + (𝑝 – 3)π‘₯ + 𝑝 = 0
π‘Ž = 1
𝑏 = (𝑝– 3)
𝑐 = 𝑝
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
0 = 𝑏2 βˆ’ 4π‘Žπ‘
(𝑝 βˆ’ 3)2 βˆ’ 4(1)(𝑝) = 0
(𝑝 βˆ’ 3)2 βˆ’ 4𝑝 = 0
𝑝2 + 9– 6𝑝– 4𝑝 = 0
𝑝2– 10𝑝 + 9 = 0
𝑝2 βˆ’ 9𝑝– 𝑝 + 9 = 0
𝑝(𝑝– 9) – 1(𝑝– 9) = 0
(𝑝 βˆ’ 9)(𝑝– 1) = 0
Then
𝑝 βˆ’ 9 = 0             π‘ βˆ’ 1 = 0
𝑝 = 9                   π‘ = 1

Question 4. The equation 3π‘₯2– 12π‘₯ + (𝑛– 5) = 0 has equal roots. Find the value of n.
Solution:

It is given that,
Quadratic equation has equal roots. So, D = 0
3π‘₯2– 12π‘₯ + (𝑛– 5) = 0
π‘Ž = 3
𝑏 = βˆ’12
𝑐 = (𝑛 βˆ’ 5)
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
0 = 𝑏2 βˆ’ 4π‘Žπ‘
(βˆ’12)2 βˆ’ 4(3)(𝑛 βˆ’ 5) = 0
144 βˆ’ 12(𝑛 βˆ’ 5) = 0
144 βˆ’ 12𝑛 + 60 = 0
144 βˆ’ 12𝑛 + 60 = 0
βˆ’12𝑛 + 204 = 0
βˆ’12𝑛 = βˆ’204
12𝑛 = 204
𝑛 = 204/12
𝑛 = 17
Hence, the value of 𝑛 is 17.

Question 5. Find the value of m, if the following equation has equal roots: (π‘šβ€“ 2)π‘₯2– (5 + π‘š)π‘₯ + 16 = 0.
Solution:

It is given that,
Quadratic equation has equal roots. So, D = 0
(π‘šβ€“ 2)π‘₯2– (5 + π‘š)π‘₯ + 16 = 0
π‘Ž = (π‘š βˆ’ 2)
𝑏 = βˆ’(5 + π‘š)
𝑐 = 16
We know that by discriminate,
D = 𝑏2 βˆ’ 4π‘Žπ‘
0 = 𝑏2 βˆ’ 4π‘Žπ‘
[βˆ’(5 + π‘š)]2 βˆ’ 4(π‘š βˆ’ 2)(16) = 0
(5)2 + 2(5)(π‘š) + π‘š2 βˆ’ 64π‘š + 128 = 0
25 + 10π‘š + π‘š2 βˆ’ 64π‘š + 128 = 0
βˆ’54π‘š + π‘š2 + 153 = 0
π‘š2 βˆ’ 54π‘š + 153 = 0
π‘š2 βˆ’ 51π‘š βˆ’ 3π‘š + 153 = 0
π‘š(π‘š βˆ’ 51) βˆ’ 3(π‘š βˆ’ 51) = 0
(π‘š βˆ’ 51)(π‘š βˆ’ 3) = 0
Then
π‘š βˆ’ 51 = 0          π‘š βˆ’ 3 = 0
π‘š = 51                 π‘š = 3

Question 6. Find the value of p for which the equation 3π‘₯2– 6π‘₯ + π‘˜ = 0 has distinct and real roots.
Solution:

It is given that,
Quadratic equation has real roots. So, 𝐷 > 0
3π‘₯2– 6π‘₯ + π‘˜ = 0
π‘Ž = 3
𝑏 = 6
𝑐 = π‘˜
We know that by discriminate,
D < 𝑏2 βˆ’ 4π‘Žπ‘
0 < 𝑏2 βˆ’ 4π‘Žπ‘
0 < (6)2 βˆ’ 4(3)(π‘˜)
0 < 36 βˆ’ 12π‘˜
12π‘˜ < 36
π‘˜ < 36/12
π‘˜ < 3
Hence, the value of K is 3.

Exercise 5C

Question 1. Solve: π‘₯Β² – 10π‘₯– 24 = 0
Solution:

It is given that,
π‘₯Β² – 10π‘₯– 24 = 0
By the splitting the middle term,
π‘₯Β² – 12π‘₯ + 2π‘₯– 24 = 0
π‘₯(π‘₯ βˆ’ 12) + 2(π‘₯ βˆ’ 12) = 0
(π‘₯ + 2)(π‘₯ βˆ’ 12) = 0
Then
π‘₯ + 2 = 0            π‘₯ βˆ’ 12 = 0
π‘š = βˆ’2               π‘₯ = 12

Question 2. Solve: π‘₯²– 16 = 0
Solution:

It is given that,
π‘₯²– 16 = 0
π‘₯²– (4)2 = 0
π‘₯²– (4)2 = 0
We know that,
(π‘Ž2 βˆ’ 𝑏2) = (π‘Ž βˆ’ 𝑏)(π‘Ž + 𝑏)
(π‘₯Β² βˆ’ 42) = (π‘₯ βˆ’ 4)(π‘₯ + 4)
(π‘₯ βˆ’ 4)(π‘₯ + 4) = 0
Then
π‘₯ βˆ’ 4 = 0           π‘₯ + 4 = 0
π‘₯ = 4                 π‘₯ = βˆ’4

Question 3. Solve: 2π‘₯Β² βˆ’ 1/2 π‘₯ = 0
Solution:

It is given that,

Question 4. Solve: π‘₯(π‘₯– 5) = 24
Solution:

It is given that,
π‘₯(π‘₯– 5) = 24
π‘₯Β² βˆ’ 5π‘₯ = 24
π‘₯Β² βˆ’ 5π‘₯ βˆ’ 24 = 0
By the splliting the middle term
π‘₯Β² βˆ’ 5π‘₯ βˆ’ 24 = 0
π‘₯Β² βˆ’ 8π‘₯ + 3π‘₯ βˆ’ 24 = 0
π‘₯(π‘₯ βˆ’ 8) + 3(π‘₯ βˆ’ 8) = 0
(π‘₯ + 3)(π‘₯ βˆ’ 8) = 0
Then
π‘₯ + 3 = 0            π‘₯ βˆ’ 8 = 0
π‘₯ = βˆ’3                π‘₯ = 8

Question 5. Solve: 9/2 π‘₯ = 5 + π‘₯Β²
Solution:

It is given that, 9/2 π‘₯ = 5 + π‘₯Β²
9π‘₯ = 2(5 + π‘₯2)
9π‘₯ = 10 + 2π‘₯Β²
0 = 2π‘₯Β² βˆ’ 9π‘₯ + 10
2π‘₯Β² βˆ’ 9π‘₯ + 10 = 0
2π‘₯Β² βˆ’ 5π‘₯ βˆ’ 4π‘₯ + 10 = 0
π‘₯(2π‘₯ βˆ’ 5) βˆ’ 2(2π‘₯ βˆ’ 5) = 0
(π‘₯ βˆ’ 2)(2π‘₯ βˆ’ 5) = 0
Then
π‘₯ βˆ’ 2 = 0            2π‘₯ βˆ’ 5 = 0
π‘₯ = 2                  π‘₯ = 5/2

Question 6. Solve 6/π‘₯ = 1 + π‘₯
Solution:

It is given that, 6/π‘₯ = 1 + π‘₯
6 = π‘₯(1 + π‘₯)
6 = π‘₯ + π‘₯2
π‘₯Β² + π‘₯ βˆ’ 6 = 0
π‘₯Β² + 3π‘₯ βˆ’ 2π‘₯ βˆ’ 6 = 0
π‘₯(π‘₯ + 3) βˆ’ 2(π‘₯ + 3) = 0
(π‘₯ βˆ’ 2)(π‘₯ + 3) = 0
Than
π‘₯ βˆ’ 2 = 0            π‘₯ + 3 = 0
π‘₯ = 2                  π‘₯ = βˆ’3

Question 7. Solve: π‘₯ = 3π‘₯ + 1/4π‘₯
Solution:

It is given that,
π‘₯ = 3π‘₯+1/4π‘₯
π‘₯ Γ— 4π‘₯ = 3π‘₯ + 1
4π‘₯Β² βˆ’ 3π‘₯ βˆ’ 1 = 0
4π‘₯Β² βˆ’ 4π‘₯ + π‘₯ βˆ’ 1 = 0
4π‘₯(π‘₯ βˆ’ 1) βˆ’ 1(π‘₯ βˆ’ 1) = 0
(4π‘₯ βˆ’ 1)(π‘₯ βˆ’ 1) = 0
than
4π‘₯ βˆ’ 1 = 0            π‘₯ βˆ’ 1 = 0
π‘₯ = 1/4                 π‘₯ = 1

Question 8. Solve π‘₯ + 1/π‘₯ = 2.5
Solution:

It is given that,
π‘₯ + 1
π‘₯ = 2.5
π‘₯Β²+1/π‘₯ = 2.5 (we can write 2.5 as 5/4)
π‘₯Β²+1/π‘₯ = 5/2
2(π‘₯Β² + 1) = 5π‘₯
2π‘₯Β² + 2 = 5π‘₯
2π‘₯Β² βˆ’ 5π‘₯ + 2 = 0
2π‘₯Β² βˆ’ 4π‘₯ βˆ’ π‘₯ + 2 = 0
2π‘₯(π‘₯ βˆ’ 2) βˆ’ 1(π‘₯ βˆ’ 2) = 0
2π‘₯(π‘₯ βˆ’ 2) βˆ’ 1(π‘₯ βˆ’ 2) = 0
(2π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2) = 0
than
2π‘₯ βˆ’ 1 = 0            π‘₯ βˆ’ 2 = 0
π‘₯ = 1/2                 π‘₯ = 2

Question 9. Solve: (2π‘₯– 3)Β² = 49
Solution:

It is given that,
(2π‘₯– 3)Β² = 49
2π‘₯– 3 = √49
2π‘₯– 3 = Β±7
Then
2π‘₯ βˆ’ 3 = 7       2π‘₯ βˆ’ 3 = βˆ’7
2π‘₯ = 7 + 3       2π‘₯ = βˆ’7 + 3
π‘₯ = 10/2            π‘₯ = βˆ’4/2
π‘₯ = 5                 π‘₯ = βˆ’2

Question 10. Solve: 2(π‘₯²– 6) = 3(π‘₯ – 4)
Solution:

It is given that,
2(π‘₯²– 6) = 3(π‘₯ – 4)
2π‘₯²– 12 = 3π‘₯ – 12
2π‘₯²– 3π‘₯ = βˆ’12 + 12
2π‘₯²– 3π‘₯ = 0
π‘₯(2π‘₯– 3) = 0
Then
π‘₯ = 0        2π‘₯– 3 = 0
π‘₯ = 3/2

Question 11. Solve: (π‘₯ + 1)(2π‘₯ + 8) = (π‘₯ + 7)(π‘₯ + 3)
Solution:

It is given that,
(π‘₯ + 1)(2π‘₯ + 8) = (π‘₯ + 7)(π‘₯ + 3)
2π‘₯Β² + 8π‘₯ + 2π‘₯ + 8 = π‘₯Β² + 3π‘₯ + 7π‘₯ + 21
2π‘₯Β² βˆ’ π‘₯Β² + 8π‘₯ + 2π‘₯ βˆ’ 3π‘₯ βˆ’ 7π‘₯ + 8 βˆ’ 21 = 0
π‘₯Β² + 10π‘₯ βˆ’ 10π‘₯ βˆ’ 13 = 0
π‘₯Β² βˆ’ 13 = 0
(π‘₯)2 + (√13)2 = 0
(π‘₯)2 + (√13)2 = 0
(π‘₯ + √13)(π‘₯ βˆ’ √13) = 0
Then
π‘₯ + √13 = 0            π‘₯ βˆ’ √13 = 0
π‘₯ = βˆ’βˆš13                π‘₯ = √13

Question 12. Solve: π‘₯²– (π‘Ž + 𝑏)π‘₯ + π‘Žπ‘ = 0
Solution:

It is given that,
π‘₯²– (π‘Ž + 𝑏)π‘₯ + π‘Žπ‘ = 0
π‘₯²– π‘Žπ‘₯ βˆ’ 𝑏π‘₯ + π‘Žπ‘ = 0
π‘₯(π‘₯– π‘Ž) βˆ’ 𝑏(π‘₯ βˆ’ π‘Ž) = 0
(π‘₯– π‘Ž)(π‘₯ βˆ’ 𝑏) = 0
Then
π‘₯ βˆ’ π‘Ž = 0          π‘₯ βˆ’ 𝑏 = 0
π‘₯ = π‘Ž                π‘₯ = 𝑏

Question 13. Solve: (π‘₯ + 3)²– 4(π‘₯ + 3)– 5 = 0
Solution:

It is given that,
(π‘₯ + 3)²– 4(π‘₯ + 3)– 5 = 0
We know that,
(π‘Ž + 𝑏)2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘
π‘₯2 + (3)2 + 2(π‘₯) (3)– 4π‘₯ βˆ’ 12– 5 = 0
π‘₯2 + 9 + 6π‘₯– 4π‘₯ βˆ’ 17 = 0
π‘₯2 + 2π‘₯ βˆ’ 8 = 0
π‘₯2 + 4π‘₯ βˆ’ 2π‘₯ βˆ’ 8 = 0
π‘₯(π‘₯ + 4) βˆ’ 2(π‘₯ + 4) = 0
(π‘₯ βˆ’ 2)(π‘₯ + 4) = 0
Than
π‘₯ βˆ’ 2 = 0           π‘₯ + 4 = 0
π‘₯ = 2                 π‘₯ = βˆ’4

Question 14. Solve: 4(2π‘₯– 3)²– (2π‘₯– 3) – 14 = 0
Solution:

It is given that,
4(2π‘₯– 3)²– (2π‘₯– 3) – 14 = 0
Let us assumed that,
2π‘₯– 3 = 𝑦
4𝑦²– 𝑦 – 14 = 0
4𝑦²– 8𝑦 + 7𝑦 – 14 = 0
4𝑦(𝑦– 2) + 7(𝑦– 2) = 0
(4𝑦 + 7)(𝑦– 2) = 0
Than
4𝑦 + 7 = 0           𝑦– 2 = 0
𝑦 = βˆ’ 7/4            π‘¦ = 2

Question 15. Solve:

Solution:
It is given that,

(3π‘₯ βˆ’ 2)(π‘₯ + 4) = (3π‘₯ βˆ’ 8) (2π‘₯ βˆ’ 3)
3π‘₯2 + 12π‘₯ βˆ’ 2π‘₯ βˆ’ 8 = 6π‘₯2 βˆ’ 9π‘₯ βˆ’ 16π‘₯ + 24
3π‘₯2 βˆ’ 6π‘₯2 + 12π‘₯ βˆ’ 2π‘₯ + 9π‘₯ + 16π‘₯ βˆ’ 8 βˆ’ 24 = 0
βˆ’3π‘₯2 + 10π‘₯ + 25π‘₯ βˆ’ 32 = 0
βˆ’3π‘₯2 + 35π‘₯ βˆ’ 32 = 0
βˆ’(3π‘₯2 βˆ’ 35π‘₯ + 32) = 0
3π‘₯2 βˆ’ 35π‘₯ + 32 = 0
3π‘₯2 βˆ’ 32π‘₯ βˆ’ 3π‘₯ + 32 = 0
π‘₯(3π‘₯ βˆ’ 32) βˆ’ 1(3π‘₯ βˆ’ 32) = 0
(π‘₯ βˆ’ 1)(3π‘₯ βˆ’ 32) = 0
Than
π‘₯ βˆ’ 1 = 0        3π‘₯– 32 = 0
𝑦 = 1               3π‘₯ = 32
                        π‘₯ = 32/3
                        π‘₯ = 10 (2/3)

Question 16. 2π‘₯2– 9π‘₯ + 10 = 0, π‘Šβ„Žπ‘’π‘›
(i) π‘₯ ∈ 𝑁
(ii) π‘₯ ∈ 𝑄
Solution:
It is given that,
2π‘₯2– 9π‘₯ + 10 = 0
2π‘₯2– 5π‘₯ βˆ’ 4π‘₯ + 10 = 0
π‘₯(2π‘₯– 5) βˆ’ 2(2π‘₯ βˆ’ 5) = 0
(π‘₯ βˆ’ 2)(2π‘₯– 5) = 0
Than
π‘₯ βˆ’ 2 = 0          2π‘₯– 5 = 0
𝑦 = 2                2π‘₯ = 5
                         π‘₯ = 5/2

Question 17. Solve

Solution:
It is given that,

2(2π‘₯2 + 18) = 5(π‘₯2 βˆ’ 9)
4π‘₯2 + 36 = 5π‘₯2 βˆ’ 45
4π‘₯2 βˆ’ 5π‘₯2 = βˆ’45 βˆ’ 36
βˆ’π‘₯2 = βˆ’81
π‘₯2 βˆ’ 81 = 0
π‘₯2 βˆ’ (9)2 = 0
(π‘₯ βˆ’ 9)(π‘₯ + 9) = 0
Than
π‘₯ βˆ’ 9 = 0 π‘₯ + 9 = 0
𝑦 = 9 π‘₯ = βˆ’9

Question 18. Solve:

Solution:
It is given that,

(3π‘₯ + 10) (2π‘₯ + 1) = 4(π‘₯2 + 5π‘₯ + 6)
6π‘₯2 + 3π‘₯ + 20π‘₯ + 10 = 4π‘₯2 + 20π‘₯ + 24
6π‘₯2 + 23π‘₯ + 10 βˆ’ 4π‘₯2 βˆ’ 20π‘₯ βˆ’ 24 = 0
2π‘₯2 + 3π‘₯ βˆ’ 14 = 0
2π‘₯2 + 7π‘₯ βˆ’ 4π‘₯ βˆ’ 14 = 0
π‘₯(2π‘₯ + 7) βˆ’ 2(2π‘₯ + 7) = 0
(π‘₯ βˆ’ 2) (2π‘₯ + 7) = 0
Than
π‘₯ βˆ’ 2 = 0            2π‘₯ + 7 = 0
π‘₯ = 2                  2π‘₯ = βˆ’7
                          π‘₯ = βˆ’ 7/2

Question 19. Solve:Q20. Solve:

Solution:
It is given that,

π‘₯(2π‘₯ + 36) = 4(π‘₯2 + 4π‘₯ βˆ’ 12)
2π‘₯2 + 36π‘₯ = 4π‘₯2 + 16π‘₯ βˆ’ 48
2π‘₯2 βˆ’ 4π‘₯2 + 36π‘₯ βˆ’ 16π‘₯ + 48 = 0
βˆ’2π‘₯2 + 20π‘₯ + 48 = 0
βˆ’2(π‘₯2 βˆ’ 10π‘₯ βˆ’ 24) = 0
π‘₯2 βˆ’ 10π‘₯ βˆ’ 24 = 0
π‘₯2 βˆ’ 12π‘₯ + 2π‘₯ βˆ’ 24 = 0
π‘₯(π‘₯ βˆ’ 12) + 2(π‘₯ βˆ’ 12) = 0
(π‘₯ + 2) (π‘₯ βˆ’ 12) = 0
Than
π‘₯ + 2 = 0             π‘₯ βˆ’ 12 = 0
𝑦 = βˆ’2                 π‘₯ = 12

Question 20. Solve:

Solution:
It is given that,

8(π‘₯2 βˆ’ 4) = 7(π‘₯2 βˆ’ 1)
8π‘₯2 βˆ’ 32 = 7π‘₯2 βˆ’ 7
8π‘₯2 βˆ’ 7π‘₯2 = βˆ’7 + 32
π‘₯2 = 25
π‘₯2 = ±√25
π‘₯ = Β± 5

Question 21. Find the quadratic equation, whose solution set is:

Solution:
(i) It is given that,
The solution set is {3, 5}
π‘₯ = 3 π‘₯ = 5
π‘₯ βˆ’ 3 = 0 π‘₯ βˆ’ 5 = 0
(π‘₯ βˆ’ 3)(π‘₯ βˆ’ 5) = 0
π‘₯2 βˆ’ 5π‘₯ βˆ’ 3π‘₯ + 15 = 0
π‘₯2 βˆ’ 8π‘₯ + 15 = 0
Hence, the required equation is π‘₯2 βˆ’ 8π‘₯ + 15 = 0.
(ii) It is given that,
The solution set is {βˆ’2, 3}
π‘₯ = βˆ’2 π‘₯ = 3
π‘₯ + 2 = 0 π‘₯ βˆ’ 3 = 0
(π‘₯ + 2)(π‘₯ βˆ’ 3) = 0
π‘₯2 βˆ’ 3π‘₯ + 2π‘₯ βˆ’ 6 = 0
π‘₯2 βˆ’ π‘₯ βˆ’ 6 = 0
Hence, the required equation is π‘₯2 βˆ’ π‘₯ βˆ’ 6 = 0.
(iii) It is given that,
The solution set is {5, βˆ’4}
π‘₯ = 5 π‘₯ = βˆ’4
π‘₯ βˆ’ 5 = 0 π‘₯ + 4 = 0
(π‘₯ βˆ’ 5)(π‘₯ + 4) = 0
π‘₯2 + 4π‘₯ βˆ’ 5π‘₯ βˆ’ 20 = 0
π‘₯2 βˆ’ π‘₯ βˆ’ 20 = 0
Hence, the required equation is π‘₯2 βˆ’ π‘₯ βˆ’ 20 = 0.
(iv) It is given that,

π‘₯ = βˆ’3 π‘₯ = βˆ’2/5
π‘₯ + 3 = 0 π‘₯ + 2
5 = 0
5π‘₯ + 2 = 0
(π‘₯ + 3)(5π‘₯ + 2) = 0
5π‘₯2 + 2π‘₯ + 15π‘₯ + 6 = 0
5π‘₯2 βˆ’ 17π‘₯ + 6 = 0
Hence, the required equation is 5π‘₯2 βˆ’ 17π‘₯ + 6 = 0.

Question 22. Solve:

Solution:
It is given that,

15(6π‘₯ βˆ’ π‘₯2 + 9) = 3(6 βˆ’ π‘₯)(12 + 2π‘₯)
5(6π‘₯ βˆ’ π‘₯2 + 9) = (6 βˆ’ π‘₯)(12 + 2π‘₯)
5(6π‘₯ βˆ’ π‘₯2 + 9) = 72 + 12π‘₯ βˆ’ 12π‘₯ βˆ’ 2π‘₯2
30π‘₯ βˆ’ 5π‘₯2 + 45 = 72 + 12π‘₯ βˆ’ 12π‘₯ βˆ’ 2π‘₯2
30π‘₯ βˆ’ 5π‘₯2 + 45 = 72 βˆ’ 2π‘₯2
30π‘₯ βˆ’ 5π‘₯2 + 2π‘₯2 + 45 βˆ’ 72 = 0
30π‘₯ βˆ’ 3π‘₯2 + 27 = 0
βˆ’3π‘₯2 + 30π‘₯ + 27 = 0
βˆ’3(π‘₯2 βˆ’ 10π‘₯ + 9) = 0
π‘₯2 βˆ’ 10π‘₯ + 9 = 0
π‘₯2 βˆ’ 9π‘₯ βˆ’ π‘₯ + 9 = 0
π‘₯(π‘₯ βˆ’ 9) βˆ’ 1(π‘₯ βˆ’ 9) = 0
(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 9) = 0
Than
π‘₯ βˆ’ 1 = 0           π‘₯ βˆ’ 9 = 0
π‘₯ = 1                 π‘₯ = 9

Question 23. Solve the equation 9π‘₯2 + 3π‘₯/4 + 2 = 0, if possible, for real values of π‘₯.
Solution:

It is given that,

We can’t solve the given equation for π‘₯ since βˆšβˆ’1143 isn’t possible.
Hence, there is no real value of π‘₯.

Question 24. Find the value of π‘₯, if π‘Ž + 1 = 0 and π‘₯2 + π‘Žπ‘₯– 6 = 0.
Solution:

It is given that,
π‘Ž + 1 = 0
π‘Ž = βˆ’1
Put the value of π‘Ž in given equation,
π‘₯2 + π‘Žπ‘₯– 6 = 0
π‘₯2 + 1π‘₯– 6 = 0
π‘₯2 + π‘₯– 6 = 0
π‘₯2 βˆ’ 3π‘₯ + 2π‘₯– 6 = 0
π‘₯(π‘₯ βˆ’ 3) + 2(π‘₯– 3) = 0
(π‘₯ + 2)(π‘₯ βˆ’ 3) = 0
Than
π‘₯ + 2 = 0           π‘₯ βˆ’ 3 = 0
π‘₯ = βˆ’2               π‘₯ = 3

Question 25. Find the value of π‘₯, if π‘Ž + 7 = 0; 𝑏 + 10 = 0 and 12π‘₯2 = π‘Žπ‘₯– 𝑏.
Solution:

It is given that,
π‘Ž + 7 = 0
π‘Ž = βˆ’7
and
𝑏 + 10 = 0
𝑏 = βˆ’10
Put the value of π‘Ž and 𝑏 in the given equation,
12π‘₯2 = π‘Žπ‘₯– 𝑏
12π‘₯2 = (βˆ’7)π‘₯– (βˆ’10)
12π‘₯2 = βˆ’7π‘₯ + 10
12π‘₯2 + 7π‘₯ βˆ’ 10 = 0
12π‘₯2 + 15π‘₯ βˆ’ 8π‘₯ βˆ’ 10 = 0
3π‘₯(4π‘₯ + 5) βˆ’ 2(4π‘₯ + 5) = 0
(3π‘₯ βˆ’ 2)(4π‘₯ + 5) = 0
Than
3π‘₯ βˆ’ 2 = 0           4π‘₯ + 5 = 0
π‘₯ = 2/3                  π‘₯ = βˆ’(5/4)

Question 26. Use the substitution 𝑦 = 2π‘₯ + 3 to solve for π‘₯, if 4(2π‘₯ + 3)2– (2π‘₯ + 3)– 14 = 0.
Solution:

It is given that,
𝑦 = 2π‘₯ + 3…………….(i)
Put the value of (2π‘₯ + 3) as 𝑦 in given equation,
4𝑦2– 𝑦– 14 = 0
4𝑦2 + 7𝑦 βˆ’ 8𝑦– 14 = 0
𝑦(4𝑦 + 7) βˆ’ 2(4𝑦 + 7) = 0
(𝑦 βˆ’ 2)(4𝑦 + 7) = 0
Than
𝑦 βˆ’ 2 = 0           4𝑦 + 7 = 0
𝑦 = 2                  π‘¦ = βˆ’(7/4)
Put the value of 𝑦 in equation (i)
𝑦 = 2
𝑦 = 2π‘₯ + 3
2 = 2π‘₯ + 3
2 βˆ’ 3 = 2π‘₯

Question 27. Without solving the quadratic equation 6π‘₯2 – π‘₯ – 2 = 0, find whether π‘₯ = 2/3 is a solution of this equation or not.
Solution:

It is given that,
π‘₯ = 2/3
Put the value π‘₯ in given equation,

Question 28. Determine whether π‘₯ = βˆ’1 is a root of the equation π‘₯2 – 3π‘₯ + 2 = 0 or not.
Solution:

It is given that,
π‘₯ = βˆ’1
Put the value π‘₯ in given equation,
π‘₯2 – 3π‘₯ + 2 = 0
(βˆ’1)2 – 3(βˆ’1) + 2 = 0
1 + 3 + 2 = 0
6 β‰  0
L.H.S. β‰  R.H.S.
Hence, π‘₯ = βˆ’1 is not a solution of the given equation.

Question 29. If π‘₯ = 2/3 is a solution of the quadratic equation 7π‘₯2 + π‘šπ‘₯– 3 = 0; Find the value of m.
Solution:

It is given that,
π‘₯ = 2/3
Put the value π‘₯ in given equation,
7π‘₯2 + π‘šπ‘₯ βˆ’ 3 = 0

Question 30. If π‘₯ = βˆ’3 and π‘₯ = 2/3 are solutions of quadratic equation π‘šπ‘₯2 + 7π‘₯ + 𝑛 = 0, find the values of m and n.
Solution:

It is given that,
π‘₯ = βˆ’3
Put the value of x in given equation,
π‘šπ‘₯2 + 7π‘₯ + 𝑛 = 0
π‘š(βˆ’3)2 + 7(βˆ’3) + 𝑛 = 0
9π‘š βˆ’ 21 + 𝑛 = 0
9π‘š + 𝑛 = 21_____________(i)
And
π‘₯ = 2/3
Put the value of x in given equation,
π‘šπ‘₯2 + 7π‘₯ + 𝑛 = 0

4π‘š + 42 + 9𝑛 = 0
4π‘š + 9𝑛 = βˆ’42_____________(ii)
From equation first get the value of 𝑛
9π‘š + 𝑛 = 21
𝑛 = 21 βˆ’ 9π‘š__(iii)
Value of n put in equation (ii)
4π‘š + 9𝑛 = βˆ’42
4π‘š + 9(21 βˆ’ 9π‘š) = βˆ’42
4π‘š + 189 βˆ’ 81π‘š = βˆ’42
4π‘š βˆ’ 81π‘š = βˆ’42 βˆ’ 189
βˆ’77π‘š = βˆ’231
π‘š = 231/77
π‘š = 3
Put the value of π‘š in equation (iii)
𝑛 = 21 βˆ’ 9π‘š
𝑛 = 21 βˆ’ 9(3)
𝑛 = 21 βˆ’ 27
𝑛 = βˆ’6

Question 31. If quadratic equation π‘₯2– (π‘š + 1) π‘₯ + 6 = 0 has one root as π‘₯ = 3; find the value of m and
the root of the equation.
Solution:

It is given that,
π‘₯ = 3
Put the value of π‘₯ in given equation,
π‘₯2– (π‘š + 1) π‘₯ + 6 = 0
(3)2– (π‘š + 1) 3 + 6 = 0
9– 3π‘š βˆ’ 3 + 6 = 0
– 3π‘š + 12 = 0
– 3π‘š = βˆ’12
π‘š = βˆ’12/βˆ’3
π‘š = 4
Put the value of π‘š in given equation,
π‘₯2– (π‘š + 1) π‘₯ + 6 = 0
π‘₯2– (4 + 1)π‘₯ + 6 = 0
π‘₯2– 5π‘₯ + 6 = 0
π‘₯2– 3π‘₯ βˆ’ 2π‘₯ + 6 = 0
π‘₯(π‘₯– 3) βˆ’ 2(π‘₯ βˆ’ 3) = 0
(π‘₯ βˆ’ 2)(π‘₯ βˆ’ 3) = 0
Than
π‘₯ βˆ’ 2 = 0           π‘₯ βˆ’ 3 = 0
π‘₯ = 2                 π‘₯ = 3

Question 32. Given that 2 is a root of the equation 3π‘₯²– 𝑝(π‘₯ + 1) = 0 and that the equation 𝑝π‘₯²– π‘žπ‘₯ + 9 = 0 has equal roots, find the values of p and q.
Solution:

It is given that,
π‘₯ = 2
Put the value of π‘₯ in given equation,
3(2)²– 𝑝(2 + 1) = 0
3(2)²– 𝑝(3) = 0
12– 3𝑝 = 0
– 3𝑝 = βˆ’12
𝑝 = 12/3
𝑝 = 4
Put the value of 𝑝 in other equation
𝑝π‘₯²– π‘žπ‘₯ + 9 = 0
4π‘₯²– π‘žπ‘₯ + 9 = 0
It is given that the roots are equal, we know
𝑏2 βˆ’ 4π‘Žπ‘ = 0
Here,
π‘Ž = 4
𝑏 = βˆ’π‘ž
𝑐 = 9
(βˆ’π‘ž)2 βˆ’ 4(4)(9) = 0
π‘ž2 βˆ’ 144 = 0
π‘ž2 = 144
π‘ž = √144
π‘ž = 12
Hence, the value of 𝑝 and π‘ž is 4 and 12 respectively.

Question 33. Solve:

Solution:
It is given that,

π‘₯2 = π‘Žπ‘ + 𝑏2 + π‘Ž2 + π‘Žπ‘
π‘₯2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘
π‘₯2 = (π‘Ž + 𝑏)2
π‘₯ = π‘Ž + 𝑏

Question 34. Solve:

Solution:
It is given that,

(600 + π‘₯)(π‘₯ βˆ’ 10) = 630π‘₯
600π‘₯ βˆ’ 6000 + π‘₯2 βˆ’ 10π‘₯ = 630π‘₯
600π‘₯ βˆ’ 6000 + π‘₯2 βˆ’ 10π‘₯ βˆ’ 630π‘₯ = 0
βˆ’6000 + π‘₯2 βˆ’ 10π‘₯ βˆ’ 30π‘₯ = 0
βˆ’6000 + π‘₯2 βˆ’ 40π‘₯ = 0
π‘₯2 βˆ’ 40π‘₯ βˆ’ 6000 = 0
π‘₯2 βˆ’ 100π‘₯ + 60π‘₯ βˆ’ 6000 = 0
π‘₯2 βˆ’ 100π‘₯ + 60π‘₯ βˆ’ 6000 = 0
π‘₯(π‘₯ βˆ’ 100) + 60(π‘₯ βˆ’ 100) = 0
(π‘₯ + 60)(π‘₯ βˆ’ 100) = 0
Than
π‘₯ + 60 = 0           π‘₯ βˆ’ 100 = 0
π‘₯ = βˆ’60               π‘₯ = 100

Question 35. If βˆ’1 and 3 are the roots of π‘₯2 + 𝑝π‘₯ + π‘ž = 0, find the values of p and q.
Solution:

It is given that,
π‘₯ = βˆ’1
Put the value of π‘₯ in given equation,
(βˆ’1)2 + 𝑝(βˆ’1) + π‘ž = 0
1 βˆ’ 𝑝 + π‘ž = 0
βˆ’π‘ + π‘ž = βˆ’1_______________ (i)
Also
π‘₯ = 3
Put the value of π‘₯ in given equation,
(3)2 + 𝑝(3) + π‘ž = 0
9 + 3𝑝 + π‘ž = 0
3𝑝 + π‘ž = βˆ’9______________(ii)
From equation (i) we get the value of π‘ž
βˆ’π‘ + π‘ž = βˆ’1
π‘ž = βˆ’1 + 𝑝__ (iii)
Put the value of π‘ž in equation (ii)
3𝑝 + (βˆ’1 + 𝑝) = βˆ’9
3𝑝 + 𝑝 = βˆ’9 + 1
4𝑝 = βˆ’8
𝑝 = βˆ’ 8/4
𝑝 = βˆ’2
Put the value of 𝑝 in equation (iii)
π‘ž = βˆ’1 + 𝑝
π‘ž = βˆ’1 + (βˆ’2)
π‘ž = βˆ’1 βˆ’ 2
π‘ž = βˆ’3
Hence, the value of 𝑝 is βˆ’2 and π‘ž is βˆ’3.

Exercise 5D

Question 1. Solve each of the following equations using the formula:
(i) π‘₯2 βˆ’ 6π‘₯ = 27
(ii) π‘₯2 βˆ’ 10π‘₯ + 21 = 0
(iii) π‘₯2 + 6π‘₯ βˆ’ 10 = 0
(iv) π‘₯2 + 2π‘₯ βˆ’ 6 = 0
(v) 3π‘₯2 + 2π‘₯ βˆ’ 1 = 0
(vi) 2π‘₯2 + 7π‘₯ + 5 = 0

Solution:
(i) π‘₯2 βˆ’ 6π‘₯ = 27
π‘₯2 βˆ’ 6π‘₯ βˆ’ 27 = 0
π‘Ž = 1
𝑏 = βˆ’6
𝑐 = βˆ’27
By the quadratic formula,

(ii) π‘₯2 βˆ’ 10π‘₯ + 21 = 0
π‘Ž = 1
𝑏 = βˆ’10
𝑐 = 21
By the quadratic formula,

(iii) π‘₯2 + 6π‘₯ βˆ’ 10 = 0
π‘Ž = 1
𝑏 = 6
𝑐 = βˆ’10
By the quadratic formula,

(iv) π‘₯2 + 2π‘₯ βˆ’ 6 = 0
π‘Ž = 1
𝑏 = 2
𝑐 = βˆ’6
By the quadratic formula,

(v) 3π‘₯2 + 2π‘₯ βˆ’ 1 = 0
π‘Ž = 3
𝑏 = 2
𝑐 = βˆ’1
By the quadratic formula,

(vi) 2π‘₯2 + 7π‘₯ + 5 = 0
π‘Ž = 2
𝑏 = 7
𝑐 = 5
By the quadratic formula,

(ix) π‘₯2 βˆ’ 6 = 2√2π‘₯
π‘₯2 βˆ’ 2√2π‘₯ βˆ’ 6 = 0
π‘Ž = 1
𝑏 = βˆ’2√2
𝑐 = βˆ’6
By the quadratic formula,

(4 βˆ’ 3π‘₯)(2π‘₯ + 3) = 5π‘₯
(4 βˆ’ 3π‘₯)(2π‘₯ + 3) = 5π‘₯
8π‘₯ + 12 βˆ’ 6π‘₯2 βˆ’ 9π‘₯ = 5π‘₯
12 βˆ’ 6π‘₯2 βˆ’ π‘₯ = 5π‘₯
12 βˆ’ 6π‘₯2 βˆ’ π‘₯ βˆ’ 5π‘₯ = 0
12 βˆ’ 6π‘₯2 βˆ’ 6π‘₯ = 0
βˆ’(6π‘₯2 + 6π‘₯ βˆ’ 12) = 0
6π‘₯2 + 6π‘₯ βˆ’ 12 = 0
6(π‘₯2 + π‘₯ βˆ’ 2) = 0
π‘₯2 + π‘₯ βˆ’ 2 = 0
π‘Ž = 1
𝑏 = 1
𝑐 = βˆ’2
By the quadratic formula,

(xii) √6π‘₯2 βˆ’ 4π‘₯ βˆ’ 2√6 = 0
π‘Ž = √6
𝑏 = βˆ’4
𝑐 = βˆ’2√6
By the quadratic formula,

3(2π‘₯2 βˆ’ 10π‘₯ + 10) = 10(π‘₯2 βˆ’ 6π‘₯ + 8)
6π‘₯2 βˆ’ 30π‘₯ + 30 = 10π‘₯2 βˆ’ 60π‘₯ + 80
6π‘₯2 βˆ’ 30π‘₯ + 30 βˆ’ 10π‘₯2 + 60π‘₯ βˆ’ 80 = 0
βˆ’4π‘₯2 βˆ’ 30π‘₯ + 50 = 0
βˆ’2(2π‘₯2 βˆ’ 15π‘₯ + 25) = 0
2π‘₯2 βˆ’ 15π‘₯ + 25 = 0
π‘Ž = 2
𝑏 = βˆ’15
𝑐 = 25
By the quadratic formula,

Question 2. Solve each of the following equations for π‘₯ and give, in each case, your answer correct to one decimal place:
(𝑖) π‘₯2– 8π‘₯ + 5 = 0
(𝑖𝑖) 5π‘₯2 + 10π‘₯– 3 = 0
Solution:

(i) π‘₯2– 8π‘₯ + 5 = 0
π‘Ž = 1
𝑏 = βˆ’8
𝑐 = 5
By the quadratic formula,

(ii) 5π‘₯2 + 10π‘₯ βˆ’ 3 = 0
π‘Ž = 5
𝑏 = 10
𝑐 = βˆ’3
By the quadratic formula,

Question 3 (i). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
(𝑖) 2π‘₯2– 10π‘₯ + 5 = 0
Solution:

It is given that,
2π‘₯2– 10π‘₯ + 5 = 0
π‘Ž = 2
𝑏 = βˆ’10
𝑐 = 5
By the quadratic formula,

Question 3 (ii). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
4π‘₯ + 6/π‘₯ + 13 = 0

Question 3 (iii). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
π‘₯2 – 3π‘₯– 9 = 0
Solution:

It is given that,
π‘₯2 – 3π‘₯– 9 = 0
π‘Ž = 1
𝑏 = βˆ’3
𝑐 = βˆ’9
By the quadratic formula,

Question 3 (iv). Solve each of the following equations for x and give, in each case, your answer correct to two decimal places:
π‘₯2– 5π‘₯– 10 = 0
Solution:

It is given that,
π‘₯2– 5π‘₯– 10 = 0
π‘Ž = 1
𝑏 = βˆ’5
𝑐 = βˆ’10
By the quadratic formula,

Question 4. Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places:
(𝑖) 3π‘₯2– 12π‘₯ – 1 = 0
(𝑖𝑖) π‘₯2– 16 π‘₯ + 6 = 0
(𝑖𝑖𝑖) 2π‘₯2 + 11π‘₯ + 4 = 0
Solution:

(𝑖) 3π‘₯2– 12π‘₯ – 1 = 0
π‘Ž = 3
𝑏 = βˆ’12
𝑐 = βˆ’1
By the quadratic formula,

(𝑖𝑖) π‘₯2– 16 π‘₯ + 6 = 0
π‘Ž = 1
𝑏 = βˆ’16
𝑐 = 6
By the quadratic formula,

(𝑖𝑖𝑖) 2π‘₯2 + 11π‘₯ + 4 = 0
π‘Ž = 2
𝑏 = 11
𝑐 = 4
By the quadratic formula,

Question 5. Solve:
(𝑖) π‘₯4 – 2π‘₯2 – 3 = 0
(𝑖𝑖) π‘₯4 – 10π‘₯2 + 9 = 0
Solution:

(𝑖) π‘₯4 – 2π‘₯2 – 3 = 0
By splitting the middle term method,
π‘₯4 – 3π‘₯2 + π‘₯2 βˆ’ 3 = 0
π‘₯2(π‘₯2 βˆ’ 3) + 1(π‘₯2 βˆ’ 3) = 0
(π‘₯2 + 1)(π‘₯2 βˆ’ 3) = 0
π‘₯2 + 1 = 0 π‘₯2 βˆ’ 3 = 0
π‘₯2 = βˆ’1 π‘₯ = ±√3
(𝑖𝑖) π‘₯2 – 10π‘₯2 + 9 = 0
By splitting the middle term method,
π‘₯4 – 9π‘₯2 βˆ’ π‘₯2 βˆ’ 9 = 0
π‘₯2(π‘₯2 βˆ’ 9) + 1(π‘₯2 βˆ’ 9) = 0
(π‘₯2 + 1)(π‘₯2 βˆ’ 9) = 0
π‘₯2 + 1 = 0 π‘₯2 βˆ’ 9 = 0
π‘₯2 = βˆ’1 π‘₯2 = 9
π‘₯ = Β±3

Question 6. Solve:
(𝑖) (π‘₯2 – π‘₯)2 + 5(π‘₯2– π‘₯) + 4 = 0
(𝑖𝑖) (π‘₯2 – 3π‘₯)2 – 16(π‘₯2– 3π‘₯)– 36 = 0
Solution:

(𝑖) (π‘₯2 – π‘₯)2 + 5(π‘₯2– π‘₯) + 4 = 0
Let us assumed that,
π‘₯2 – π‘₯ = 𝑦
𝑦2 + 5𝑦 + 4 = 0
By the splitting the middle term
𝑦2 + 4𝑦 + 𝑦 + 4 = 0
𝑦(𝑦 + 4) + 1(𝑦 + 4) = 0
(𝑦 + 1)(𝑦 + 4) = 0
𝑦 + 1 =          0 𝑦 + 4 = 0
𝑦 = βˆ’1             π‘¦ = βˆ’4
Put the value of y as βˆ’1
π‘₯2 – π‘₯ = βˆ’1
π‘₯2 – π‘₯ + 1 = 0
By the quadratic formula,
π‘Ž = 1
𝑏 = βˆ’1
𝑐 = 1

(𝑖𝑖) (π‘₯2 – 3π‘₯)2 – 16(π‘₯2– 3π‘₯)– 36 = 0
Let us assumed that,
π‘₯2 – 3π‘₯ = 𝑦
𝑦2 βˆ’ 16𝑦 βˆ’ 36 = 0
By the splitting the middle term
𝑦2 – 16𝑦– 36 = 0
𝑦2– 18𝑦 + 2𝑦– 36 = 0
𝑦(𝑦– 18) + 2(𝑦– 18) = 0
(𝑦 + 2)(𝑦– 18) = 0
𝑦 + 2 = 0 𝑦– 18 = 0
𝑦 = βˆ’2 𝑦 = 18
Put the value of y as βˆ’2
π‘₯2 – 3π‘₯ = βˆ’2
π‘₯2 – 3π‘₯ + 2 = 0
By the quadratic formula,
π‘Ž = 1
𝑏 = βˆ’3
𝑐 = 2

Question 7. Solve:-

Question 8. Solve the equation 2π‘₯ βˆ’ (1/π‘₯) = 7. Write your answer correct to two decimal places.
Solution:

It is given that,

Question 9. Solve the following equation and give your answer correct to 3 significant figures:
5π‘₯²– 3π‘₯ – 4 = 0
Solution:

It is given that,
5π‘₯²– 3π‘₯ – 4 = 0
By quadratic formula
π‘Ž = 5
𝑏 = βˆ’3
𝑐 = βˆ’4

Question 10. Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(π‘₯– 1)2– 3π‘₯ + 4 = 0
Solution:

It is given that,
(π‘₯– 1)2– 3π‘₯ + 4 = 0
π‘₯2 βˆ’ 2(π‘₯)(1) + 1– 3π‘₯ + 4 = 0
π‘₯2 βˆ’ 5π‘₯ + 5 = 0
By quadratic formula
π‘Ž = 1
𝑏 = βˆ’5
𝑐 = 5

Question 11. Solve the quadratic equation π‘₯²– 3(π‘₯ + 3) = 0; Give your answer correct to two significant figures.
Solution:

It is given that,
π‘₯²– 3(π‘₯ + 3) = 0
π‘₯²– 3π‘₯ βˆ’ 9 = 0
By quadratic formula
π‘Ž = 1
𝑏 = βˆ’3
𝑐 = βˆ’9

Exercise 5E

Question 1. Solve:

4π‘₯2 + 6π‘₯ + π‘₯ βˆ’ 3 + 3π‘₯ + 9 = 0
4π‘₯2 + 10π‘₯ + 6 = 0
By the splitting the middle term
4π‘₯2 + 4π‘₯ + 6π‘₯ + 6 = 0
4π‘₯(π‘₯ + 1) + 6(π‘₯ + 1) = 0
(4π‘₯ + 6)(π‘₯ + 1) = 0
4π‘₯ + 6 = 0           π‘₯ + 1 = 0
4π‘₯ = βˆ’6               π‘₯ = βˆ’1
π‘₯ = βˆ’6/4
π‘₯ = βˆ’3/2

Question 2. Solve: (2π‘₯ + 3)2 = 81
Solution:

It is given that,
(2π‘₯ + 3)2 = 81
2π‘₯ + 3 = √81
2π‘₯ + 3 = Β±9
2π‘₯ + 3 = 9           2π‘₯ + 3 = βˆ’9
2π‘₯ = 9 βˆ’ 3           2π‘₯ = βˆ’9 βˆ’ 3
2π‘₯ = 6                 2π‘₯ = βˆ’12
π‘₯ = 6/2                 π‘₯ = βˆ’12/2
π‘₯ = 3                    π‘₯ = βˆ’6

Question 3. Solve: π‘ŽΒ²π‘₯²– 𝑏² = 0
Solution:

It is given that,
π‘ŽΒ²π‘₯²– 𝑏² = 0
(π‘Žπ‘₯)2– 𝑏² = 0
We know that,
π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)
(π‘Žπ‘₯)2– 𝑏² = 0
(π‘Žπ‘₯ + 𝑏)(π‘Žπ‘₯ βˆ’ 𝑏) = 0
π‘Žπ‘₯ + 𝑏 = 0        π‘Žπ‘₯ βˆ’ 𝑏 = 0
π‘Žπ‘₯ = βˆ’π‘           π‘Žπ‘₯ = 𝑏
π‘₯ = βˆ’π‘/π‘Ž           π‘₯ = 𝑏/π‘Ž

Question 4. Solve:

Question 5. Solve: π‘₯ + 4/π‘₯ = βˆ’4; π‘₯ β‰  0
Solution:

It is given that,
π‘₯ + 4/π‘₯ = βˆ’4
By the LCM of denominator,
π‘₯2 + 4/π‘₯ = βˆ’4
π‘₯2 + 4 = βˆ’4π‘₯
π‘₯2 + 4π‘₯ + 4 = 0
By splitting the middle term
π‘₯2 + 4π‘₯ + 4 = 0
π‘₯2 + 2π‘₯ + 2π‘₯ + 4 = 0
π‘₯(π‘₯ + 2) + 2(π‘₯ + 2) = 0
(π‘₯ + 2)(π‘₯ + 2) = 0
π‘₯ + 2 = 0
π‘₯ = βˆ’2
Hence, the value of π‘₯ is βˆ’2.

Question 6. Solve: 2π‘₯4– 5π‘₯Β² + 3 = 0
Solution:

It is given that,
2π‘₯4– 5π‘₯Β² + 3 = 0
2π‘₯4– 3π‘₯Β² βˆ’ 2π‘₯Β² + 3 = 0
π‘₯Β²(2π‘₯2– 3) βˆ’ 1(2π‘₯2 + 3) = 0
(π‘₯Β² βˆ’ 1)(2π‘₯2– 3) = 0
π‘₯Β² βˆ’ 1 = 0 2π‘₯2– 3 = 0

Q7. Solve: π‘₯4– 2π‘₯²– 3 = 0.
Solution:

It is given that,
π‘₯4– 2π‘₯²– 3 = 0
π‘₯4– 3π‘₯Β² + π‘₯²– 3 = 0
π‘₯2(π‘₯2– 3) + 1(π‘₯2– 3) = 0
(π‘₯2 + 1)(π‘₯2– 3) = 0
π‘₯2 + 1 = 0           π‘₯2– 3 = 0
π‘₯Β² = βˆ’1               π‘₯2 = 3
π‘₯ = βˆšβˆ’1             π‘₯ = ±√3π‘₯2

Question 8. Solve:

Put the values,
9(𝑦2 βˆ’ 2) βˆ’ 9𝑦 βˆ’ 52 = 0
9𝑦2 βˆ’ 18 βˆ’ 9𝑦 βˆ’ 52 = 0
9𝑦2 βˆ’ 9𝑦 βˆ’ 70 = 0
9𝑦2 βˆ’ 30𝑦 + 21𝑦 βˆ’ 70 = 0
3𝑦(3𝑦 βˆ’ 10) + 7(3𝑦 βˆ’ 10) = 0
(3𝑦 + 7)(3𝑦 βˆ’ 10) = 0
3𝑦 + 7 = 0           3𝑦 βˆ’ 10 = 0
3𝑦 = βˆ’7               3𝑦 = 10
𝑦 = βˆ’7/3               π‘¦ = 10/3
Put the value of y as -7/3

3(π‘₯2 + 1) = 10π‘₯
3π‘₯2 + 3 = 10π‘₯
3π‘₯2 βˆ’ 10π‘₯ + 3 = 0
By the splitting the middle term
3π‘₯2 βˆ’ 9π‘₯ βˆ’ π‘₯ + 3 = 0
3π‘₯(π‘₯ βˆ’ 3) βˆ’ 1(π‘₯ βˆ’ 3) = 0
(3π‘₯ βˆ’ 1)(π‘₯ βˆ’ 3) = 0
3π‘₯ βˆ’ 1 = 0           π‘₯ βˆ’ 3 = 0
3π‘₯ = 1                 π‘₯ = 3
π‘₯ = 1/3
Hence, the value of π‘₯ is 3.

Question 9. Solve:

Put the values,
2(𝑦2 βˆ’ 2) βˆ’ 𝑦 = 11
2𝑦2 βˆ’ 4 βˆ’ 𝑦 = 11
2𝑦2 βˆ’ 4 βˆ’ 𝑦 βˆ’ 11 = 0
2𝑦2 βˆ’ 𝑦 βˆ’ 15 = 0
By splitting term,
2𝑦2 βˆ’ 6𝑦 + 5𝑦 βˆ’ 15 = 0
2𝑦(𝑦 βˆ’ 3) + 5(𝑦 βˆ’ 3) = 0
(2𝑦 + 5)(𝑦 βˆ’ 3) = 0
2𝑦 + 5 = 0           π‘¦ βˆ’ 3 = 0
𝑦 = βˆ’ 5/2            π‘¦ = 3
Put the value of 𝑦 as 3

π‘₯2 + 1 = βˆ’ 5
2 Γ— π‘₯
2(π‘₯2 + 1) = βˆ’5π‘₯
2π‘₯2 + 2 = βˆ’5π‘₯
2π‘₯2 + 5π‘₯ + 2 = 0
2π‘₯2 + 4π‘₯ + π‘₯ + 2 = 0
2π‘₯(π‘₯ + 2) + 1(π‘₯ + 2) = 0
(2π‘₯ + 1)(π‘₯ + 2) = 0
2π‘₯ + 1 = 0           π‘₯ + 2 = 0
2π‘₯ = βˆ’1               π‘₯ = βˆ’2
π‘₯ = βˆ’ 1/2

Question 10. Solve:

Question 11. Solve: (π‘₯Β² + 5π‘₯ + 4)(π‘₯Β² + 5π‘₯ + 6) = 120
Solution:

It is given that,
(π‘₯Β² + 5π‘₯ + 4)(π‘₯Β² + 5π‘₯ + 6) = 120
Let us assumed that,
π‘₯Β² + 5π‘₯ = 𝑦 …………..(i)
(𝑦 + 4)(𝑦 + 6) = 120
𝑦2 + 6𝑦 + 4𝑦 + 24 = 120
𝑦2 + 10𝑦 + 24 = 120
𝑦2 + 10𝑦 + 24 βˆ’ 120 = 0
𝑦2 + 10𝑦 βˆ’ 96 = 0
𝑦2 + 16𝑦 βˆ’ 6𝑦 βˆ’ 96 = 0
𝑦(𝑦 + 16) βˆ’ 6(𝑦 + 16) = 0
(𝑦 βˆ’ 6)(𝑦 + 16) = 0
𝑦 βˆ’ 6 = 0           π‘¦ + 16 = 0
𝑦 = 6                 π‘¦ = βˆ’16
Put the value of 𝑦 as 6 in equation (i)
π‘₯Β² + 5π‘₯ = 6
π‘₯Β² + 5π‘₯ βˆ’ 6 = 0
π‘₯Β² + 6π‘₯ βˆ’ π‘₯ βˆ’ 6 = 0
π‘₯(π‘₯ + 6) βˆ’ 1(π‘₯ + 6) = 0
(π‘₯ + 6)(π‘₯ βˆ’ 1) = 0
(π‘₯ + 6)(π‘₯ βˆ’ 1) = 0
π‘₯ + 6 = 0           π‘₯ βˆ’ 1 = 0
𝑦 = βˆ’6               π‘₯ = 1
Put the value of 𝑦 as -16 in equation (i)
π‘₯Β² + 5π‘₯ = βˆ’16
π‘₯Β² + 5π‘₯ + 16 = 0
By quadratic formula
π‘Ž = 1
𝑏 = 5
𝑐 = 16

Question 12. Solve each of the following equations, giving answer upto two decimal places.
(𝑖) π‘₯2 – 5π‘₯ βˆ’ 10 = 0
(𝑖𝑖) 3π‘₯2 – π‘₯– 7 = 0
Solution:

(i) π‘₯2 – 5π‘₯ βˆ’ 10 = 0
By quadratic formula
π‘Ž = 1
𝑏 = βˆ’5
𝑐 = βˆ’10

(𝑖𝑖) 3π‘₯2 – π‘₯– 7 = 0
By quadratic formula
π‘Ž = 3
𝑏 = βˆ’1
𝑐 = βˆ’7

Question 13. Solve:

𝑦2 βˆ’ 7𝑦 + 12 = 0
By splitting the middle term,
𝑦2 βˆ’ 4𝑦 βˆ’ 3𝑦 + 12 = 0
𝑦(𝑦 βˆ’ 4) βˆ’ 3(𝑦 βˆ’ 4) = 0
(𝑦 βˆ’ 3)(𝑦 βˆ’ 4) = 0
𝑦 βˆ’ 3 = 0           π‘¦ βˆ’ 4 = 0
𝑦 = 3                 π‘¦ = 4
Put the value of 𝑦 as 3 in equation (i)
π‘₯/π‘₯+2 = 3
π‘₯ = 3(π‘₯ + 2)
π‘₯ = 3π‘₯ + 6
π‘₯ βˆ’ 3π‘₯ = 6
βˆ’2π‘₯ = 6
π‘₯ = βˆ’ 6/2
π‘₯ = βˆ’3
Put the value of 𝑦 as 4 in equation (i)
π‘₯/π‘₯+2 = 4
π‘₯ = 4(π‘₯ + 2)
π‘₯ = 4π‘₯ + 8
π‘₯ βˆ’ 4π‘₯ = 8
βˆ’3π‘₯ = 8
π‘₯ = βˆ’ 8/3

Question 14. Solve:
(𝑖) π‘₯2– 11π‘₯ – 12 = 0; π‘€β„Žπ‘’π‘› π‘₯ ∈ 𝑁
(𝑖𝑖) π‘₯2 – 4π‘₯ – 12 = 0; π‘€β„Žπ‘’π‘› π‘₯ ∈ 𝐼
(𝑖𝑖𝑖) 2π‘₯2 – 9π‘₯ + 10 = 0; π‘€β„Žπ‘’π‘› π‘₯ ∈ 𝑄
Solution:

(i) π‘₯2– 11π‘₯ – 12 = 0
By splitting the middle term,
π‘₯2– 12π‘₯ + π‘₯ – 12 = 0
π‘₯(π‘₯– 12) + 1(π‘₯ – 12) = 0
(π‘₯ + 1)(π‘₯– 12) = 0
π‘₯ + 1 = 0 π‘₯– 12 = 0
π‘₯ = βˆ’1 π‘₯ = 12
Hence, the value of π‘₯ is 12 since π‘₯ ∈ 𝑁.
(ii) π‘₯2– 4π‘₯ – 12 = 0
By splitting the middle term,
π‘₯2 + 2π‘₯ βˆ’ 6π‘₯ – 12 = 0
π‘₯(π‘₯ + 2) βˆ’ 6(π‘₯ + 12) = 0
(π‘₯ βˆ’ 6)(π‘₯ + 2) = 0
π‘₯ βˆ’ 6 = 0          π‘₯ + 2 = 0
π‘₯ = 6                π‘₯ = βˆ’2
Hence, the value of π‘₯ is 6 and -2 since π‘₯ ∈ 𝐼
(iii) 2π‘₯2 – 9π‘₯ + 10 = 0
By splitting the middle term,
2π‘₯2 – 4π‘₯ βˆ’ 5π‘₯ + 10 = 0
2π‘₯(π‘₯ – 2) βˆ’ 5(π‘₯ βˆ’ 2) = 0
(2π‘₯ βˆ’ 5)(π‘₯ βˆ’ 2) = 0
2π‘₯ βˆ’ 5 = 0           π‘₯ βˆ’ 2 = 0
π‘₯ = 5/2               π‘₯ = 2
Hence, the value of π‘₯ is 5/2 and 2 since π‘₯ ∈ 𝑄.

Question 15. Solve: (π‘Ž + 𝑏)Β²π‘₯Β² – (π‘Ž + 𝑏)π‘₯ – 6 = 0; π‘Ž + 𝑏 β‰  0.
Solution:

It is given that,
(π‘Ž + 𝑏)Β²π‘₯Β² – (π‘Ž + 𝑏)π‘₯ – 6 = 0
By splitting the middle term,
(π‘Ž + 𝑏)2π‘₯2– 3(π‘Ž + 𝑏)π‘₯ + 2(π‘Ž + 𝑏)π‘₯ – 6 = 0
(π‘Ž + 𝑏)π‘₯[(π‘Ž + 𝑏)π‘₯– 3] + 2[(π‘Ž + 𝑏)π‘₯ – 3] = 0
[(π‘Ž + 𝑏)π‘₯ + 2][(π‘Ž + 𝑏)π‘₯ βˆ’ 3] = 0
(π‘Ž + 𝑏)π‘₯ + 2 = 0           (π‘Ž + 𝑏)π‘₯ βˆ’ 3 = 0
(π‘Ž + 𝑏)π‘₯ = βˆ’2               (π‘Ž + 𝑏)π‘₯ = 3
π‘₯ = βˆ’2/(π‘Ž+𝑏)               π‘₯ = 3/(π‘Ž+𝑏)

Question 16. Solve:

Question 17. Solve:
  (i) π‘₯(π‘₯ + 1) + (π‘₯ + 2)(π‘₯ + 3) = 42

Solution:
(i) π‘₯(π‘₯ + 1) + (π‘₯ + 2)(π‘₯ + 3) = 42
π‘₯2 + π‘₯ + π‘₯2 + 3π‘₯ + 2π‘₯ + 6 = 42
2π‘₯2 + 6π‘₯ + 6 βˆ’ 42 = 0
2π‘₯2 + 6π‘₯ βˆ’ 36 = 0
2π‘₯2 + 12π‘₯ βˆ’ 6π‘₯ βˆ’ 36 = 0
2π‘₯(π‘₯ + 6) βˆ’ 6(π‘₯ + 6) = 0
(2π‘₯ βˆ’ 6)(π‘₯ + 6) = 0
2π‘₯ βˆ’ 6 = 0          π‘₯ + 6 = 0
π‘₯ = 6/2               π‘₯ = βˆ’6
π‘₯ = 3

Question 18. For each equation, given below, find the value of π‘š so that the equation has equal roots.
Also, find the solution of each equation:
(i) (π‘š βˆ’ 3)π‘₯2 βˆ’ 4π‘₯ + 1 = 0
(ii) 3π‘₯2 + 12π‘₯ + (π‘š + 7) = 0
(iii) π‘₯2 βˆ’ (π‘š + 2)π‘₯ + (π‘š + 5) = 0
Solution:

(i) (π‘š βˆ’ 3)π‘₯2 βˆ’ 4π‘₯ + 1 = 0
It is given that, the above equation equal roots so 𝐷 = 0
We know that,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝑏2 βˆ’ 4π‘Žπ‘ = 0
Here,
π‘Ž = (π‘š βˆ’ 3)
𝑏 = βˆ’4
𝑐 = 1
Put the values,
(βˆ’4)2 βˆ’ 4(π‘š βˆ’ 3)(1) = 0
16 βˆ’ 4π‘š + 12 = 0
βˆ’4π‘š + 28 = 0
βˆ’4π‘š = βˆ’28
π‘š = 7
Put the value of π‘š in above equation,
(7 βˆ’ 3)π‘₯2 βˆ’ 4π‘₯ + 1 = 0
4π‘₯2 βˆ’ 4π‘₯ + 1 = 0
4π‘₯2 βˆ’ 2π‘₯ βˆ’ 2π‘₯ + 1 = 0
2π‘₯(2π‘₯ βˆ’ 1) βˆ’ 1(2π‘₯ βˆ’ 1) = 0
(2π‘₯ βˆ’ 1)(2π‘₯ βˆ’ 1) = 0
2π‘₯ βˆ’ 1 = 0
π‘₯ = 1/2
Hence, the value of π‘₯ is 1/2.
(ii) 3π‘₯2 + 12π‘₯ + (π‘š + 7) = 0
It is given that, the above equation equal roots so 𝐷 = 0
We know that,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝑏2 βˆ’ 4π‘Žπ‘ = 0
Here,
π‘Ž = 3
𝑏 = 12
𝑐 = (π‘š + 7)
Put the values,
(12)2 βˆ’ 4(3)(π‘š + 7) = 0
144 βˆ’ 12(π‘š + 7) = 0
144 βˆ’ 12π‘š βˆ’ 84 = 0
βˆ’12π‘š + 60 = 0
βˆ’12π‘š = βˆ’60
π‘š = 5
Put the value of π‘š in above equation,
3π‘₯2 + 12π‘₯ + (5 + 7) = 0
3π‘₯2 + 12π‘₯ + 12 = 0
3(π‘₯2 + 4π‘₯ + 4) = 0
π‘₯2 + 4π‘₯ + 4 = 0
π‘₯2 + 2π‘₯ + 2π‘₯ + 4 = 0
π‘₯(π‘₯ + 2) + 2(π‘₯ + 2) = 0
π‘₯(π‘₯ + 2) + 2(π‘₯ + 2) = 0
π‘₯ + 2 = 0
π‘₯ = βˆ’2
Hence, the value of π‘₯ is βˆ’2.
(iii) π‘₯2 βˆ’ (π‘š + 2)π‘₯ + (π‘š + 5) = 0
It is given that, the above equation equal roots so 𝐷 = 0
We know that,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝑏2 βˆ’ 4π‘Žπ‘ = 0
Here,
π‘Ž = 1
𝑏 = βˆ’(π‘š + 2)
𝑐 = (π‘š + 5)
Put the values,
[βˆ’(π‘š + 2)]2 βˆ’ 4(1)(π‘š + 5) = 0
[π‘š2 + 2 Γ— π‘š Γ— (2) + 4]2 βˆ’ 4(π‘š + 5) = 0
π‘š2 + 4π‘š + 4 βˆ’ 4π‘š βˆ’ 20 = 0
π‘š2 βˆ’ 16 = 0
π‘š2 = 16
π‘š = Β±4
Put the value of π‘š as 4 in above equation,
π‘₯2 βˆ’ (4 + 2)π‘₯ + (4 + 5) = 0
π‘₯2 βˆ’ 6π‘₯ + 9 = 0
π‘₯2 βˆ’ 3π‘₯ βˆ’ 3π‘₯ + 9 = 0
π‘₯(π‘₯ βˆ’ 3) βˆ’ 3(π‘₯ βˆ’ 3) = 0
(π‘₯ βˆ’ 3)(π‘₯ βˆ’ 3) = 0
π‘₯ βˆ’ 3 = 0
π‘₯ = 3
Put the value of π‘š as -4 in above equation,
π‘₯2 βˆ’ (βˆ’4 + 2)π‘₯ + (βˆ’4 + 5) = 0
π‘₯2 + 2π‘₯ + 1 = 0
π‘₯2 βˆ’ π‘₯ βˆ’ π‘₯ + 1 = 0
π‘₯(π‘₯ βˆ’ 1) βˆ’ 1(π‘₯ βˆ’ 1) = 0
(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 1) = 0
π‘₯ βˆ’ 1 = 0
π‘₯ = 1
Hence, the value of π‘₯ is 3 or 1.

Question 19. Without solving the following quadratic equation, find the value of p for which the roots are equal. 𝑝π‘₯2 βˆ’ 4π‘₯ + 3 = 0
Solution:

It is given that, the above equation equal roots so 𝐷 = 0
We know that,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝑏2 βˆ’ 4π‘Žπ‘ = 0
𝑝π‘₯2 βˆ’ 4π‘₯ + 3 = 0
Here,
π‘Ž = 𝑝
𝑏 = βˆ’4
𝑐 = 3
Put the values,
(βˆ’4)2 βˆ’ 4(𝑝)(3) = 0
16 βˆ’ 12𝑝 = 0
βˆ’12𝑝 = βˆ’16
𝑝 = βˆ’16/βˆ’12
𝑝 = βˆ’16/βˆ’12
𝑝 = 4/3

Question 20. Without solving the following quadratic equation, find the value of β€˜m’ for which the given
equation has real and equal roots.
π‘₯Β² + 2(π‘šβ€“ 1)π‘₯ + (π‘š + 5) = 0
Solution:

It is given that, the above equation equal roots so 𝐷 = 0
We know that,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝑏2 βˆ’ 4π‘Žπ‘ = 0
π‘₯Β² + 2(π‘šβ€“ 1)π‘₯ + (π‘š + 5) = 0
Here,
π‘Ž = 1
𝑏 = 2(π‘š βˆ’ 1)
𝑐 = (π‘š + 5)
Put the values,
[2(π‘š βˆ’ 1)]2 βˆ’ 4(1)(π‘š + 5) = 0
[4(π‘š2 βˆ’ 2(π‘š)(1) + (1)2]2 βˆ’ 4π‘š βˆ’ 20 = 0
4(π‘š2 βˆ’ 2π‘š + 1) βˆ’ 4π‘š βˆ’ 20 = 0
4π‘š2 βˆ’ 8π‘š + 4 βˆ’ 4π‘š βˆ’ 20 = 0
4π‘š2 βˆ’ 12π‘š βˆ’ 16 = 0
4(π‘š2 βˆ’ 3π‘š βˆ’ 4) = 0
π‘š2 βˆ’ 3π‘š βˆ’ 4 = 0
π‘š2 βˆ’ 4π‘š + π‘š βˆ’ 4 = 0
π‘š(π‘š βˆ’ 4) + 1(π‘š βˆ’ 4) = 0
(π‘š + 1)(π‘š βˆ’ 4) = 0
π‘š + 1 = 0           π‘š βˆ’ 4 = 0
π‘š = βˆ’1               π‘š = 4
Hence, the value of m is -1 and 4.

Exercise 5F

Question 1. Solve:
(i) (π‘₯ + 5)(π‘₯– 5) = 24
(ii) 3π‘₯2 βˆ’ 2√6π‘₯ + 2 = 0
(ii) 3√2π‘₯2 βˆ’ 5π‘₯ + √26 = 0
Solution:

(i) (π‘₯ + 5)(π‘₯ – 5) = 24
π‘₯2 – (5)2 = 24
We know that,
(π‘Ž – 𝑏)(π‘Ž + 𝑏) = π‘Ž2– 𝑏2
π‘₯2– 25 = 24
π‘₯2 = 49
π‘₯ = Β± 7
(ii) 3π‘₯2 – 2√6π‘₯ + 2 = 0
3π‘₯2– √6π‘₯– √6π‘₯ + 2 = 0
√3π‘₯(√3π‘₯– √2)– √2(√3π‘₯ βˆ’ √2) = 0
(√3π‘₯– √2)(√3π‘₯ βˆ’ √2) = 0
(√3π‘₯– √2)(√3π‘₯ βˆ’ √2) = 0
√3π‘₯– √2 = 0
π‘₯ = √2/√3
(iii) 3√2π‘₯2 βˆ’ 5π‘₯ βˆ’ √2 = 0
3√2π‘₯2 βˆ’ 6π‘₯ + π‘₯ βˆ’ √2 = 0
3√2π‘₯(π‘₯ βˆ’ √2) + 1(π‘₯ βˆ’ √2) = 0
(3√2π‘₯ + 1)(π‘₯ βˆ’ √2) = 0
3√2π‘₯ + 1 =           0 π‘₯ βˆ’ √2 = 0
3√2π‘₯ = βˆ’1               π‘₯ = √2
π‘₯ = βˆ’ 1/3√2

Question 2. One root of the quadratic equation 8π‘₯2 + π‘šπ‘₯ + 15 is 3/4. Find the value of m. Also, find the other root of the equation.
Solution:

It is given that,
The one root of equation is 3/4.
Quadratic equation is 8π‘₯2 + π‘šπ‘₯ + 15 = 0
Put the value of π‘₯ as 3/4 in given equation,

Put the value of π‘š in given equation,
8π‘₯2 + π‘šπ‘₯ + 15 = 0
8π‘₯2 + (βˆ’26)π‘₯ + 15 = 0
8π‘₯2 βˆ’ 26π‘₯ + 15 = 0
8π‘₯2 βˆ’ 20π‘₯ βˆ’ 6π‘₯ + 15 = 0
4π‘₯(2π‘₯ βˆ’ 5) βˆ’ 3(2π‘₯ βˆ’ 5) = 0
(4π‘₯ βˆ’ 3)(2π‘₯ βˆ’ 5) = 0
4π‘₯ βˆ’ 3 = 0           2π‘₯ βˆ’ 5 = 0
π‘₯ = 3/4                  π‘₯ = 5/2
Hence, the other root of given equation is 5/2.

Question 3. One root of the quadratic equation π‘₯2 – 3π‘₯– 2π‘Žπ‘₯ – 6π‘Ž = 0 is βˆ’3, find its other root.
Solution:

It is given that,
The one root of equation is βˆ’3.
Quadratic equation is π‘₯2 – 3π‘₯– 2π‘Žπ‘₯ – 6π‘Ž = 0
Put the value of π‘₯ as βˆ’3 in given equation,
π‘₯2 – 3π‘₯ – 2π‘Žπ‘₯ – 6π‘Ž = 0
π‘₯(π‘₯ + 3) – 2π‘Ž(π‘₯ + 3) = 0
(π‘₯– 2π‘Ž)(π‘₯ + 3) = 0
π‘₯ βˆ’ 2π‘Ž = 0           π‘₯ + 3 = 0
π‘₯ = 2π‘Ž                 π‘₯ = βˆ’3
Hence, the other root of the given equation is 2a.

Question 4. If 𝑝– 15 = 0 and 2π‘₯2 + 15π‘₯ + 15 = 0; find the values of π‘₯.
Solution:

It is given that,
𝑝– 15 = 0
𝑝 = 15
Put the value of 𝑝 in given equation,
2π‘₯2 + 15π‘₯ + 15 = 0
2π‘₯ + 10π‘₯ + 5π‘₯ + 15 = 0
2π‘₯(π‘₯ + 5) + 5(π‘₯ + 5)
(2π‘₯ + 5)(π‘₯ + 5) = 0
2π‘₯ + 5 = 0           π‘₯ + 5 = 0
π‘₯ = βˆ’5/2            π‘₯ = βˆ’5
Hence, the value of π‘₯ is βˆ’5/2 and -5.

Question 5. Find the solution of the equation 2π‘₯2 βˆ’ π‘šπ‘₯ – 25𝑛 = 0; if π‘š + 5 = 0 and 𝑛 – 1 = 0.
Solution:

It is given that,
π‘š + 5 = 0
π‘š = βˆ’5
And
𝑛 – 1 = 0
𝑛 = 1
Put the value of π‘š and 𝑛 in given equation
2π‘₯2 βˆ’ π‘šπ‘₯ – 25𝑛 = 0
2π‘₯2 βˆ’ (βˆ’5)π‘₯ – 25(1) = 0
2π‘₯2 + 5π‘₯ – 25 = 0
2π‘₯ + 10π‘₯– 5π‘₯– 25 = 0
2π‘₯(π‘₯ + 5) βˆ’ 5(π‘₯ + 5) = 0
(π‘₯ + 5)(2π‘₯– 5) = 0
π‘₯ + 5 = 0 2π‘₯ βˆ’ 5 = 0
π‘₯ = βˆ’5 π‘₯ = 5/2
Hence, the value of π‘₯ is -5 and 5/2.π‘₯2

Question 6. If m and n are roots of the equation (1/π‘₯) βˆ’ (1/π‘₯βˆ’2) = 3 where π‘₯ β‰  0 and π‘₯ β‰  2; find π‘š Γ— 𝑛.
Solution:

Question 7. Solve, using formula: π‘₯2 + π‘₯ – (π‘Ž + 2)(π‘Ž + 1) = 0
Solution:

It is given that,
π‘₯2 + π‘₯ – (π‘Ž + 2)(π‘Ž + 1) = 0
By the quadratic formula
π‘Ž = 1
𝑏 = 1
𝑐 = – (π‘Ž + 2)(π‘Ž + 1)

Question 8. Solve the quadratic equation 8π‘₯2– 14π‘₯ + 3 = 0
(i) When π‘₯ ∈ 𝐼 (integers)
(ii) When π‘₯ ∈ 𝑄 (rational numbers)
Solution:

It is given that,
8π‘₯2– 14π‘₯ + 3 = 0
8π‘₯2– 12π‘₯ βˆ’ 2π‘₯ + 3 = 0
4π‘₯(2π‘₯ – 3) – (2π‘₯– 3) = 0
(4π‘₯– 1)(2π‘₯– 3) = 0
4π‘₯ βˆ’ 1 = 0         2π‘₯ βˆ’ 3 = 0
π‘₯ = 1/4                 π‘₯ = 3/2
(i) When π‘₯ ∈ 𝐼 (integers) the given 8π‘₯2– 14π‘₯ + 3 = 0 has no equal roots.
(ii) When π‘₯ ∈ 𝑄 (rational numbers) the given 8π‘₯2– 14π‘₯ + 3 = 0 are π‘₯ = 1/4 , 3/2
.
Question 9. Find the value of m for which the equation (π‘š + 4 )2 + (π‘š + 1)π‘₯ + 1 = 0 has real and equal roots.
Solution:

Given quadratic equation is (m + 4 )2 + (m + 1)x + 1 = 0
The quadratic equation has real and equal roots if its discriminant is zero.
β‡’ D = 𝑏2 – 4ac = 0
β‡’ (m + 1)2 -4(m + 4)(1) = 0
β‡’ m2 + 2m + 1 – 4m – 16 = 0
β‡’ m2 – 2m – 15 = 0
β‡’ m2 – 5m + 3m – 15 = 0
β‡’ m(m – 5) +3(m =5) = 0
β‡’ (m – 5)(m + 3) = 0
β‡’ m = 5 or m = -3

Question 10. Find the values of m for which equation 3π‘₯2 + mπ‘₯ + 2 = 0 has equal roots. Also, find the roots of the given equation.
Solution:

It is given that, the above equation equal roots so 𝐷 = 0
We know that,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝑏2 βˆ’ 4π‘Žπ‘ = 0
3π‘₯Β² + π‘šπ‘₯ + 2 = 0
Here,
π‘Ž = 3
𝑏 = π‘š
𝑐 = 2
Put the values,
m2 βˆ’ 4(2)(3) = 0
m2 βˆ’ 24 = 0
m2 = 24
π‘š = √24
π‘š = Β±2√6
Put the value of π‘š as 2√6 in given equation,
3π‘₯Β² + 2√6π‘₯ + 2 = 0
(√3π‘₯)Β² + 2(√3π‘₯)(√2) + (√2)2 = 0
(√3π‘₯ + √2)Β² = 0
√3π‘₯ + √2 = 0
√3π‘₯ = βˆ’βˆš2

Question 11. Find the value of k for which equation 4π‘₯2 + 8π‘₯ – π‘˜ = 0 has real roots.
Solution:

It is given that, the above equation equal roots so 𝐷 β‰₯ 0
We know that,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝑏2 βˆ’ 4π‘Žπ‘ β‰₯ 0
4π‘₯Β² + 8π‘₯ βˆ’ π‘˜ β‰₯ 0
Here,
π‘Ž = 4
𝑏 = 8
𝑐 = βˆ’π‘˜
Put the values,
(8)2 βˆ’ 4(4)(βˆ’π‘˜) β‰₯ 0
64 + 16π‘˜ β‰₯ 0
16π‘˜ β‰₯ βˆ’64
π‘˜ β‰₯ βˆ’64/16
π‘˜ β‰₯ βˆ’4
Hence, the value of π‘˜ is βˆ’4 when given equation has real roots.

Question 12. Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(𝑖) 3π‘₯2 – 5π‘₯ + 2 = 0
(𝑖𝑖) π‘₯2 + 4π‘₯ + 5 = 0
Solution:

(i) 3π‘₯2 – 5π‘₯ + 2 = 0
Here,
π‘Ž = 3
𝑏 = βˆ’5
𝑐 = 2
Put the values,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝐷 = (βˆ’5)2 – 4(3)(2)
𝐷 = 25 – 24
𝐷 = 1
We know that,
𝐷 > 0, so the roots of the quadratic equation are real and distinct.
By quadratic formula

(ii) π‘₯2 + 4π‘₯ + 5 = 0
Here,
π‘Ž = 1
𝑏 = 4
𝑐 = 5
Put the values,
𝐷 = 𝑏2 βˆ’ 4π‘Žπ‘
𝐷 = (4)2 βˆ’ 4(1)(5)
𝐷 = 16 βˆ’ 20
𝐷 = βˆ’4
Hence, 𝐷 > 0, so the roots of the quadratic equation does not exist.

Question 13. Solve:

48π‘₯ = 324– π‘₯2
π‘₯2 + 48π‘₯ βˆ’ 324 =
π‘₯2 + 48π‘₯ – 324 = 0
π‘₯2 + 54π‘₯ – 6π‘₯ – 324 = 0
π‘₯(π‘₯ + 54) βˆ’ 6(π‘₯ + 54) = 0
(π‘₯ + 54)(π‘₯ – 6) = 0
π‘₯ = βˆ’54 π‘œπ‘Ÿ π‘₯ = 6
Hence, the value of π‘₯ > 0.

Selina ICSE Class 10 Maths Solutions Chapter 5 Quadratic Equations