Question 1. Find which of the following equations are quadratic:
(π) (3π₯β 1)2 = 5(π₯ + 8)
(ππ) 5π₯2 β 8π₯ = β3(7β 2π₯)
(πππ) (π₯β 4)(3π₯ + 1) = (3π₯β 1)(π₯ + 2)
(ππ£) π₯2 + 5π₯β 5 = (π₯β 3)2
(π£) 7π₯3β 2π₯2+ 10 = (2π₯β 5)2
(π£π) (π₯ β 1)2 + (π₯ + 2)2 + 3(π₯ + 1) = 0
Solution:
(π) (3π₯β 1)2 = 5(π₯ + 8)
We know that,
(π β π)2 = π2 β 2ππ + π2
(3π₯)2 β 2(3π₯)(1) + (1)2 = 5π₯ + 40
9π₯2 β 6π₯ + 1 = 5π₯ + 40
9π₯2 β 6π₯ β 5π₯ = 40 β 1
9π₯2 β 6π₯ β 5π₯ = 39
9π₯2 β 11π₯ β 39 = 0
Above equation is the form of ππ₯2 + ππ₯ + π = 0
Hence, the given equation is a quadratic equation.
(ππ) 5π₯2 β 8π₯ = β3(7β 2π₯)
5π₯2 β 8π₯ = β21 + 6π₯
5π₯2 β 8π₯ β 6π₯ = β21
5π₯2 β 14π₯ + 21 = 0
5π₯2β 14π₯ + 21 = 0
Above equation is the form of ππ₯2 + ππ₯ + π = 0
Hence, the given equation is a quadratic equation.
(πππ) (π₯ β 4)(3π₯ + 1) = (3π₯β 1)(π₯ + 2)
(π₯ β 4)(3π₯ + 1) = (3π₯β 1)(π₯ + 2)
3π₯2 + π₯ β 12π₯ β 4 = 3π₯2 + 6π₯ β π₯ β 2
3π₯2 β 11π₯β 4 = 3π₯2 + 5π₯ β 2
β11π₯β 5π₯ = β 2 + 4
β16π₯ = 2
β16π₯ β 2 = 0
16π₯ + 2 = 0
Above equation is not in the form of ππ₯2 + ππ₯ + π = 0
Hence, the given equation is not a quadratic equation.
(ππ£) π₯2 + 5π₯β 5 = (π₯β 3)2
We know that,
(π β π)2 = π2 β 2ππ + π2
π₯2 + 5π₯β 5 = π₯2 β 2(π₯)(3) + 32
π₯2 + 5π₯β 5 = π₯2 β 6π₯ + 9
5π₯ + 6π₯ = 9 + 5
11π₯ = 14
11π₯β 14 = 0
Above equation is not in the form of ππ₯2 + ππ₯ + π = 0
Hence, the given equation is not a quadratic equation.
(π£) 7π₯3β 2π₯2 + 10 = (2π₯β 5)2
We know that,
(π β π)2 = π2 β 2ππ + π2
7π₯3 β 2π₯2 + 10 = (2π₯)2 β 2(2π₯)(5) + (5)2
7π₯3 β 2π₯2 + 10 = 4π₯2 β 20π₯ + 25
7π₯3 β 2π₯2 β 4π₯2 + 10 + 20π₯ β 25 = 0
7π₯3β 6π₯2 + 20π₯β 15 = 0
Above equation is not in the form of ππ₯2 + ππ₯ + π = 0
Hence, the given equation is not a quadratic equation.
(π£π) (π₯ β 1)2 + (π₯ + 2)2 + 3(π₯ + 1) = 0
We know that,
(π β π)2 = π2 β 2ππ + π2
(π + π)2 = π2 + 2ππ + π2
π₯2 β 2(π₯)(1) + (1)2 + π₯2 + 2(π₯)(2) + 22 + 3π₯ + 3 = 0
π₯2 β 2π₯ + 1 + π₯2 + 4π₯ + 4 + 3π₯ + 3 = 0
2π₯2 + 5π₯ + 8 = 0
Above equation is the form of ππ₯2 + ππ₯ + π = 0
Hence, the given equation is a quadratic equation.
Question 2. (i) Is x = 5 a solution of the quadratic equation π₯2β 2π₯β 15 = 0?
Solution:
It is given that,
π₯2 β 2π₯ β 15 = 0
π₯ = 5
Put the value of x in given equation
L.H.S = (5)2 β 2(5) β 15
25 β 10 β 15 = 0
L.H.S = R.H.S
Thus, π₯ = 5 is a solution of the quadratic equation π₯2 β 2π₯ β 15 = 0
Question 2. (ii) Is x = -3 a solution of the quadratic equation 2π₯2 β 7π₯ + 9 = 0?
Solution:
It is given that,
2π₯2 β 7π₯ + 9 = 0
π₯ = β3
Put the value of x in given equation
L.H.S = 2(β3)2 β 7(β3) + 9
2(9) + 21 + 9
18 + 21 + 9 = 48
L.H.S. β R.H.S
Thus, π₯ = β3 is not a solution of the quadratic equation 2π₯2 β 7π₯ + 9 = 0
Question 3. If β(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 = β(2/3)
Put the value of x in given equation

Question 4. 2/3 and 1 are the solutions of equation ππ₯2 + ππ₯ + 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 + 6π + 54 = 0β¦β¦β¦β¦β¦β¦β¦β¦(i)
Put the value of π₯ = 1 in the given equation.
π(1)2 + π(1) + 6 = 0
π + π + 6 = 0
π + π + 6 = 0β¦β¦β¦β¦β¦β¦β¦β¦(ii)
Get the value of π from equation (ii)
π = β6 β πβ¦β¦β¦β¦β¦(iii)
Put the value of π in equation (i)
4(β6 β π) + 6π + 54 = 0
β24 β 4π + 6π + 54 = 0
2π + 30 = 0
2π = β30
π = β30/2
π = β15
Put the value of π in equation (iii)
π = β6 β (β15)
π = β6 + 15
π = 9
Hence, the value of π is -15 and π is 9.
Question 5. If 3 and -3 are the solutions of equation ππ₯2 + ππ₯β 9 = 0. Find the values of π and π.
Solution:
It is given that,
ππ₯2 + ππ₯β 9 = 0
Put the value of π₯ = 3 in the given equation.
π(3)2 + π(3)β 9 = 0
9π + 3πβ 9 = 0
3(3π + πβ 3) = 0
3π + πβ 3 = 0β¦β¦β¦β¦β¦β¦..(i)
Put the value of π₯ = β3 in the given equation.
π(β3)2 + π(β3)β 9 = 0
9π β 3πβ 9 = 0
3(3π β πβ 3) = 0
3π β πβ 3 = 0β¦β¦β¦β¦β¦β¦..(ii)
Get the value of π from equation (ii)
π = π+3/3 β¦β¦β¦β¦β¦(iii)
Put the value of π in equation (i)
3 (π+3/3 ) + πβ 3 = 0
π + 3 + πβ 3 = 0
2π = 0
π = 0
Put the value of π in equation (iii)
π = 0+3/3
π = 3/3
π = 1
Hence, the value of π is 1 and π is 0.
Exercise 5B
Question 1. Without solving, comment upon the nature of roots of each of the following equations:
(π) 7π₯2β 9π₯ + 2 = 0
(ππ) 6π₯2β 13π₯ + 4 = 0
(πππ) 25π₯2β 10π₯ + 1 = 0
(ππ£) π₯2 + 2β3π₯ β 9 = 0
(π£) π₯2β ππ₯β π2 = 0
(π£π) 2π₯2 + 8π₯ + 9 = 0
Solution:
(i) It is given that,
7π₯2β 9π₯ + 2 = 0
π = 7
π = 9
π = 2
We know that by discriminate,
D = π2 β 4ππ
D = (9)2 β 4(7)(2)
D = 81 β 56
D = 25
Here π· > 0 then given equation have two real and unequal roots.
(ii) It is given that,
6π₯2β 13π₯ + 4 = 0
π = 6
π = β13
π = 4
We know that by discriminate,
D = π2 β 4ππ
D = (β13)2 β 4(6)(4)
D = 169 β 96
D = 73
Here π· > 0 then given equation have two real and unequal roots.
(iii) It is given that,
25π₯2β 10π₯ + 1 = 0
π = 25
π = β10
π = 1
We know that by discriminate,
D = π2 β 4ππ
D = (β10)2 β 4(25)(1)
D = 100 β 100
D = 0
Here π· = 0 then given equation have two real and unequal roots.
(iv) It is given that,
π₯2β 2β3π₯ β 9 = 0
π = 1
π = β2β3
π = β9
We know that by discriminate,
D = π2 β 4ππ
D = (β2β3)2 β 4(1)(β9)
D = 12 + 36
D = 48
Here π· > 0 then given equation have two real and unequal roots.
(v) It is given that,
π₯2β ππ₯ β π2 = 0
π = 1
π = βπ
π = βπ2
We know that by discriminate,
D = π2 β 4ππ
D = (βπ)2 β 4(1)(βπ2)
D = π2 + 4π2
Here π· > 0 then given equation have two real and unequal roots.
(vi) It is given that,
2π₯2β 8π₯ β 9 = 0
π = 2
π = 8
π = 9
We know that by discriminate,
D = π2 β 4ππ
D = (8)2 β 4(2)(9)
D = π2 + 4π2
Here π· > 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₯2β (πβ 2)π₯ + 1 = 0
Solution:
It is given that,
Quadratic equation has equal roots. So, D = 0
4π₯2β (πβ 2)π₯ + 1 = 0
π = 4
π = β(π β 2)
π = 1
We know that by discriminate,
D = π2 β 4ππ
0 = [β(π β 2)]2 β 4(4)(1)
0 = π2 β 2(π)(2) + 4 β 16
0 = π2 β 4π + 4 β 16
0 = π2 β 4π β 12
By splitting the middle term
π2 β 6π + 2π β 12 = 0
π(π β 6) + 2(π β 6) = 0
(π + 2)(π β 6) = 0
Then
π + 2 = 0 π β 6 = 0
π = β2 π = 6
Question 3. Find the value of βpβ, if the following quadratic equations have equal roots: π₯2 + (π β 3)π₯ + π = 0
Solution:
It is given that,
Quadratic equation has equal roots. So, D = 0
π₯2 + (π β 3)π₯ + π = 0
π = 1
π = (πβ 3)
π = π
We know that by discriminate,
D = π2 β 4ππ
0 = π2 β 4ππ
(π β 3)2 β 4(1)(π) = 0
(π β 3)2 β 4π = 0
π2 + 9β 6πβ 4π = 0
π2β 10π + 9 = 0
π2 β 9πβ π + 9 = 0
π(πβ 9) β 1(πβ 9) = 0
(π β 9)(πβ 1) = 0
Then
π β 9 = 0 π β 1 = 0
π = 9 π = 1
Question 4. The equation 3π₯2β 12π₯ + (πβ 5) = 0 has equal roots. Find the value of n.
Solution:
It is given that,
Quadratic equation has equal roots. So, D = 0
3π₯2β 12π₯ + (πβ 5) = 0
π = 3
π = β12
π = (π β 5)
We know that by discriminate,
D = π2 β 4ππ
0 = π2 β 4ππ
(β12)2 β 4(3)(π β 5) = 0
144 β 12(π β 5) = 0
144 β 12π + 60 = 0
144 β 12π + 60 = 0
β12π + 204 = 0
β12π = β204
12π = 204
π = 204/12
π = 17
Hence, the value of π is 17.
Question 5. Find the value of m, if the following equation has equal roots: (πβ 2)π₯2β (5 + π)π₯ + 16 = 0.
Solution:
It is given that,
Quadratic equation has equal roots. So, D = 0
(πβ 2)π₯2β (5 + π)π₯ + 16 = 0
π = (π β 2)
π = β(5 + π)
π = 16
We know that by discriminate,
D = π2 β 4ππ
0 = π2 β 4ππ
[β(5 + π)]2 β 4(π β 2)(16) = 0
(5)2 + 2(5)(π) + π2 β 64π + 128 = 0
25 + 10π + π2 β 64π + 128 = 0
β54π + π2 + 153 = 0
π2 β 54π + 153 = 0
π2 β 51π β 3π + 153 = 0
π(π β 51) β 3(π β 51) = 0
(π β 51)(π β 3) = 0
Then
π β 51 = 0 π β 3 = 0
π = 51 π = 3
Question 6. Find the value of p for which the equation 3π₯2β 6π₯ + π = 0 has distinct and real roots.
Solution:
It is given that,
Quadratic equation has real roots. So, π· > 0
3π₯2β 6π₯ + π = 0
π = 3
π = 6
π = π
We know that by discriminate,
D < π2 β 4ππ
0 < π2 β 4ππ
0 < (6)2 β 4(3)(π)
0 < 36 β 12π
12π < 36
π < 36/12
π < 3
Hence, the value of K is 3.
Exercise 5C
Question 1. Solve: π₯Β² β 10π₯β 24 = 0
Solution:
It is given that,
π₯Β² β 10π₯β 24 = 0
By the splitting the middle term,
π₯Β² β 12π₯ + 2π₯β 24 = 0
π₯(π₯ β 12) + 2(π₯ β 12) = 0
(π₯ + 2)(π₯ β 12) = 0
Then
π₯ + 2 = 0 π₯ β 12 = 0
π = β2 π₯ = 12
Question 2. Solve: π₯Β²β 16 = 0
Solution:
It is given that,
π₯Β²β 16 = 0
π₯Β²β (4)2 = 0
π₯Β²β (4)2 = 0
We know that,
(π2 β π2) = (π β π)(π + π)
(π₯Β² β 42) = (π₯ β 4)(π₯ + 4)
(π₯ β 4)(π₯ + 4) = 0
Then
π₯ β 4 = 0 π₯ + 4 = 0
π₯ = 4 π₯ = β4
Question 3. Solve: 2π₯Β² β 1/2 π₯ = 0
Solution:
It is given that,

Question 4. Solve: π₯(π₯β 5) = 24
Solution:
It is given that,
π₯(π₯β 5) = 24
π₯Β² β 5π₯ = 24
π₯Β² β 5π₯ β 24 = 0
By the splliting the middle term
π₯Β² β 5π₯ β 24 = 0
π₯Β² β 8π₯ + 3π₯ β 24 = 0
π₯(π₯ β 8) + 3(π₯ β 8) = 0
(π₯ + 3)(π₯ β 8) = 0
Then
π₯ + 3 = 0 π₯ β 8 = 0
π₯ = β3 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 β 9π₯ + 10
2π₯Β² β 9π₯ + 10 = 0
2π₯Β² β 5π₯ β 4π₯ + 10 = 0
π₯(2π₯ β 5) β 2(2π₯ β 5) = 0
(π₯ β 2)(2π₯ β 5) = 0
Then
π₯ β 2 = 0 2π₯ β 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₯ β 6 = 0
π₯Β² + 3π₯ β 2π₯ β 6 = 0
π₯(π₯ + 3) β 2(π₯ + 3) = 0
(π₯ β 2)(π₯ + 3) = 0
Than
π₯ β 2 = 0 π₯ + 3 = 0
π₯ = 2 π₯ = β3
Question 7. Solve: π₯ = 3π₯ + 1/4π₯
Solution:
It is given that,
π₯ = 3π₯+1/4π₯
π₯ Γ 4π₯ = 3π₯ + 1
4π₯Β² β 3π₯ β 1 = 0
4π₯Β² β 4π₯ + π₯ β 1 = 0
4π₯(π₯ β 1) β 1(π₯ β 1) = 0
(4π₯ β 1)(π₯ β 1) = 0
than
4π₯ β 1 = 0 π₯ β 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 β 5π₯ + 2 = 0
2π₯Β² β 4π₯ β π₯ + 2 = 0
2π₯(π₯ β 2) β 1(π₯ β 2) = 0
2π₯(π₯ β 2) β 1(π₯ β 2) = 0
(2π₯ β 1)(π₯ β 2) = 0
than
2π₯ β 1 = 0 π₯ β 2 = 0
π₯ = 1/2 π₯ = 2
Question 9. Solve: (2π₯β 3)Β² = 49
Solution:
It is given that,
(2π₯β 3)Β² = 49
2π₯β 3 = β49
2π₯β 3 = Β±7
Then
2π₯ β 3 = 7 2π₯ β 3 = β7
2π₯ = 7 + 3 2π₯ = β7 + 3
π₯ = 10/2 π₯ = β4/2
π₯ = 5 π₯ = β2
Question 10. Solve: 2(π₯Β²β 6) = 3(π₯ β 4)
Solution:
It is given that,
2(π₯Β²β 6) = 3(π₯ β 4)
2π₯Β²β 12 = 3π₯ β 12
2π₯Β²β 3π₯ = β12 + 12
2π₯Β²β 3π₯ = 0
π₯(2π₯β 3) = 0
Then
π₯ = 0 2π₯β 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 β π₯Β² + 8π₯ + 2π₯ β 3π₯ β 7π₯ + 8 β 21 = 0
π₯Β² + 10π₯ β 10π₯ β 13 = 0
π₯Β² β 13 = 0
(π₯)2 + (β13)2 = 0
(π₯)2 + (β13)2 = 0
(π₯ + β13)(π₯ β β13) = 0
Then
π₯ + β13 = 0 π₯ β β13 = 0
π₯ = ββ13 π₯ = β13
Question 12. Solve: π₯Β²β (π + π)π₯ + ππ = 0
Solution:
It is given that,
π₯Β²β (π + π)π₯ + ππ = 0
π₯Β²β ππ₯ β ππ₯ + ππ = 0
π₯(π₯β π) β π(π₯ β π) = 0
(π₯β π)(π₯ β π) = 0
Then
π₯ β π = 0 π₯ β π = 0
π₯ = π π₯ = π
Question 13. Solve: (π₯ + 3)Β²β 4(π₯ + 3)β 5 = 0
Solution:
It is given that,
(π₯ + 3)Β²β 4(π₯ + 3)β 5 = 0
We know that,
(π + π)2 = π2 + π2 + 2ππ
π₯2 + (3)2 + 2(π₯) (3)β 4π₯ β 12β 5 = 0
π₯2 + 9 + 6π₯β 4π₯ β 17 = 0
π₯2 + 2π₯ β 8 = 0
π₯2 + 4π₯ β 2π₯ β 8 = 0
π₯(π₯ + 4) β 2(π₯ + 4) = 0
(π₯ β 2)(π₯ + 4) = 0
Than
π₯ β 2 = 0 π₯ + 4 = 0
π₯ = 2 π₯ = β4
Question 14. Solve: 4(2π₯β 3)Β²β (2π₯β 3) β 14 = 0
Solution:
It is given that,
4(2π₯β 3)Β²β (2π₯β 3) β 14 = 0
Let us assumed that,
2π₯β 3 = π¦
4π¦Β²β π¦ β 14 = 0
4π¦Β²β 8π¦ + 7π¦ β 14 = 0
4π¦(π¦β 2) + 7(π¦β 2) = 0
(4π¦ + 7)(π¦β 2) = 0
Than
4π¦ + 7 = 0 π¦β 2 = 0
π¦ = β 7/4 π¦ = 2

Question 15. Solve:

Solution:
It is given that,

(3π₯ β 2)(π₯ + 4) = (3π₯ β 8) (2π₯ β 3)
3π₯2 + 12π₯ β 2π₯ β 8 = 6π₯2 β 9π₯ β 16π₯ + 24
3π₯2 β 6π₯2 + 12π₯ β 2π₯ + 9π₯ + 16π₯ β 8 β 24 = 0
β3π₯2 + 10π₯ + 25π₯ β 32 = 0
β3π₯2 + 35π₯ β 32 = 0
β(3π₯2 β 35π₯ + 32) = 0
3π₯2 β 35π₯ + 32 = 0
3π₯2 β 32π₯ β 3π₯ + 32 = 0
π₯(3π₯ β 32) β 1(3π₯ β 32) = 0
(π₯ β 1)(3π₯ β 32) = 0
Than
π₯ β 1 = 0 3π₯β 32 = 0
π¦ = 1 3π₯ = 32
π₯ = 32/3
π₯ = 10 (2/3)
Question 16. 2π₯2β 9π₯ + 10 = 0, πβππ
(i) π₯ β π
(ii) π₯ β π
Solution:
It is given that,
2π₯2β 9π₯ + 10 = 0
2π₯2β 5π₯ β 4π₯ + 10 = 0
π₯(2π₯β 5) β 2(2π₯ β 5) = 0
(π₯ β 2)(2π₯β 5) = 0
Than
π₯ β 2 = 0 2π₯β 5 = 0
π¦ = 2 2π₯ = 5
π₯ = 5/2
Question 17. Solve

Solution:
It is given that,

2(2π₯2 + 18) = 5(π₯2 β 9)
4π₯2 + 36 = 5π₯2 β 45
4π₯2 β 5π₯2 = β45 β 36
βπ₯2 = β81
π₯2 β 81 = 0
π₯2 β (9)2 = 0
(π₯ β 9)(π₯ + 9) = 0
Than
π₯ β 9 = 0 π₯ + 9 = 0
π¦ = 9 π₯ = β9
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 β 4π₯2 β 20π₯ β 24 = 0
2π₯2 + 3π₯ β 14 = 0
2π₯2 + 7π₯ β 4π₯ β 14 = 0
π₯(2π₯ + 7) β 2(2π₯ + 7) = 0
(π₯ β 2) (2π₯ + 7) = 0
Than
π₯ β 2 = 0 2π₯ + 7 = 0
π₯ = 2 2π₯ = β7
π₯ = β 7/2
Question 19. Solve:Q20. Solve:

Solution:
It is given that,

π₯(2π₯ + 36) = 4(π₯2 + 4π₯ β 12)
2π₯2 + 36π₯ = 4π₯2 + 16π₯ β 48
2π₯2 β 4π₯2 + 36π₯ β 16π₯ + 48 = 0
β2π₯2 + 20π₯ + 48 = 0
β2(π₯2 β 10π₯ β 24) = 0
π₯2 β 10π₯ β 24 = 0
π₯2 β 12π₯ + 2π₯ β 24 = 0
π₯(π₯ β 12) + 2(π₯ β 12) = 0
(π₯ + 2) (π₯ β 12) = 0
Than
π₯ + 2 = 0 π₯ β 12 = 0
π¦ = β2 π₯ = 12
Question 20. Solve:

Solution:
It is given that,

8(π₯2 β 4) = 7(π₯2 β 1)
8π₯2 β 32 = 7π₯2 β 7
8π₯2 β 7π₯2 = β7 + 32
π₯2 = 25
π₯2 = Β±β25
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₯ β 3 = 0 π₯ β 5 = 0
(π₯ β 3)(π₯ β 5) = 0
π₯2 β 5π₯ β 3π₯ + 15 = 0
π₯2 β 8π₯ + 15 = 0
Hence, the required equation is π₯2 β 8π₯ + 15 = 0.
(ii) It is given that,
The solution set is {β2, 3}
π₯ = β2 π₯ = 3
π₯ + 2 = 0 π₯ β 3 = 0
(π₯ + 2)(π₯ β 3) = 0
π₯2 β 3π₯ + 2π₯ β 6 = 0
π₯2 β π₯ β 6 = 0
Hence, the required equation is π₯2 β π₯ β 6 = 0.
(iii) It is given that,
The solution set is {5, β4}
π₯ = 5 π₯ = β4
π₯ β 5 = 0 π₯ + 4 = 0
(π₯ β 5)(π₯ + 4) = 0
π₯2 + 4π₯ β 5π₯ β 20 = 0
π₯2 β π₯ β 20 = 0
Hence, the required equation is π₯2 β π₯ β 20 = 0.
(iv) It is given that,

π₯ = β3 π₯ = β2/5
π₯ + 3 = 0 π₯ + 2
5 = 0
5π₯ + 2 = 0
(π₯ + 3)(5π₯ + 2) = 0
5π₯2 + 2π₯ + 15π₯ + 6 = 0
5π₯2 β 17π₯ + 6 = 0
Hence, the required equation is 5π₯2 β 17π₯ + 6 = 0.
Question 22. Solve:

Solution:
It is given that,

15(6π₯ β π₯2 + 9) = 3(6 β π₯)(12 + 2π₯)
5(6π₯ β π₯2 + 9) = (6 β π₯)(12 + 2π₯)
5(6π₯ β π₯2 + 9) = 72 + 12π₯ β 12π₯ β 2π₯2
30π₯ β 5π₯2 + 45 = 72 + 12π₯ β 12π₯ β 2π₯2
30π₯ β 5π₯2 + 45 = 72 β 2π₯2
30π₯ β 5π₯2 + 2π₯2 + 45 β 72 = 0
30π₯ β 3π₯2 + 27 = 0
β3π₯2 + 30π₯ + 27 = 0
β3(π₯2 β 10π₯ + 9) = 0
π₯2 β 10π₯ + 9 = 0
π₯2 β 9π₯ β π₯ + 9 = 0
π₯(π₯ β 9) β 1(π₯ β 9) = 0
(π₯ β 1)(π₯ β 9) = 0
Than
π₯ β 1 = 0 π₯ β 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 ββ1143 isn’t possible.
Hence, there is no real value of π₯.
Question 24. Find the value of π₯, if π + 1 = 0 and π₯2 + ππ₯β 6 = 0.
Solution:
It is given that,
π + 1 = 0
π = β1
Put the value of π in given equation,
π₯2 + ππ₯β 6 = 0
π₯2 + 1π₯β 6 = 0
π₯2 + π₯β 6 = 0
π₯2 β 3π₯ + 2π₯β 6 = 0
π₯(π₯ β 3) + 2(π₯β 3) = 0
(π₯ + 2)(π₯ β 3) = 0
Than
π₯ + 2 = 0 π₯ β 3 = 0
π₯ = β2 π₯ = 3
Question 25. Find the value of π₯, if π + 7 = 0; π + 10 = 0 and 12π₯2 = ππ₯β π.
Solution:
It is given that,
π + 7 = 0
π = β7
and
π + 10 = 0
π = β10
Put the value of π and π in the given equation,
12π₯2 = ππ₯β π
12π₯2 = (β7)π₯β (β10)
12π₯2 = β7π₯ + 10
12π₯2 + 7π₯ β 10 = 0
12π₯2 + 15π₯ β 8π₯ β 10 = 0
3π₯(4π₯ + 5) β 2(4π₯ + 5) = 0
(3π₯ β 2)(4π₯ + 5) = 0
Than
3π₯ β 2 = 0 4π₯ + 5 = 0
π₯ = 2/3 π₯ = β(5/4)
Question 26. Use the substitution π¦ = 2π₯ + 3 to solve for π₯, if 4(2π₯ + 3)2β (2π₯ + 3)β 14 = 0.
Solution:
It is given that,
π¦ = 2π₯ + 3β¦β¦β¦β¦β¦.(i)
Put the value of (2π₯ + 3) as π¦ in given equation,
4π¦2β π¦β 14 = 0
4π¦2 + 7π¦ β 8π¦β 14 = 0
π¦(4π¦ + 7) β 2(4π¦ + 7) = 0
(π¦ β 2)(4π¦ + 7) = 0
Than
π¦ β 2 = 0 4π¦ + 7 = 0
π¦ = 2 π¦ = β(7/4)
Put the value of π¦ in equation (i)
π¦ = 2
π¦ = 2π₯ + 3
2 = 2π₯ + 3
2 β 3 = 2π₯

Question 27. Without solving the quadratic equation 6π₯2 β π₯ β 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₯ = β1 is a root of the equation π₯2 β 3π₯ + 2 = 0 or not.
Solution:
It is given that,
π₯ = β1
Put the value π₯ in given equation,
π₯2 β 3π₯ + 2 = 0
(β1)2 β 3(β1) + 2 = 0
1 + 3 + 2 = 0
6 β 0
L.H.S. β R.H.S.
Hence, π₯ = β1 is not a solution of the given equation.
Question 29. If π₯ = 2/3 is a solution of the quadratic equation 7π₯2 + ππ₯β 3 = 0; Find the value of m.
Solution:
It is given that,
π₯ = 2/3
Put the value π₯ in given equation,
7π₯2 + ππ₯ β 3 = 0

Question 30. If π₯ = β3 and π₯ = 2/3 are solutions of quadratic equation ππ₯2 + 7π₯ + π = 0, find the values of m and n.
Solution:
It is given that,
π₯ = β3
Put the value of x in given equation,
ππ₯2 + 7π₯ + π = 0
π(β3)2 + 7(β3) + π = 0
9π β 21 + π = 0
9π + π = 21_____________(i)
And
π₯ = 2/3
Put the value of x in given equation,
ππ₯2 + 7π₯ + π = 0

4π + 42 + 9π = 0
4π + 9π = β42_____________(ii)
From equation first get the value of π
9π + π = 21
π = 21 β 9π__(iii)
Value of n put in equation (ii)
4π + 9π = β42
4π + 9(21 β 9π) = β42
4π + 189 β 81π = β42
4π β 81π = β42 β 189
β77π = β231
π = 231/77
π = 3
Put the value of π in equation (iii)
π = 21 β 9π
π = 21 β 9(3)
π = 21 β 27
π = β6
Question 31. If quadratic equation π₯2β (π + 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₯2β (π + 1) π₯ + 6 = 0
(3)2β (π + 1) 3 + 6 = 0
9β 3π β 3 + 6 = 0
β 3π + 12 = 0
β 3π = β12
π = β12/β3
π = 4
Put the value of π in given equation,
π₯2β (π + 1) π₯ + 6 = 0
π₯2β (4 + 1)π₯ + 6 = 0
π₯2β 5π₯ + 6 = 0
π₯2β 3π₯ β 2π₯ + 6 = 0
π₯(π₯β 3) β 2(π₯ β 3) = 0
(π₯ β 2)(π₯ β 3) = 0
Than
π₯ β 2 = 0 π₯ β 3 = 0
π₯ = 2 π₯ = 3
Question 32. Given that 2 is a root of the equation 3π₯Β²β π(π₯ + 1) = 0 and that the equation ππ₯Β²β ππ₯ + 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)Β²β π(2 + 1) = 0
3(2)Β²β π(3) = 0
12β 3π = 0
β 3π = β12
π = 12/3
π = 4
Put the value of π in other equation
ππ₯Β²β ππ₯ + 9 = 0
4π₯Β²β ππ₯ + 9 = 0
It is given that the roots are equal, we know
π2 β 4ππ = 0
Here,
π = 4
π = βπ
π = 9
(βπ)2 β 4(4)(9) = 0
π2 β 144 = 0
π2 = 144
π = β144
π = 12
Hence, the value of π and π is 4 and 12 respectively.
Question 33. Solve:

Solution:
It is given that,

π₯2 = ππ + π2 + π2 + ππ
π₯2 = π2 + π2 + 2ππ
π₯2 = (π + π)2
π₯ = π + π
Question 34. Solve:

Solution:
It is given that,

(600 + π₯)(π₯ β 10) = 630π₯
600π₯ β 6000 + π₯2 β 10π₯ = 630π₯
600π₯ β 6000 + π₯2 β 10π₯ β 630π₯ = 0
β6000 + π₯2 β 10π₯ β 30π₯ = 0
β6000 + π₯2 β 40π₯ = 0
π₯2 β 40π₯ β 6000 = 0
π₯2 β 100π₯ + 60π₯ β 6000 = 0
π₯2 β 100π₯ + 60π₯ β 6000 = 0
π₯(π₯ β 100) + 60(π₯ β 100) = 0
(π₯ + 60)(π₯ β 100) = 0
Than
π₯ + 60 = 0 π₯ β 100 = 0
π₯ = β60 π₯ = 100
Question 35. If β1 and 3 are the roots of π₯2 + ππ₯ + π = 0, find the values of p and q.
Solution:
It is given that,
π₯ = β1
Put the value of π₯ in given equation,
(β1)2 + π(β1) + π = 0
1 β π + π = 0
βπ + π = β1_______________ (i)
Also
π₯ = 3
Put the value of π₯ in given equation,
(3)2 + π(3) + π = 0
9 + 3π + π = 0
3π + π = β9______________(ii)
From equation (i) we get the value of π
βπ + π = β1
π = β1 + π__ (iii)
Put the value of π in equation (ii)
3π + (β1 + π) = β9
3π + π = β9 + 1
4π = β8
π = β 8/4
π = β2
Put the value of π in equation (iii)
π = β1 + π
π = β1 + (β2)
π = β1 β 2
π = β3
Hence, the value of π is β2 and π is β3.
Exercise 5D
Question 1. Solve each of the following equations using the formula:
(i) π₯2 β 6π₯ = 27
(ii) π₯2 β 10π₯ + 21 = 0
(iii) π₯2 + 6π₯ β 10 = 0
(iv) π₯2 + 2π₯ β 6 = 0
(v) 3π₯2 + 2π₯ β 1 = 0
(vi) 2π₯2 + 7π₯ + 5 = 0

Solution:
(i) π₯2 β 6π₯ = 27
π₯2 β 6π₯ β 27 = 0
π = 1
π = β6
π = β27
By the quadratic formula,

(ii) π₯2 β 10π₯ + 21 = 0
π = 1
π = β10
π = 21
By the quadratic formula,

(iii) π₯2 + 6π₯ β 10 = 0
π = 1
π = 6
π = β10
By the quadratic formula,

(iv) π₯2 + 2π₯ β 6 = 0
π = 1
π = 2
π = β6
By the quadratic formula,

(v) 3π₯2 + 2π₯ β 1 = 0
π = 3
π = 2
π = β1
By the quadratic formula,

(vi) 2π₯2 + 7π₯ + 5 = 0
π = 2
π = 7
π = 5
By the quadratic formula,



(ix) π₯2 β 6 = 2β2π₯
π₯2 β 2β2π₯ β 6 = 0
π = 1
π = β2β2
π = β6
By the quadratic formula,

(4 β 3π₯)(2π₯ + 3) = 5π₯
(4 β 3π₯)(2π₯ + 3) = 5π₯
8π₯ + 12 β 6π₯2 β 9π₯ = 5π₯
12 β 6π₯2 β π₯ = 5π₯
12 β 6π₯2 β π₯ β 5π₯ = 0
12 β 6π₯2 β 6π₯ = 0
β(6π₯2 + 6π₯ β 12) = 0
6π₯2 + 6π₯ β 12 = 0
6(π₯2 + π₯ β 2) = 0
π₯2 + π₯ β 2 = 0
π = 1
π = 1
π = β2
By the quadratic formula,


(xii) β6π₯2 β 4π₯ β 2β6 = 0
π = β6
π = β4
π = β2β6
By the quadratic formula,



3(2π₯2 β 10π₯ + 10) = 10(π₯2 β 6π₯ + 8)
6π₯2 β 30π₯ + 30 = 10π₯2 β 60π₯ + 80
6π₯2 β 30π₯ + 30 β 10π₯2 + 60π₯ β 80 = 0
β4π₯2 β 30π₯ + 50 = 0
β2(2π₯2 β 15π₯ + 25) = 0
2π₯2 β 15π₯ + 25 = 0
π = 2
π = β15
π = 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:
(π) π₯2β 8π₯ + 5 = 0
(ππ) 5π₯2 + 10π₯β 3 = 0
Solution:
(i) π₯2β 8π₯ + 5 = 0
π = 1
π = β8
π = 5
By the quadratic formula,

(ii) 5π₯2 + 10π₯ β 3 = 0
π = 5
π = 10
π = β3
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:
(π) 2π₯2β 10π₯ + 5 = 0
Solution:
It is given that,
2π₯2β 10π₯ + 5 = 0
π = 2
π = β10
π = 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 β 3π₯β 9 = 0
Solution:
It is given that,
π₯2 β 3π₯β 9 = 0
π = 1
π = β3
π = β9
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₯2β 5π₯β 10 = 0
Solution:
It is given that,
π₯2β 5π₯β 10 = 0
π = 1
π = β5
π = β10
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:
(π) 3π₯2β 12π₯ β 1 = 0
(ππ) π₯2β 16 π₯ + 6 = 0
(πππ) 2π₯2 + 11π₯ + 4 = 0
Solution:
(π) 3π₯2β 12π₯ β 1 = 0
π = 3
π = β12
π = β1
By the quadratic formula,

(ππ) π₯2β 16 π₯ + 6 = 0
π = 1
π = β16
π = 6
By the quadratic formula,

(πππ) 2π₯2 + 11π₯ + 4 = 0
π = 2
π = 11
π = 4
By the quadratic formula,

Question 5. Solve:
(π) π₯4 β 2π₯2 β 3 = 0
(ππ) π₯4 β 10π₯2 + 9 = 0
Solution:
(π) π₯4 β 2π₯2 β 3 = 0
By splitting the middle term method,
π₯4 β 3π₯2 + π₯2 β 3 = 0
π₯2(π₯2 β 3) + 1(π₯2 β 3) = 0
(π₯2 + 1)(π₯2 β 3) = 0
π₯2 + 1 = 0 π₯2 β 3 = 0
π₯2 = β1 π₯ = Β±β3
(ππ) π₯2 β 10π₯2 + 9 = 0
By splitting the middle term method,
π₯4 β 9π₯2 β π₯2 β 9 = 0
π₯2(π₯2 β 9) + 1(π₯2 β 9) = 0
(π₯2 + 1)(π₯2 β 9) = 0
π₯2 + 1 = 0 π₯2 β 9 = 0
π₯2 = β1 π₯2 = 9
π₯ = Β±3
Question 6. Solve:
(π) (π₯2 β π₯)2 + 5(π₯2β π₯) + 4 = 0
(ππ) (π₯2 β 3π₯)2 β 16(π₯2β 3π₯)β 36 = 0
Solution:
(π) (π₯2 β π₯)2 + 5(π₯2β π₯) + 4 = 0
Let us assumed that,
π₯2 β π₯ = π¦
π¦2 + 5π¦ + 4 = 0
By the splitting the middle term
π¦2 + 4π¦ + π¦ + 4 = 0
π¦(π¦ + 4) + 1(π¦ + 4) = 0
(π¦ + 1)(π¦ + 4) = 0
π¦ + 1 = 0 π¦ + 4 = 0
π¦ = β1 π¦ = β4
Put the value of y as β1
π₯2 β π₯ = β1
π₯2 β π₯ + 1 = 0
By the quadratic formula,
π = 1
π = β1
π = 1

(ππ) (π₯2 β 3π₯)2 β 16(π₯2β 3π₯)β 36 = 0
Let us assumed that,
π₯2 β 3π₯ = π¦
π¦2 β 16π¦ β 36 = 0
By the splitting the middle term
π¦2 β 16π¦β 36 = 0
π¦2β 18π¦ + 2π¦β 36 = 0
π¦(π¦β 18) + 2(π¦β 18) = 0
(π¦ + 2)(π¦β 18) = 0
π¦ + 2 = 0 π¦β 18 = 0
π¦ = β2 π¦ = 18
Put the value of y as β2
π₯2 β 3π₯ = β2
π₯2 β 3π₯ + 2 = 0
By the quadratic formula,
π = 1
π = β3
π = 2


Question 7. Solve:-






Question 8. Solve the equation 2π₯ β (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₯Β²β 3π₯ β 4 = 0
Solution:
It is given that,
5π₯Β²β 3π₯ β 4 = 0
By quadratic formula
π = 5
π = β3
π = β4

Question 10. Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(π₯β 1)2β 3π₯ + 4 = 0
Solution:
It is given that,
(π₯β 1)2β 3π₯ + 4 = 0
π₯2 β 2(π₯)(1) + 1β 3π₯ + 4 = 0
π₯2 β 5π₯ + 5 = 0
By quadratic formula
π = 1
π = β5
π = 5

Question 11. Solve the quadratic equation π₯Β²β 3(π₯ + 3) = 0; Give your answer correct to two significant figures.
Solution:
It is given that,
π₯Β²β 3(π₯ + 3) = 0
π₯Β²β 3π₯ β 9 = 0
By quadratic formula
π = 1
π = β3
π = β9

Exercise 5E
Question 1. Solve:

4π₯2 + 6π₯ + π₯ β 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₯ = β6 π₯ = β1
π₯ = β6/4
π₯ = β3/2
Question 2. Solve: (2π₯ + 3)2 = 81
Solution:
It is given that,
(2π₯ + 3)2 = 81
2π₯ + 3 = β81
2π₯ + 3 = Β±9
2π₯ + 3 = 9 2π₯ + 3 = β9
2π₯ = 9 β 3 2π₯ = β9 β 3
2π₯ = 6 2π₯ = β12
π₯ = 6/2 π₯ = β12/2
π₯ = 3 π₯ = β6
Question 3. Solve: πΒ²π₯Β²β πΒ² = 0
Solution:
It is given that,
πΒ²π₯Β²β πΒ² = 0
(ππ₯)2β πΒ² = 0
We know that,
π2 β π2 = (π + π)(π β π)
(ππ₯)2β πΒ² = 0
(ππ₯ + π)(ππ₯ β π) = 0
ππ₯ + π = 0 ππ₯ β π = 0
ππ₯ = βπ ππ₯ = π
π₯ = βπ/π π₯ = π/π
Question 4. Solve:

Question 5. Solve: π₯ + 4/π₯ = β4; π₯ β 0
Solution:
It is given that,
π₯ + 4/π₯ = β4
By the LCM of denominator,
π₯2 + 4/π₯ = β4
π₯2 + 4 = β4π₯
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₯ = β2
Hence, the value of π₯ is β2.
Question 6. Solve: 2π₯4β 5π₯Β² + 3 = 0
Solution:
It is given that,
2π₯4β 5π₯Β² + 3 = 0
2π₯4β 3π₯Β² β 2π₯Β² + 3 = 0
π₯Β²(2π₯2β 3) β 1(2π₯2 + 3) = 0
(π₯Β² β 1)(2π₯2β 3) = 0
π₯Β² β 1 = 0 2π₯2β 3 = 0

Q7. Solve: π₯4β 2π₯Β²β 3 = 0.
Solution:
It is given that,
π₯4β 2π₯Β²β 3 = 0
π₯4β 3π₯Β² + π₯Β²β 3 = 0
π₯2(π₯2β 3) + 1(π₯2β 3) = 0
(π₯2 + 1)(π₯2β 3) = 0
π₯2 + 1 = 0 π₯2β 3 = 0
π₯Β² = β1 π₯2 = 3
π₯ = ββ1 π₯ = Β±β3π₯2
Question 8. Solve:

Put the values,
9(π¦2 β 2) β 9π¦ β 52 = 0
9π¦2 β 18 β 9π¦ β 52 = 0
9π¦2 β 9π¦ β 70 = 0
9π¦2 β 30π¦ + 21π¦ β 70 = 0
3π¦(3π¦ β 10) + 7(3π¦ β 10) = 0
(3π¦ + 7)(3π¦ β 10) = 0
3π¦ + 7 = 0 3π¦ β 10 = 0
3π¦ = β7 3π¦ = 10
π¦ = β7/3 π¦ = 10/3
Put the value of y as -7/3

3(π₯2 + 1) = 10π₯
3π₯2 + 3 = 10π₯
3π₯2 β 10π₯ + 3 = 0
By the splitting the middle term
3π₯2 β 9π₯ β π₯ + 3 = 0
3π₯(π₯ β 3) β 1(π₯ β 3) = 0
(3π₯ β 1)(π₯ β 3) = 0
3π₯ β 1 = 0 π₯ β 3 = 0
3π₯ = 1 π₯ = 3
π₯ = 1/3
Hence, the value of π₯ is 3.
Question 9. Solve:

Put the values,
2(π¦2 β 2) β π¦ = 11
2π¦2 β 4 β π¦ = 11
2π¦2 β 4 β π¦ β 11 = 0
2π¦2 β π¦ β 15 = 0
By splitting term,
2π¦2 β 6π¦ + 5π¦ β 15 = 0
2π¦(π¦ β 3) + 5(π¦ β 3) = 0
(2π¦ + 5)(π¦ β 3) = 0
2π¦ + 5 = 0 π¦ β 3 = 0
π¦ = β 5/2 π¦ = 3
Put the value of π¦ as 3

π₯2 + 1 = β 5
2 Γ π₯
2(π₯2 + 1) = β5π₯
2π₯2 + 2 = β5π₯
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₯ = β1 π₯ = β2
π₯ = β 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₯ = π¦ β¦β¦β¦β¦..(i)
(π¦ + 4)(π¦ + 6) = 120
π¦2 + 6π¦ + 4π¦ + 24 = 120
π¦2 + 10π¦ + 24 = 120
π¦2 + 10π¦ + 24 β 120 = 0
π¦2 + 10π¦ β 96 = 0
π¦2 + 16π¦ β 6π¦ β 96 = 0
π¦(π¦ + 16) β 6(π¦ + 16) = 0
(π¦ β 6)(π¦ + 16) = 0
π¦ β 6 = 0 π¦ + 16 = 0
π¦ = 6 π¦ = β16
Put the value of π¦ as 6 in equation (i)
π₯Β² + 5π₯ = 6
π₯Β² + 5π₯ β 6 = 0
π₯Β² + 6π₯ β π₯ β 6 = 0
π₯(π₯ + 6) β 1(π₯ + 6) = 0
(π₯ + 6)(π₯ β 1) = 0
(π₯ + 6)(π₯ β 1) = 0
π₯ + 6 = 0 π₯ β 1 = 0
π¦ = β6 π₯ = 1
Put the value of π¦ as -16 in equation (i)
π₯Β² + 5π₯ = β16
π₯Β² + 5π₯ + 16 = 0
By quadratic formula
π = 1
π = 5
π = 16

Question 12. Solve each of the following equations, giving answer upto two decimal places.
(π) π₯2 β 5π₯ β 10 = 0
(ππ) 3π₯2 β π₯β 7 = 0
Solution:
(i) π₯2 β 5π₯ β 10 = 0
By quadratic formula
π = 1
π = β5
π = β10

(ππ) 3π₯2 β π₯β 7 = 0
By quadratic formula
π = 3
π = β1
π = β7

Question 13. Solve:

π¦2 β 7π¦ + 12 = 0
By splitting the middle term,
π¦2 β 4π¦ β 3π¦ + 12 = 0
π¦(π¦ β 4) β 3(π¦ β 4) = 0
(π¦ β 3)(π¦ β 4) = 0
π¦ β 3 = 0 π¦ β 4 = 0
π¦ = 3 π¦ = 4
Put the value of π¦ as 3 in equation (i)
π₯/π₯+2 = 3
π₯ = 3(π₯ + 2)
π₯ = 3π₯ + 6
π₯ β 3π₯ = 6
β2π₯ = 6
π₯ = β 6/2
π₯ = β3
Put the value of π¦ as 4 in equation (i)
π₯/π₯+2 = 4
π₯ = 4(π₯ + 2)
π₯ = 4π₯ + 8
π₯ β 4π₯ = 8
β3π₯ = 8
π₯ = β 8/3
Question 14. Solve:
(π) π₯2β 11π₯ β 12 = 0; π€βππ π₯ β π
(ππ) π₯2 β 4π₯ β 12 = 0; π€βππ π₯ β πΌ
(πππ) 2π₯2 β 9π₯ + 10 = 0; π€βππ π₯ β π
Solution:
(i) π₯2β 11π₯ β 12 = 0
By splitting the middle term,
π₯2β 12π₯ + π₯ β 12 = 0
π₯(π₯β 12) + 1(π₯ β 12) = 0
(π₯ + 1)(π₯β 12) = 0
π₯ + 1 = 0 π₯β 12 = 0
π₯ = β1 π₯ = 12
Hence, the value of π₯ is 12 since π₯ β π.
(ii) π₯2β 4π₯ β 12 = 0
By splitting the middle term,
π₯2 + 2π₯ β 6π₯ β 12 = 0
π₯(π₯ + 2) β 6(π₯ + 12) = 0
(π₯ β 6)(π₯ + 2) = 0
π₯ β 6 = 0 π₯ + 2 = 0
π₯ = 6 π₯ = β2
Hence, the value of π₯ is 6 and -2 since π₯ β πΌ
(iii) 2π₯2 β 9π₯ + 10 = 0
By splitting the middle term,
2π₯2 β 4π₯ β 5π₯ + 10 = 0
2π₯(π₯ β 2) β 5(π₯ β 2) = 0
(2π₯ β 5)(π₯ β 2) = 0
2π₯ β 5 = 0 π₯ β 2 = 0
π₯ = 5/2 π₯ = 2
Hence, the value of π₯ is 5/2 and 2 since π₯ β π.
Question 15. Solve: (π + π)Β²π₯Β² β (π + π)π₯ β 6 = 0; π + π β 0.
Solution:
It is given that,
(π + π)Β²π₯Β² β (π + π)π₯ β 6 = 0
By splitting the middle term,
(π + π)2π₯2β 3(π + π)π₯ + 2(π + π)π₯ β 6 = 0
(π + π)π₯[(π + π)π₯β 3] + 2[(π + π)π₯ β 3] = 0
[(π + π)π₯ + 2][(π + π)π₯ β 3] = 0
(π + π)π₯ + 2 = 0 (π + π)π₯ β 3 = 0
(π + π)π₯ = β2 (π + π)π₯ = 3
π₯ = β2/(π+π) π₯ = 3/(π+π)
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 β 42 = 0
2π₯2 + 6π₯ β 36 = 0
2π₯2 + 12π₯ β 6π₯ β 36 = 0
2π₯(π₯ + 6) β 6(π₯ + 6) = 0
(2π₯ β 6)(π₯ + 6) = 0
2π₯ β 6 = 0 π₯ + 6 = 0
π₯ = 6/2 π₯ = β6
π₯ = 3

Question 18. For each equation, given below, find the value of π so that the equation has equal roots.
Also, find the solution of each equation:
(i) (π β 3)π₯2 β 4π₯ + 1 = 0
(ii) 3π₯2 + 12π₯ + (π + 7) = 0
(iii) π₯2 β (π + 2)π₯ + (π + 5) = 0
Solution:
(i) (π β 3)π₯2 β 4π₯ + 1 = 0
It is given that, the above equation equal roots so π· = 0
We know that,
π· = π2 β 4ππ
π2 β 4ππ = 0
Here,
π = (π β 3)
π = β4
π = 1
Put the values,
(β4)2 β 4(π β 3)(1) = 0
16 β 4π + 12 = 0
β4π + 28 = 0
β4π = β28
π = 7
Put the value of π in above equation,
(7 β 3)π₯2 β 4π₯ + 1 = 0
4π₯2 β 4π₯ + 1 = 0
4π₯2 β 2π₯ β 2π₯ + 1 = 0
2π₯(2π₯ β 1) β 1(2π₯ β 1) = 0
(2π₯ β 1)(2π₯ β 1) = 0
2π₯ β 1 = 0
π₯ = 1/2
Hence, the value of π₯ is 1/2.
(ii) 3π₯2 + 12π₯ + (π + 7) = 0
It is given that, the above equation equal roots so π· = 0
We know that,
π· = π2 β 4ππ
π2 β 4ππ = 0
Here,
π = 3
π = 12
π = (π + 7)
Put the values,
(12)2 β 4(3)(π + 7) = 0
144 β 12(π + 7) = 0
144 β 12π β 84 = 0
β12π + 60 = 0
β12π = β60
π = 5
Put the value of π 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₯ = β2
Hence, the value of π₯ is β2.
(iii) π₯2 β (π + 2)π₯ + (π + 5) = 0
It is given that, the above equation equal roots so π· = 0
We know that,
π· = π2 β 4ππ
π2 β 4ππ = 0
Here,
π = 1
π = β(π + 2)
π = (π + 5)
Put the values,
[β(π + 2)]2 β 4(1)(π + 5) = 0
[π2 + 2 Γ π Γ (2) + 4]2 β 4(π + 5) = 0
π2 + 4π + 4 β 4π β 20 = 0
π2 β 16 = 0
π2 = 16
π = Β±4
Put the value of π as 4 in above equation,
π₯2 β (4 + 2)π₯ + (4 + 5) = 0
π₯2 β 6π₯ + 9 = 0
π₯2 β 3π₯ β 3π₯ + 9 = 0
π₯(π₯ β 3) β 3(π₯ β 3) = 0
(π₯ β 3)(π₯ β 3) = 0
π₯ β 3 = 0
π₯ = 3
Put the value of π as -4 in above equation,
π₯2 β (β4 + 2)π₯ + (β4 + 5) = 0
π₯2 + 2π₯ + 1 = 0
π₯2 β π₯ β π₯ + 1 = 0
π₯(π₯ β 1) β 1(π₯ β 1) = 0
(π₯ β 1)(π₯ β 1) = 0
π₯ β 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. ππ₯2 β 4π₯ + 3 = 0
Solution:
It is given that, the above equation equal roots so π· = 0
We know that,
π· = π2 β 4ππ
π2 β 4ππ = 0
ππ₯2 β 4π₯ + 3 = 0
Here,
π = π
π = β4
π = 3
Put the values,
(β4)2 β 4(π)(3) = 0
16 β 12π = 0
β12π = β16
π = β16/β12
π = β16/β12
π = 4/3
Question 20. Without solving the following quadratic equation, find the value of βmβ for which the given
equation has real and equal roots.
π₯Β² + 2(πβ 1)π₯ + (π + 5) = 0
Solution:
It is given that, the above equation equal roots so π· = 0
We know that,
π· = π2 β 4ππ
π2 β 4ππ = 0
π₯Β² + 2(πβ 1)π₯ + (π + 5) = 0
Here,
π = 1
π = 2(π β 1)
π = (π + 5)
Put the values,
[2(π β 1)]2 β 4(1)(π + 5) = 0
[4(π2 β 2(π)(1) + (1)2]2 β 4π β 20 = 0
4(π2 β 2π + 1) β 4π β 20 = 0
4π2 β 8π + 4 β 4π β 20 = 0
4π2 β 12π β 16 = 0
4(π2 β 3π β 4) = 0
π2 β 3π β 4 = 0
π2 β 4π + π β 4 = 0
π(π β 4) + 1(π β 4) = 0
(π + 1)(π β 4) = 0
π + 1 = 0 π β 4 = 0
π = β1 π = 4
Hence, the value of m is -1 and 4.
Exercise 5F
Question 1. Solve:
(i) (π₯ + 5)(π₯β 5) = 24
(ii) 3π₯2 β 2β6π₯ + 2 = 0
(ii) 3β2π₯2 β 5π₯ + β26 = 0
Solution:
(i) (π₯ + 5)(π₯ β 5) = 24
π₯2 β (5)2 = 24
We know that,
(π β π)(π + π) = π2β π2
π₯2β 25 = 24
π₯2 = 49
π₯ = Β± 7
(ii) 3π₯2 β 2β6π₯ + 2 = 0
3π₯2β β6π₯β β6π₯ + 2 = 0
β3π₯(β3π₯β β2)β β2(β3π₯ β β2) = 0
(β3π₯β β2)(β3π₯ β β2) = 0
(β3π₯β β2)(β3π₯ β β2) = 0
β3π₯β β2 = 0
π₯ = β2/β3
(iii) 3β2π₯2 β 5π₯ β β2 = 0
3β2π₯2 β 6π₯ + π₯ β β2 = 0
3β2π₯(π₯ β β2) + 1(π₯ β β2) = 0
(3β2π₯ + 1)(π₯ β β2) = 0
3β2π₯ + 1 = 0 π₯ β β2 = 0
3β2π₯ = β1 π₯ = β2
π₯ = β 1/3β2
Question 2. One root of the quadratic equation 8π₯2 + ππ₯ + 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 + ππ₯ + 15 = 0
Put the value of π₯ as 3/4 in given equation,

Put the value of π in given equation,
8π₯2 + ππ₯ + 15 = 0
8π₯2 + (β26)π₯ + 15 = 0
8π₯2 β 26π₯ + 15 = 0
8π₯2 β 20π₯ β 6π₯ + 15 = 0
4π₯(2π₯ β 5) β 3(2π₯ β 5) = 0
(4π₯ β 3)(2π₯ β 5) = 0
4π₯ β 3 = 0 2π₯ β 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 β 3π₯β 2ππ₯ β 6π = 0 is β3, find its other root.
Solution:
It is given that,
The one root of equation is β3.
Quadratic equation is π₯2 β 3π₯β 2ππ₯ β 6π = 0
Put the value of π₯ as β3 in given equation,
π₯2 β 3π₯ β 2ππ₯ β 6π = 0
π₯(π₯ + 3) β 2π(π₯ + 3) = 0
(π₯β 2π)(π₯ + 3) = 0
π₯ β 2π = 0 π₯ + 3 = 0
π₯ = 2π π₯ = β3
Hence, the other root of the given equation is 2a.
Question 4. If πβ 15 = 0 and 2π₯2 + 15π₯ + 15 = 0; find the values of π₯.
Solution:
It is given that,
πβ 15 = 0
π = 15
Put the value of π 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₯ = β5/2 π₯ = β5
Hence, the value of π₯ is β5/2 and -5.
Question 5. Find the solution of the equation 2π₯2 β ππ₯ β 25π = 0; if π + 5 = 0 and π β 1 = 0.
Solution:
It is given that,
π + 5 = 0
π = β5
And
π β 1 = 0
π = 1
Put the value of π and π in given equation
2π₯2 β ππ₯ β 25π = 0
2π₯2 β (β5)π₯ β 25(1) = 0
2π₯2 + 5π₯ β 25 = 0
2π₯ + 10π₯β 5π₯β 25 = 0
2π₯(π₯ + 5) β 5(π₯ + 5) = 0
(π₯ + 5)(2π₯β 5) = 0
π₯ + 5 = 0 2π₯ β 5 = 0
π₯ = β5 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₯) β (1/π₯β2) = 3 where π₯ β 0 and π₯ β 2; find π Γ π.
Solution:


Question 7. Solve, using formula: π₯2 + π₯ β (π + 2)(π + 1) = 0
Solution:
It is given that,
π₯2 + π₯ β (π + 2)(π + 1) = 0
By the quadratic formula
π = 1
π = 1
π = β (π + 2)(π + 1)

Question 8. Solve the quadratic equation 8π₯2β 14π₯ + 3 = 0
(i) When π₯ β πΌ (integers)
(ii) When π₯ β π (rational numbers)
Solution:
It is given that,
8π₯2β 14π₯ + 3 = 0
8π₯2β 12π₯ β 2π₯ + 3 = 0
4π₯(2π₯ β 3) β (2π₯β 3) = 0
(4π₯β 1)(2π₯β 3) = 0
4π₯ β 1 = 0 2π₯ β 3 = 0
π₯ = 1/4 π₯ = 3/2
(i) When π₯ β πΌ (integers) the given 8π₯2β 14π₯ + 3 = 0 has no equal roots.
(ii) When π₯ β π (rational numbers) the given 8π₯2β 14π₯ + 3 = 0 are π₯ = 1/4 , 3/2
.
Question 9. Find the value of m for which the equation (π + 4 )2 + (π + 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 = π2 β 4ac = 0
β (m + 1)2 -4(m + 4)(1) = 0
β m2 + 2m + 1 β 4m β 16 = 0
β m2 β 2m β 15 = 0
β m2 β 5m + 3m β 15 = 0
β m(m β 5) +3(m =5) = 0
β (m β 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 π· = 0
We know that,
π· = π2 β 4ππ
π2 β 4ππ = 0
3π₯Β² + ππ₯ + 2 = 0
Here,
π = 3
π = π
π = 2
Put the values,
m2 β 4(2)(3) = 0
m2 β 24 = 0
m2 = 24
π = β24
π = Β±2β6
Put the value of π as 2β6 in given equation,
3π₯Β² + 2β6π₯ + 2 = 0
(β3π₯)Β² + 2(β3π₯)(β2) + (β2)2 = 0
(β3π₯ + β2)Β² = 0
β3π₯ + β2 = 0
β3π₯ = ββ2

Question 11. Find the value of k for which equation 4π₯2 + 8π₯ β π = 0 has real roots.
Solution:
It is given that, the above equation equal roots so π· β₯ 0
We know that,
π· = π2 β 4ππ
π2 β 4ππ β₯ 0
4π₯Β² + 8π₯ β π β₯ 0
Here,
π = 4
π = 8
π = βπ
Put the values,
(8)2 β 4(4)(βπ) β₯ 0
64 + 16π β₯ 0
16π β₯ β64
π β₯ β64/16
π β₯ β4
Hence, the value of π is β4 when given equation has real roots.
Question 12. Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(π) 3π₯2 β 5π₯ + 2 = 0
(ππ) π₯2 + 4π₯ + 5 = 0
Solution:
(i) 3π₯2 β 5π₯ + 2 = 0
Here,
π = 3
π = β5
π = 2
Put the values,
π· = π2 β 4ππ
π· = (β5)2 β 4(3)(2)
π· = 25 β 24
π· = 1
We know that,
π· > 0, so the roots of the quadratic equation are real and distinct.
By quadratic formula

(ii) π₯2 + 4π₯ + 5 = 0
Here,
π = 1
π = 4
π = 5
Put the values,
π· = π2 β 4ππ
π· = (4)2 β 4(1)(5)
π· = 16 β 20
π· = β4
Hence, π· > 0, so the roots of the quadratic equation does not exist.
Question 13. Solve:

48π₯ = 324β π₯2
π₯2 + 48π₯ β 324 =
π₯2 + 48π₯ β 324 = 0
π₯2 + 54π₯ β 6π₯ β 324 = 0
π₯(π₯ + 54) β 6(π₯ + 54) = 0
(π₯ + 54)(π₯ β 6) = 0
π₯ = β54 ππ π₯ = 6
Hence, the value of π₯ > 0.
