ML Aggarwal Solutions Class 10 Maths Chapter 5 Quadratic Equations in One Variable

1.  Check whether the following are quadratic equations:
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable












 
 

Solution:
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-

According to given equation
Yes, the given equation is a quadratic equation since it has power of 2.
(ii) (2x + 1) (3x – 2) = 6(x + 1) (x – 2)
We have given that, 
 (2x + 1)(3x – 2)= 6(x + 1)(x – 2)
Let us solve the given expression, we get  
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-1
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-2
 
Solution:
We have given that: 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-3 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-4
 
Let us substitute the given values in the expression and check, we get 
When, x = 5 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-5
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-6
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-7
 
 
3. In each of the following, determine whether the given numbers are solutions of the given equation or not:
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-8
 
 
 
 
 
Solution:
We have given that: 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-9 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-10 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-11
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-12
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-13
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-14
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-15
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-16
Now, substitute the value of p in equation (2), we get
q = 21 – 9p
= 21 – 9(3)
= 21 – 27
= −6
∴ Value of p is 3 and q is −6.
Hence, the value q=−6 is the root of the equation.
 
EXERCISE 5.2

Solve the following equations (1 to 24) by factorization:

ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-17

ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-18

Let us simplify the given expression, we get 
x2  – 5x = 0
x(x – 5) = 0
x = 0 or x – 5 = 0
x = 0 or x = 5
Hence, the value x = 0,5 is the root of the equation.
 
2. (i) (x – 3) (2x + 5) = 0
(ii) x (2x + 1) = 6
Solution:
(i) (x – 3) (2x + 5) = 0
We have given that, 
(x – 3) (2x + 5) = 0
Let us simplify the given expression, we get 
(x – 3) = 0 or (2x + 5) = 0
x = 3 or 2x = -5
x = 3 or x = -5/2
Hence, the value x = 3,-5/2 is the root of the equation.
(ii) x (2x + 1) = 6
We have given that, 
x (2x + 1) = 6
Let us simplify the given expression,
2x2  + x – 6 = 0
Let factorize the expression, we get 
2x2  + 4x – 3x – 6 = 0
2x(x + 2) – 3(x + 2) = 0
(2x – 3) (x + 2) = 0
So now,
(2x – 3) = 0 or (x + 2) = 0
2x = 3 or x = -2
x = 3/2 or x = -2
Hence, the value is x = 3/2,-2.
 
3. (i) x² – 3x – 10 = 0
(ii) x(2x + 5) = 3
Solution:
 (i) x2 – 3x – 10 = 0
We have given that 
x2 – 3x – 10 = 0
Let us simplify the given expression, we get 
x2  – 5x + 2x – 10 = 0
x(x – 5) + 2(x – 5) = 0
(x + 2) (x – 5) =0
So now,
(x + 2) = 0 or (x – 5) =0
x = -2 or x = 5
Hence, the value x = -2,5 is the root of the equation.
(ii) x(2x + 5) = 3
We have given that, 
x(2x + 5) = 3
Let us simplify the given expression, we get 
2x2  + 5x – 3 = 0
Now, let factorize the expression
2x2  + 6x – x – 3 = 0
2x(x + 3) -1(x + 3) = 0
(2x – 1) (x + 3) = 0
So now,
(2x – 1) = 0 or (x + 3) = 0
2x = 1 or x = -3
x = ½ or x = -3
Hence, the value of x = ½ ,-3 
 
4. (i) 3x2  – 5x – 12 = 0
(ii) 21x2  – 8x – 4 = 0
Solution:
(i) 3x2  – 5x – 12 = 0
We have given that, 
3x2  – 5x – 12 = 0
Let us simplify the given expression, we get 
3x2  – 9x + 4x – 12 = 0
3x(x – 3) + 4(x – 3) = 0
(3x + 4) (x – 3) =0
Now, 
(3x + 4) = 0 or (x – 3) =0
3x = -4 or x = 3
x = -4/3 or x = 3
Hence, the value of x = -4/3,3
(ii) 21x2  – 8x – 4 = 0
We have given that, 
21x2  – 8x – 4 = 0
Let us simplify the given expression, we get 
21x2  – 14x + 6x – 4 = 0
7x(3x – 2) + 2(3x – 2) = 0
(7x + 2) (3x – 2) = 0
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-19
 
5. (i) 3x2 = x + 4
(ii) x(6x – 1) = 35
Solution:
We have given that:
(i)  3x2 = x + 4
Let us simplify the given expression,
 3x2  – x – 4 = 0
Now, let us factorize
 3x2  – 4x + 3x – 4 = 0
x(3x – 4) + 1(3x – 4) = 0
(x + 1) (3x – 4) = 0
So now,
(x + 1) = 0 or (3x – 4) = 0
x = -1 or 3x = 4
x = -1 or x =4/3
∴ Value of x = -1,4/3
Hence, the value of  x = -1,4/3.
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-20
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-21
(ii) 2/3x2  -1/3x = 1
Let us simplify the given expression,
2x2 – x = 3
2x2  – x – 3 = 0
Let us factorize the given expression,
2x2  – 3x + 2x – 3 = 0
x(2x – 3) + 1(2x – 3) = 0
(x + 1) (2x – 3) = 0
So now,
(x + 1) = 0 or (2x – 3) = 0
x = -1 or 2x = 3
x = -1 or x =3/2
∴ Value of x = -1,3/2
Hence, the value x = -1,3/2.
 
7. (i) (x – 4)2 + 52 = 132
(ii) 3(x – 2)2 = 147
Solution:
We have given that:
(i) (x – 4)2  + 52 = 132
Firstly let us expand the given expression,
x2  – 8x + 16 ++ 25 = 169
x2  – 8x + 41 – 169 = 0
x2  – 8x – 128 = 0
Let us factorize the expression, we get 
x2– 16x + 8x – 128 = 0
x(x – 16) + 8(x – 1) = 0
(x + 8) (x – 16) = 0
So now,
(x + 8) = 0 or (x – 16) = 0
x = -8 or x = 16
∴ Value of x = -8,16
Hence, the value is x = -8,16  .
(ii) We have given that: 3(x – 2)2 = 147
Firstly let us expand the given expression,
3(x2 – 4x + 4) = 147
3x2 – 12x + 12 = 147
3x2 – 12x +12 – 147 = 0
3x2 – 12x – 135 = 0
Divide by 3, we get
x2– 4x – 45 = 0
Let us factorize the expression, we get 
x2 – 9x + 5x – 45 = 0
x(x – 9) + 5(x – 9) = 0
(x + 5) (x – 9) = 0
So now,
(x + 5) = 0 or (x – 9) = 0
x = -5 or x = 9
∴ Value of x = -5, 9
Hence, the value is x = -5,9.
 
 
8. (i) 1/7(3x – 5)2 = 28
(ii) 3(y2 – 6) = y(y + 7) – 3
Solution:
We have given that:
(i) 1/7(3x – 5)2  = 28
Let us simplify the expression,
(3x – 5)2  = 28 × 7
(3x – 5)2  = 196
Now let us expand,
9x2  – 30x + 25 = 196
9x2  – 30x + 25 – 196 = 0
9x2  – 30x – 171 = 0
Divide by 3, we get
3x2  – 10x – 57 = 0
Let us factorize the expression,
3x2  – 19x + 9x – 57 = 0
x(3x – 19) + 3(3x – 19) = 0
(x + 3) (3x – 19) = 0
So now,
(x + 3) = 0 or (3x – 19) = 0
x = -3 or 3x = 19
x = -3 or x = 19/3
Hence, the value is x = -3,19/3
(ii) We have given that: 3(y2 – 6) = y(y + 7) – 3
Let us simplify the expression, we get 
3y2  – 18 = y2  + 7y – 3
3y2  – 18 – y2  – 7y + 3 = 0
2y2  – 7y – 15 = 0
Let us factorize the expression, we get 
2y2  – 10y + 3y – 15 = 0
2y(y – 5) + 3(y – 5) = 0
(2y + 3) (y – 5) = 0
So now,
(2y + 3) = 0 or (y – 5) = 0
2y = -3 or y = 5
y = -3/2 or y = 5
Hence, the value is y=-3/2,5 
 
9. x2  – 4x – 12 = 0, when x ∈ N
Solution:
Let us factorize the expression,
We have given that:
x2  – 4x – 12 = 0
x2  – 6x + 2x – 12 = 0
x(x – 6) + 2(x – 6) = 0
(x + 2) (x – 6) = 0
So now,
(x + 2) = 0 or (x – 6) = 0
x = -2 or x = 6
Hence, Value of x = 6 (Since, -2 is not a natural number).
 
11. 2x2 – 8x – 24 = 0 when x ∈ I
Solution:
Let us simplify the expression,
We have given that:
2x2 – 8x – 24 = 0
Divide the expression by 2, we get
x2  – 4x – 12 = 0
Now, let us factorize the expression,
x2  – 6x + 2x – 12 = 0
x(x- 6) + 2 (x – 6) = 0
(x + 2) (x – 6) = 0
So now,
(x + 2) = 0 or (x – 6) = 0
x = -2 or x = 6
Hence, the value of is x = -2,6.
 
12. 5x2 – 8x – 4 = 0 when x ∈ Q
Solution:
Let us factorize the expression,
We have given that:
5x2 – 8x – 4 = 0
5x2  – 10x + 2x – 4 = 0
5x(x – 2) + 2 (x – 2) = 0
(5x + 2) (x – 2) = 0
So now,
(5x + 2) = 0 or (x – 2) = 0
5x = -2 or x = 2
x = -2/5 or x = 2
Hence, the value x = -2/5,2.
 
13. 2x2 – 9x + 10 = 0, when
(i) x ∈ N
(ii) x ∈ Q
Solution:
Let us factorize the expression,
We have given that:
2x2 – 9x + 10 = 0
2x2  – 4x – 5x + 10 = 0
2x(x – 2) – 5(x – 2) = 0
(2x – 5) (x – 2) = 0
So now,
(2x – 5) = 0 or (x – 2) = 0
2x = 5 or x = 2
x = 5/2 or x = 2
(i) When,x ∈ N then, x = 2
(ii) When, x ∈ Q then, x = 2,5/2 
 
14. (i) a²x² + 2ax + 1 = 0,a ≠ 0
(ii) x² – (p + q)x + pq = 0
Solution:
We have given that:
(i) a²x² + 2ax + 1 = 0,a ≠ 0
Let us factorize the expression, we get 
a²x² + 2ax + 1 = 0
a²x² + ax + ax + 1 = 0
ax(ax + 1) + 1(ax + 1) = 0
(ax + 1) (ax + 1) = 0
So now,
(ax + 1) = 0 or (ax + 1) = 0
ax = -1 or ax = -1
x = -1/a or x = -1/a
Hence, the value of x is -1/a,-1/a 
(ii) x² – (p + q)x + pq = 0
We have given that 
x² – (p + q)x + pq = 0
Let us simplify the expression, we get 
x² – (p + q)x + pq = 0
x²  – px – qx + pq = 0
x(x – p) – q(x – p) = 0
(x – q) (x – p) = 0
So now,
(x – q) = 0 or (x – p) = 0
x = q or x = p
Hence, the value x is p,q.
 
15. a²x² + (a² + b²)x + b² = 0,a≠0
Solution:
We have given that:
a²x² + (a² + b²)x + b² = 0
Let us simplify the expression, we get 
a²x² + a2x + b² x + b²  = 0
a² x(x + 1) + b² (x + 1) = 0
(a² x + b²) (x + 1) = 0
So now,

ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-22

EXERCISE 5.3

Solve the following (1 to 8) equations by using the formula:
1. (i) 2x² – 7x + 6 = 0
(ii) 2x² – 6x + 3 = 0
Solution:
We have given that:
(i) 2x² – 7x + 6 = 0
Let us consider, 
a = 2,b = -7,c = 6 
So, by using the formula,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-23
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-24
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-25
 
2. (i) x2  + 7x – 7 = 0
(ii) (2x + 3) (3x – 2) + 2 = 0
Solution:
We have given that:
(i) x2  + 7x – 7 = 0
Let us consider,
a = 1,b = 7,c = -7 
So, by using the formula,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-26
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-27
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-28
 
 
3. (i) 256x² – 32x + 1 = 0
(ii) 25 + 30x + 7 = 0
Solution:
We have given that:
(i) 256x² – 32x + 1 = 0
Let us consider,
a = 256,b = -32,c = 1 
So, by using the formula,
 

ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-29

 
 
 ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-30
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ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-37
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ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-43
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-43
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-44
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ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-46
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-47
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-60
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-48
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-49
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-50
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-50
 
7. (i) x-1/x = 3,x ≠ 0
(ii) 1/ x + 1/(x-2) = 3,x ≠ 0,2
Solution:
We have given that:
(i) x-1/x = 3,x ≠ 0
Let us simplify the given expression, By taking LCM
We get
x2  – 1 = 3x 
x2  – 3x – 1 = 0 
Let us consider,
a = 1,b = -3,c = -1 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-51
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-52
 
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-54
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-55
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-56
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-57
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-58
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-59
 
10. Solve the following equation by using quadratic equations for x.
(i) x² – 5x – 10 = 0
(ii) 5x(x + 2) = 3
Solution:
(i)We have given that:
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-61
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-65
 
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-62 
Hence, the value of x is 0.264 or −2.264
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-64 
 
 ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-63
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-66
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-67
 
12. Solve the following equation: x-18/x = 6. Give your answer correct to two x significant figures. 
Solution:
Given equation:
x-18/x = 6 
By taking LCM
x2  – 18 = 6x
x2  – 6x – 18 = 0
Let us consider,
a = 1,b = -6,c = -18
So, by using the formula,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-69
 
EXERCISE 5.4
1. Find the discriminate of the following equations and hence find the nature of roots:
(i) 3x² – 5x – 2 = 0
(ii) 2x² – 3x + 5 = 0
(iii) 7x² + 8x + 2 = 0
(iv) 3x² + 2x – 1 = 0
(v) 16x² – 40x + 25 = 0
(vi) 2x² + 15x + 30 = 0
Solution:
(i) We have given that:
3x² – 5x – 2 = 0
Let us consider,
a = 3,b = -5,c = -2 
By using the formula,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-70
= -31
So,
Discriminate, D = -31
D < 0 
Hence, nature of Roots are not real.
(iii) We have given that:
 7x² + 8x + 2 = 0
Let us consider,
a = 7,b = 8,c = 2 
By using the formula, we get 
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-71
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-72
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ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-76
 
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ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-79
 
 
3. Find the nature of the roots of the following quadratic equations:
(i) x² – 1/2x – 1/2 = 0
(ii) x² – 2√3x – 1 = 0 If real roots exist, find them.
Solution:
(i) We have given that:
x² – 1/2x – 1/2 = 0 
Let us consider,
a = 1,b = -1/2,c = -1/2 
By using the formula, we get 
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-80
So,
Discriminate, D = 16
Hence, the value D =16 is the root of the equation.
D > 0 
Hence, the nature of Roots are real and unequal.
 
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= 4m – 8m + 4 – 4m – 20
= 4m2  – 12m – 16
Since, roots are equal.
D = 0 
4m2 – 12m – 16 = 0 
Divide by 4, we get
m2  – 3m – 4 = 0 
Now let us factorize,
m2  – 4m + m – 4 = 0
m(m – 4) + 1 (m – 4) = 0
(m – 4) (m + 1) = 0
So,
(m – 4) = 0 or (m + 1) = 0
m = 4 or m = -1
∴ m = 4,-1
Hence, the value m=4/-1 is the root of the equation. 
 
 
7. Find the values of k for which each of the following quadratic equation has equal roots:
(i) 9x2 + kx + 1 = 0
(ii) x2 – 2kx + 7k – 12 = 0
Also, find the roots for those values of k in each case.
Solution:
(i) We have given that:
9x2 + kx + 1 = 0 
Let us consider,
a = 9,b = k,c = 1 
By using the formula,
D = (b2)– 4ac 
= (k) – 4 (9) (1)
= k2  – 36
Since, roots are equal.
D = 0 
k2  – 36 = 0
(k + 6) (k – 6) = 0
So,
(k + 6) = 0 or (k – 6) = 0
k = -6 or k = 6
∴ k = 6,-6
Now, let us substitute in the equation
When k = 6, we get 
9x2 + kx + 1 = 0
9x2  + 6x + 1 = 0
(3x)2  + 2(3x)(1) + 12  = 0
(3x + 1)2  = 0
3x + 1 = 0
3x = -1
x = -1/3,-1/3
When k = -6, we get 
9x2 + kx + 1 = 0
9x2  – 6x + 1 = 0
(3x)2  – 2(3x)(1) + 12 = 0
(3x – 1)2  = 0
3x – 1 = 0
3x = 1
x = 1/3,1/3
Hence, the value x = 1/3 is the root of the equation.
(ii) We have given that:
x2 – 2kx + 7k – 12 = 0
Let us consider,
a = 1,b = -2k,c = (7k – 12) 
By using the formula, we get 
D = b2  – 4ac
= (-2k)2  – 4 (1) (7k – 12)
= 4k2  – 28k + 48
Since, roots are equal.
D = 0 
4k2   – 28k + 48 = 0 
Divide by 4, we get
k2  – 7k + 12 = 0 
Now let us factorize,
k2   – 3k – 4k + 12 = 0
k(k – 3) – 4 (k – 3) = 0
(k – 3) (k – 4) = 0
So,
(k – 3) = 0 or (k – 4) = 0
k = 3 or k = 4
∴ k = 3,4
Now, let us substitute in the equation
When k = 3, we get 
By using the formula,
 
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8. Find the value(s) of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.
Solution:
Given:
We have given that:
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-89
 
Let us factorize, we get 
-7p2+ 28p – 4p + 16 = 0 
-7p(p – 4) – 4 (p – 4) = 0 
(p – 4) (-7p – 4) = 0 
So,
(p – 4) = 0 or (-7p – 4) = 0
p = 4 or -7p = 4
p = 4 or p = -4/7
Hence, the value of  p is 4,-4/7
 
9. If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.
Solution:
We have given that:
– 5 is a root of the quadratic equation 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-90
 
Since, roots are equal.
49 – 28k = 0
28k = 49
k = 49/28
= 7/4
Hence, the value of k is 7/4
 
10. Find the value(s) of p for which the equation 2x² + 3x + p = 0 has real roots.
Solution:
We have given that:
2x² + 3x + p = 0 
Let us consider,
a = 2,b = 3,c = p 
By using the formula, 
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-91
 
11. Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.
Solution:
We have given that:
x² + kx + 4 = 0 
Let us consider,
a = 1,b = k,c = 4 
By using the formula,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-92
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-93
 
Exercise -5.5
1. (i) Find two consecutive natural numbers such that the sum of their squares is 61.
(ii) Find two consecutive integers such that the sum of their squares is 61.
Solution:
(i) Find two consecutive natural numbers such that the sum of their squares is 61.
According to the question,
Let us consider first natural number be ‘x’
Second natural number be ‘x + 1’
So according to the question,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-96
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-94
(ii) Find two consecutive integers such that the sum of their squares is 61.
According to the question,
Let us consider first integer number be ‘x’
Second integer number be ‘x + 1’
So according to the question,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-95
Now,
If x = -6, then
First integer number = -6
Second integer number = -6 + 1 = -5
If x = 5, then
First integer number = 5
Second integer number = 5 + 1 = 6
 
2. (i) If the product of two positive consecutive even integers is 288, find the integers.
(ii) If the product of two consecutive even integers is 224, find the integers.
(iii) Find two consecutive even natural numbers such that the sum of their squares is 340.
(iv) Find two consecutive odd integers such that the sum of their squares is 394.
Solution:
(i) If the product of two positive consecutive even integers is 288, find the integers.
According to the question,
Let us consider first positive even integer number be ‘2x’
 
 ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-97
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-98
 
 
 ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-99
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-100
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-101
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-102
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-103
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-104
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-104
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-106
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-107
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-108
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-109
 
x(2x + 19) – 8 (2x + 19) = 0
(2x + 19) (x – 8) = 0
So,
(2x + 19) = 0 or (x – 8) = 0 
2x = -19 or x = 8 
x = -19/2 or x = 8  
∴ The value of x = 8 [since -19/2 is a negative value]
So,
First integer = x = 8
 
Second integer = x + 1 = 8 + 1 = 9
 
Third integer = x + 2 = 8 + 2 = 10
 
Hence, the numbers are 8,9,10.
 
8. (i) Find three successive even natural numbers, the sum of whose squares is 308.
(ii) Find three consecutive odd integers, the sum of whose squares is 83.
Solution:
We have given that:
(i) Find three successive even natural numbers, the sum of whose squares is 308.
Let us consider first even natural number be ‘2x’
Second even number be ‘2x + 2’
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-110
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-111
 
9. In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by 1/14. Find the fraction.
Solution:
According to the question, 
Let the numerator be 'x'
Denominator be 'x+3'
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-112
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-116
 
 
 
10. The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by 4/35. Find the fraction.
Solution:
We have given that:
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-115 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-113

ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-114

 
 
11. A two digit number contains the bigger at ten’s place. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Solution:
According to the question,
 ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-117
 
 
12. A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number. 
Solution:
According to the question,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-118
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-119
 
 
13. A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.
Solution:
According to the question,
We have given that,
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-120
 
14. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square meters, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x. 
Solution:

ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-121
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-123

15. (i) Harish made a rectangular garden, with its length 5 meters more than its width. The next year, he increased the length by 3 meters and decreased the width by 2 meters. If the area of the second garden was 119 sqm, was the second garden larger or smaller ?
(ii) The length of a rectangle exceeds its breadth by 5 m. If the breadth was doubled and the length reduced by 9 m, the area of the rectangle would have increased by 140 m². Find its dimensions.
Solution:
We have given that:
(i) In first case:
According to the question, 
 
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-126
ML Aggarwal Solutions for Class 10 Maths Chapter 5 Quadratic Equations in One Variable-127
 
 
 
 
 
 
 
 
 
ML Aggarwal Solutions Class 10 Maths Chapter 1 Goods and Service Tax (GST)
ML Aggarwal Solutions Class 10 Maths Chapter 2 Banking
ML Aggarwal Solutions Class 10 Maths Chapter 3 Shares and Dividends
ML Aggarwal Solutions Class 10 Maths Chapter 4 Linear Inequations
ML Aggarwal Solutions Class 10 Maths Chapter 5 Quadratic Equations in One Variable
ML Aggarwal Solutions Class 10 Maths Chapter 6 Factorization
ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion
ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion
ML Aggarwal Solutions Class 10 Maths Chapter 8 Matrices
ML Aggarwal Solutions Class 10 Maths Chapter 9 Arithmetic and Geometric Progression
ML Aggarwal Solutions Class 10 Maths Chapter 10 Reflection
ML Aggarwal Solutions Class 10 Maths Chapter 11 Section Formula
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line
ML Aggarwal Solutions Class 10 Maths Chapter 13 Similarity
ML Aggarwal Solutions Class 10 Maths Chapter 14 Locus
ML Aggarwal Solutions Class 10 Maths Chapter 15 Circles
ML Aggarwal Solutions Class 10 Maths Chapter 16 Constructions
ML Aggarwal Solutions Class 10 Maths Chapter 17 Mensuration
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities
ML Aggarwal Solutions Class 10 Maths Chapter 19 Trigonometric Tables
ML Aggarwal Solutions Class 10 Maths Chapter 20 Heights and Distances
ML Aggarwal Solutions Class 10 Maths Chapter 21 Measures Of Central Tendency
ML Aggarwal Solutions Class 10 Maths Chapter 22 Probability
NCERT Exemplar Solutions Class 10 Maths Areas related to Circles
NCERT Exemplar Solutions Class 10 Maths Arithmetic Progression
NCERT Exemplar Solutions Class 10 Maths Circles
NCERT Exemplar Solutions Class 10 Maths Construction
NCERT Exemplar Solutions Class 10 Maths Coordinate Geometry
NCERT Exemplar Solutions Class 10 Maths Linear Equations
NCERT Exemplar Solutions Class 10 Maths Polynomials
NCERT Exemplar Solutions Class 10 Maths Quadratic Equation
NCERT Exemplar Solutions Class 10 Maths Real Numbers
NCERT Exemplar Solutions Class 10 Maths Surface Area and Volume
NCERT Exemplar Solutions Class 10 Maths Triangles
NCERT Exemplar Solutions Class 10 Maths Trigonometry