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

1.  Check whether the following are quadratic equations:

Solution:

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

Solution:
We have given that:

Let us substitute the given values in the expression and check, we get
When, x = 5

3. In each of the following, determine whether the given numbers are solutions of the given equation or not:

Solution:
We have given that:

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:

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

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.

(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,

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,

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,

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,

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

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:

Hence, the value of x is 0.264 or −2.264

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,

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,

= -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

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

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.

= 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,

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:

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

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,

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,

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,

(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,

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’

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’

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'

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:

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,

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,

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,

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:

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,