Crucial questions along with their solutions from chapter 4, quadratic equations, for Class 10 Mathematics are provided below. These questions are valuable resources for students preparing for the CBSE board exam 2022-2023. They have been curated in alignment with the NCERT book guidelines. Our team of experts has meticulously crafted these questions after extensive research, adhering to the latest exam pattern. By practicing these questions, students can familiarize themselves with the types of questions likely to appear in their Mathematics paper. These resources cover important questions for all chapters of Class 10 Mathematics, enabling students to effectively prepare for their exams.
Also Check: CBSE Class 10 Maths Important Question Chapter 8 Trigonometry
Important Questions with Solutions for Quadratic Equations
Q.1: 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:
Given that -5 is a root of the quadratic equation 2x2 + px – 15 = 0.
⇒ 2(-5)2 + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
⇒ 5p = 35
⇒ p = 7
Also, the quadratic equation p(x2 + x) + k = 0 has equal roots.
Substituting p = 7 in p(x2 + x) + k = 0,
7(x2 + x) + k = 0
7x2 + 7x + k = 0
Comparing with the standard form ax2 + bx + c = 0,
a = 7, b = 7, c = k
For equal roots, discriminant is equal to 0.
b2 – 4ac = 0
(7)2 – 4(7)(k) = 0
49 – 28k = 0
28k = 49
k = 7/4
Therefore, the value of k is 7/4.
Also Check: Important Questions for Class 10 Maths Chapter 14 Statistics
Q.2: Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. What is the speed of the train?
Solution:
(i) Let us consider,
The breadth of the rectangular plot is x m.
Thus, the length of the plot = (2x + 1) m
As we know,
Area of rectangle = length × breadth = 528 m2
Putting the value of length and breadth of the plot in the formula, we get,
(2x + 1) × x = 528
⇒ 2x2 + x = 528
⇒ 2x2 + x – 528 = 0
Hence, 2x2 + x – 528 = 0, is the required equation which represents the given situation.
(ii) Let us consider,
speed of train = x km/h
And
Time taken to travel 480 km = 480 (x) km/h
As per second situation, the speed of train = (x – 8) km/h
As given, the train will take 3 hours more to cover the same distance.
Therefore, time taken to travel 480 km = (480/x) + 3 km/h
As we know,
Speed × Time = Distance
Therefore,
(x – 8)[(480/x) + 3] = 480
⇒ 480 + 3x – (3840/x) – 24 = 480
⇒ 3x – (3840/x) = 24
⇒ 3x2 – 24x – 3840 = 0
⇒ x2 – 8x – 1280 = 0
Hence, x2 – 8x – 1280 = 0 is the required representation of the problem mathematically.
Also Check: Surface Areas and Volumes Class 10 Extra Questions Maths Chapter 13
Q.3: Solve for x: [1/(x + 1)] + [3/(5x + 1)] = 5/(x + 4); x ≠ -1, -⅕, -4
Solution:
Given,
[1/(x + 1)] + [3/(5x + 1)] = 5/(x + 4); x ≠ -1, -⅕, -4Let us take the LCM of denominators and cross multiply the terms.
[1(5x + 1) + 3(x + 1)]/ [(x + 1)(5x + 1)] = 5/(x + 4) [5x + 1 + 3x + 3]/ [5x2 + x + 5x + 1] = 5/(x + 4)(8x + 4)(x + 4) = 5(5x2 + 6x + 1)
8x2 + 32x + 4x + 16 = 25x2 + 30x + 5
25x2 + 30x + 5 – 8x2 – 36x – 16 = 0
17x2 – 6x – 11 = 0
17x2 – 17x + 11x – 11 = 0
17x(x – 1) + 11(x – 1) = 0
(17x + 11)(x – 1) = 0
17x + 11 = 0, x – 1 = 0
x = -11/17, x = 1
Also Check: Polynomials Class 10 Extra Questions Maths Chapter 2
Q.4: Find the roots of quadratic equations by factorisation:
(i) √2 x2 + 7x + 5√2=0
(ii) 100x2 – 20x + 1 = 0
Solution:
(i) √2 x2 + 7x + 5√2=0
Considering the L.H.S. first,
⇒ √2 x2 + 5x + 2x + 5√2
⇒ x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)
The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (√2x + 5)(x + √2) = 0
Therefore, √2x + 5 = 0 or x + √2 = 0
⇒ x = -5/√2 or x = -√2
(ii) Given, 100x2 – 20x + 1=0
Considering the L.H.S. first,
⇒ 100x2 – 10x – 10x + 1
⇒ 10x(10x – 1) -1(10x – 1)
⇒ (10x – 1)2
The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0
Therefore,
(10x – 1) = 0
or (10x – 1) = 0
⇒ x =1/10 or x =1/10
Also Check: Some Applications of Trigonometry Class 10 Extra Questions Maths Chapter 9
Q.5: Find two consecutive positive integers, the sum of whose squares is 365.
Solution:
Let us say, the two consecutive positive integers be x and x + 1.
Therefore, as per the given statement,
x2 + (x + 1)2 = 365
⇒ x2 + x2 + 1 + 2x = 365
⇒ 2x2 + 2x – 364 = 0
⇒ x2 + x – 182 = 0
⇒ x2 + 14x – 13x – 182 = 0
⇒ x(x + 14) -13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
Thus, either, x + 14 = 0 or x – 13 = 0,
⇒ x = – 14 or x = 13
since, the integers are positive, so x can be 13, only.
So, x + 1 = 13 + 1 = 14
Therefore, the two consecutive positive integers will be 13 and 14.
Q.6: In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the original duration of the flight.
Solution:
Let the duration of the flight be x hours.
According to the given,
(600/x) – [600/(x + 1/2) = 200
(600/x) – [1200/(2x + 1)] = 200
[600(2x + 1) – 1200x]/ [x(2x + 1)] = 200(1200x + 600 – 1200x)/ [x(2x + 1)] = 200
600 = 200x(2x + 1)
x(2x + 1) = 3
2x2 + x – 3 = 0
2x2 + 3x – 2x – 3 = 0
x(2x + 3) – 1(2x + 3) = 0
(2x + 3)(x – 1) = 0
2x + 3 = 0, x – 1 = 0
x = -3/2, x = 1
Time cannot be negative.
Therefore, x = 1
Hence, the original duration of the flight is 1 hr.
Q.7: If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k.
Solution:
Given that x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0.
⇒ (3)2 – 2k(3) – 6 = 0
⇒ 9 – 6k – 6 = 0
⇒ 3 – 6k = 0
⇒ 6k = 3
⇒ k = 1/2
Therefore, the value of k is ½.
Q.8: The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let us say, the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
Diagonal of the rectangle = √[x2+(x+30)2] As given, the length of the diagonal is = x + 60 m
⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x -90)
⇒ (x – 90)(x + 30) = 0
⇒ x = 90, -30
Q.9 : Solve the quadratic equation 2x2 – 7x + 3 = 0 by using quadratic formula.
Solution:
2x2 – 7x + 3 = 0
On comparing the given equation with ax2 + bx + c = 0, we get,
a = 2, b = -7 and c = 3
By using quadratic formula, we get,
x = [-b±√(b2 – 4ac)]/2a
⇒ x = [7±√(49 – 24)]/4
⇒ x = [7±√25]/4
⇒ x = [7±5]/4
Therefore,
⇒ x = 7+5/4 or x = 7-5/4
⇒ x = 12/4 or 2/4
∴ x = 3 or ½
Q.10: Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Solution:
(i) 2x2 + kx + 3 = 0
Comparing the given equation with ax2 + bx + c = 0, we get,
a = 2, b = k and c = 3
As we know, Discriminant = b2 – 4ac
= (k)2 – 4(2) (3)
= k2 – 24
For equal roots, we know,
Discriminant = 0
k2 – 24 = 0
k2 = 24
k = ±√24 = ±2√6
(ii) kx(x – 2) + 6 = 0
or kx2 – 2kx + 6 = 0
Comparing the given equation with ax2 + bx + c = 0, we get
a = k, b = – 2k and c = 6
We know, Discriminant = b2 – 4ac
= ( – 2k)2 – 4 (k) (6)
= 4k2 – 24k
For equal roots, we know,
b2 – 4ac = 0
4k2 – 24k = 0
4k (k – 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘.
Therefore, if this equation has two equal roots, k should be 6 only.
Q.11: The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Sum of the areas of two squares is 468 m².
∵ x² + y² = 468 . ………..(1) [ ∵ area of square = side²]
→ The difference of their perimeters is 24 m.
∵ 4x – 4y = 24 [ ∵ Perimeter of square = 4 × side] ⇒ 4( x – y ) = 24
⇒ x – y = 24/4
⇒ x – y = 6
∴ y = x – 6 ……….(2)
From equation (1) and (2),
∵ x² + ( x – 6 )² = 468
⇒ x² + x² – 12x + 36 = 468
⇒ 2x² – 12x + 36 – 468 = 0
⇒ 2x² – 12x – 432 = 0
⇒ 2( x² – 6x – 216 ) = 0
⇒ x² – 6x – 216 = 0
⇒ x² – 18x + 12x – 216 = 0
⇒ x( x – 18 ) + 12( x – 18 ) = 0
⇒ ( x + 12 ) ( x – 18 ) = 0
⇒ x + 12 = 0 and x – 18 = 0
⇒ x = – 12m [ rejected ] and x = 18m
∴ x = 18 m
Put the value of ‘x’ in equation (2),
∵ y = x – 6
⇒ y = 18 – 6
∴ y = 12 m
Hence, sides of two squares are 18m and 12m respectively.
Q.12: Find the discriminant of the equation 3x2– 2x +1/3= 0 and hence find the nature of its roots. Find them, if they are real.
Solution:
Given,
3x2– 2x +1/3= 0
Here, a = 3, b = – 2 and c = 1/3
Since, Discriminant = b2 – 4ac
= (– 2)2 – 4 × 3 × 1/3
= 4 – 4 = 0.
Hence, the given quadratic equation has two equal real roots.
The roots are -b/2a and -b/2a.
2/6 and 2/6
or
1/3, 1/3
