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# Important Questions Class 10 Maths Chapter 4 Quadratic Equations

Important questions with solutions of chapter 4 quadratic equations for class 10 Maths are available here. Students preparing for the CBSE board exam 2022-2023 can practice these questions to score good marks. These questions have been formulated as per the NCERT book. These important questions are prepared by experts after thorough research and are based on the latest exam pattern. Students can expect these types of questions in Maths paper. They can access the important questions for all the chapters of Class 10 Maths here and practice well for the exam.

The questions here are based on quadratic equations, how to solve quadratic equations and find the roots by factorisation methods. Solving these questions will help students in revision as well. A total of 15 questions, along with the detailed solutions, have been provided below. Students must go through it to boost their Maths preparation. Also, practice the additional important questions to excel in problem-solving skills of quadratic equations.

## Important Questions with Solutions for Quadratic Equations

Let us solve some of the questions which are important with respect to the class 10th Maths board exam here.

Q.1: 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.

Q.2: 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

Q.3: 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.4: Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x2 – 7x +3 = 0

(ii) 2x2 + x – 4 = 0

Solution:

(i) 2x2 – 7x + 3 = 0
⇒ 2x2 – 7x = – 3
Dividing by 2 on both sides, we get
⇒x2–7x/2 = -3/2
⇒x2 – 2 × x × 7/4 = -3/2
On adding (7/4)2 to both sides of above equation, we get
⇒ (x)2 – 2 × x × 7/4 + (7/4)2 = (7/4)2 – (3/2)
⇒ (x – 7/4)2 = (49/16) -(3/2)
⇒ (x – 7/4)2 = 25/16
⇒ (x – 7/4) = ± 5/4
⇒ x = 7/4 ±5/4
⇒ x = 7/4 +5/4 or x = 7/4-5/4

⇒ x =12/4 or x =2/4
⇒ x = 3 or 1/2

(ii) 2x2 + x – 4 = 0
⇒ 2x2 + x = 4
Dividing both sides of the above equation by 2, we get
⇒ x2 + x/2 = 2

⇒ (x)2 + 2 × x × 1/4 = 2
Now on adding (1/4)2 to both sides of the equation, we get,

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2
⇒ (x + 1/4)2 = 33/16
⇒ x + 1/4 = ± √33/4
⇒ x = ± √33/4 – 1/4
⇒ x = ± √33 – 1/4

Therefore, either x = √33-1/4 or x = -√33-1/4.

Q.5: 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.6 : 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.7: 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.8: 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.9: Is it possible to design a rectangular park of perimeter 80 and area 400 sq.m.? If so find its length and breadth.

Solution:

Let the length and breadth of the park be L and B.

Perimeter of the rectangular park = 2 (L + B) = 80

So, L + B = 40

Or, B = 40 – L

Area of the rectangular park = L × B = L(40 – L) = 40L – L2 = 400

L2 – 40 L + 400 = 0,

Comparing the equation with ax2 + bx + c = 0, we get

a = 1, b = -40, c = 400

Since, Discriminant = b2 – 4ac

=>(-40)2 – 4 × 400

=> 1600 – 1600

= 0

Thus, b2 – 4ac = 0

Therefore, this equation has equal real roots. Hence, the situation is possible.

Root of the equation,

L = –b/2a

L = (40)/2(1) = 40/2 = 20

Therefore, length of rectangular park, L = 20 m

And breadth of the park, B = 40 – L = 40 – 20 = 20 m.

Q.10: 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

Q.11: 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.12: 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.13: Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.

Solution:

px2 – 14x + 8 = 0

Let α and 6α be the roots of the given quadratic equation.

Sum of the roots = -coefficient of x/coefficient of x2

α + 6α = -(-14)/p

7α = 14/p

α  = 2/p….(i)

Product of roots = constant term/coefficient of x2

(α)(6α) = 8/p

2 = 8/p

Substituting α = 2/p from (i),

6 × (2/p)2 = 8/p

24/p2 = 8/p

3/p = 1

p = 3

Therefore, the value of p is 3.

Q.14: 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, -4

Let 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

Q.15: 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.

### Class 10 Maths Chapter 4 Quadratic Equations Questions for Practice

1. Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0.
2. Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.
3. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/hr from the usual speed. Find its usual speed.
4. If ad ≠ bc, then prove that the equation (a2 + b2)x2 + 2(ac + bd)x + (c2 + d2) = 0 has no real roots.
5. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?
6. Solve for x: [1/(x – 1)(x – 2)] + [1/(x – 2)(x – 3)] = ⅔; x ≠ 1, 2, 3
7. If the quadratic equation px2 – 2√5 px + 15 = 0 has two equal roots, then find the value of p.
8. Solve the following quadratic equation for x:
4x2 + 4bx – (a2 – b2) = 0
9. Solve for x: √3 x2 – 2√2 x – 2√3 = 0
10. Solve the quadratic equation 2x2 + ax – a2 = 0 for x.

Related Concepts:

1. MALAVIKA M

THANKS ,IT IS VERY USEFUL FOR ME

2. Dark Knight

Ya,it is very useful

3. fathima raihan

i need extra questions. but it’s just the same question from the text

4. Divyansh bhati

Importent question in chapter 1 to 6

5. AYAN TRIVEDI

Really, these are the best questions