Important questions with solutions of chapter 4 quadratic equations for class 10 Maths are available here. Students who are preparing for CBSE board exam for the session 2020-2021, can practice these questions to score good marks. These questions have been formulated as per NCERT book. Students can also access important questions for all the chapters of Class 10 Maths here.

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 the revision for the exams.

## Important Questions with Solutions for Quadratic Equations

Let us solve some of the questions which are important with respect to 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 m ^{2}. 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 m^{2}

Putting the value of length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2x^{2} + x = 528

⇒ 2x^{2} + x – 528 = 0

Hence, 2x^{2} + 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

⇒ 3x^{2} – 24x – 3840 = 0

⇒ x^{2} – 8x – 1280 = 0

Hence, x^{2} – 8x – 1280 = 0 is the required representation of the problem mathematically.

**Q.2: Find the roots of quadratic equations by factorisation:**

**(i) √2 x ^{2} + 7x + 5√2=0**

**(ii) 100x ^{2} – 20x + 1 = 0**

Solution:

(i) √2 x^{2} + 7x + 5√2=0

Considering the L.H.S. first,

⇒ √2 x^{2} + 5x + 2x + 5√2

⇒ x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of this equation, √2 x^{2} + 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, 100x^{2} – 20x + 1=0

Considering the L.H.S. first,

⇒ 100x^{2} – 10x – 10x + 1

⇒ 10x(10x – 1) -1(10x – 1)

⇒ (10x – 1)^{2}

The roots of this equation, 100x^{2} – 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, 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,

x^{2} + (x + 1)^{2} = 365

⇒ x^{2} + x^{2} + 1 + 2x = 365

⇒ 2x^{2} + 2x – 364 = 0

⇒ x^{2} + x – 182 = 0

⇒ x^{2} + 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) 2x ^{2} – 7x +3 = 0**

**(ii) 2x ^{2} + x – 4 = 0**

Solution:

(i) 2x^{2} – 7x + 3 = 0

⇒ 2x^{2} – 7x = – 3

Dividing by 2 on both sides, we get

⇒x^{2}–7x/2 = -3/2

⇒x^{2} – 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) 2x^{2} + x – 4 = 0

⇒ 2x^{2} + x = 4

Dividing both sides of the above equation by 2, we get

⇒ x^{2} + 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 = √[x^{2}+(x+30)^{2}]
As given, the length of the diagonal is = x + 60 m

⇒ x^{2} + (x + 30)^{2} = (x + 60)^{2}

⇒ x^{2} + x^{2} + 900 + 60x = x^{2} + 3600 + 120x

⇒ x^{2} – 60x – 2700 = 0

⇒ x^{2} – 90x + 30x – 2700 = 0

⇒ x(x – 90) + 30(x -90)

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

**Q.6 : Solve the quadratic equation 2 x^{2} – 7x + 3 = 0 by using quadratic formula.**

Solution: 2x^{2} – 7x + 3 = 0

On comparing the given equation with ax^{2} + bx + c = 0, we get,

a = 2, b = -7 and c = 3

By using quadratic formula, we get,

x = [-b±√(b^{2} – 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: Sum of the areas of two squares is 468 m ^{2}. 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) 2x ^{2} + kx + 3 = 0**

**(ii) kx (x – 2) + 6 = 0**

Solution:

**(i) 2x ^{2} + kx + 3 = 0**

Comparing the given equation with ax^{2} + bx + c = 0, we get,

a = 2, b = k and c = 3

As we know, Discriminant = b^{2} – 4ac

= (k)^{2} – 4(2) (3)

= k^{2} – 24

For equal roots, we know,

Discriminant = 0

k^{2} – 24 = 0

k^{2} = 24

k = ±√24 = ±2√6

**(ii) kx(x – 2) + 6 = 0**

or kx^{2} – 2kx + 6 = 0

Comparing the given equation with ax^{2} + bx + c = 0, we get

a = k, b = – 2k and c = 6

We know, Discriminant = b^{2} – 4ac

= ( – 2k)^{2} – 4 (k) (6)

= 4k^{2} – 24k

For equal roots, we know,

b^{2} – 4ac = 0

4k^{2} – 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 ‘x^{2}‘ 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 – L^{2} = 400

L^{2} – 40 L + 400 = 0,

which is a quadratic equation.

Comparing the equation with ax^{2} + bx + c = 0, we get

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

Since, Discriminant = b^{2} – 4ac

=>(-40)^{2} – 4 × 400

=> 1600 – 1600

= 0

Thus, b^{2} – 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 3x ^{2}– 2x +1/3= 0 and hence find the nature of its roots. Find them, if they are real.**

Solution:

Given,

3x^{2}– 2x +1/3= 0

Here, a = 3, b = – 2 and c = 1/3

Since, Discriminant = b^{2} – 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

**Also check: **

THANKS ,IT IS VERY USEFUL FOR ME

Ya,it is very useful

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

Importent question in chapter 1 to 6

Really, these are the best questions

Very helpful, great job👍👌👏