Quadratic Equations Class 10 Mathematics

20 Questions | 60 Marks

[3 marks] Represent the following situation using a quadratic equation: A rectangular park is designed such that its length is 5 metres more than twice its breadth. If the area of the park is 250 square metres, find an equation to determine its dimensions.

[2 marks] Check if the equation

(x + 2)^{3} = x^{3} + 8

is a quadratic equation. Justify your answer.

[3 marks] Find the roots of the quadratic equation

3 x^{2} - 10 x + 3 = 0

by factorisation.

[2 marks] Find the sum and product of the roots of the quadratic equation

2 x^{2} + 7 x - 4 = 0

[3 marks] For what value of 'k' will the quadratic equation

k x^{2} - 14 x + 8 = 0

have equal and real roots?

[2 marks] Find the discriminant of the quadratic equation

2 x^{2} + 4 x - 5 = 0

and hence determine the nature of its roots.

[3 marks] The sum of squares of two consecutive positive integers is 61. Formulate a quadratic equation to find these integers.

[5 marks] A train travels a distance of 300 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 2 hours less for the same journey. Formulate a quadratic equation to find the original speed of the train.

[3 marks] Solve the quadratic equation

x^{2} - (a + b) x + a b = 0

for x.

[2 marks] A rectangular garden has an area of 120 sq m. If the length is 2 m more than twice the breadth, find the dimensions using a quadratic equation.

[3 marks] Find the roots of the quadratic equation

\sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0

[2 marks] Find the value of k for which the quadratic equation

x^{2} - k x + 4 = 0

has equal roots.

[5 marks] A person bought some articles for Rs. 600. If 10 more articles had been bought for the same amount, each article would have cost Rs. 2 less. Formulate a quadratic equation to find the number of articles bought.

[3 marks] Find the roots of the quadratic equation

4 x^{2} + 4 \sqrt{3} x + 3 = 0

[2 marks] The discriminant of the quadratic equation

a x^{2} + b x + c = 0

is given by

D = b^{2} - 4 a c

. If

D < 0

, what can be said about the roots of the equation?

[3 marks] A rectangular park is surrounded by a path of uniform width. The dimensions of the park are 20 m by 10 m. If the area of the path is 144 sq m, find the width of the path. (Assume the path is outside the park).

[2 marks] For what value of k, the equation

k x^{2} - 6 x - 1 = 0

has real roots?

[3 marks] Two positive integers differ by 3. The sum of their squares is 117. Find the integers by forming and solving a quadratic equation.

[5 marks] A rectangular lawn is bordered by a gravel path of uniform width. The lawn is 10 m by 8 m. The area of the path is 120 sq m. Find the width of the path.

[2 marks] If the roots of the quadratic equation

x^{2} - p x + q = 0

are

2

and

3

, find the values of

p

and

q

Answers

Let the breadth of the rectangular park be x metres. Then the length is

(2 x + 5)

metres. The area is given by length

\times

breadth. So, the quadratic equation is

x (2 x + 5) = 250

, which simplifies to

2 x^{2} + 5 x - 250 = 0

(x + 2)^{3} = x^{3} + 8

x^{3} + 3 x^{2} (2) + 3 x (2^{2}) + 2^{3} = x^{3} + 8

x^{3} + 6 x^{2} + 12 x + 8 = x^{3} + 8

6 x^{2} + 12 x = 0

This is a quadratic equation because the highest power of x is 2.

3 x^{2} - 10 x + 3 = 0

We need to find two numbers whose product is

3 \times 3 = 9

and whose sum is

- 10

. These numbers are

- 9

and

- 1

3 x^{2} - 9 x - x + 3 = 0

3 x (x - 3) - 1 (x - 3) = 0

(3 x - 1) (x - 3) = 0

So,

3 x - 1 = 0

x - 3 = 0

x = \frac{1}{3}

x = 3

.The roots are

\frac{1}{3}

and

3

For the quadratic equation

a x^{2} + b x + c = 0

, the sum of roots is

- \frac{b}{a}

and the product of roots is

\frac{c}{a}

.For

2 x^{2} + 7 x - 4 = 0

a = 2

b = 7

c = - 4

.Sum of roots

= - \frac{7}{2}

Product of roots

= \frac{- 4}{2} = - 2

For equal and real roots, the discriminant

b^{2} - 4 a c = 0

.In the equation

k x^{2} - 14 x + 8 = 0

a = k

b = - 14

c = 8

(- 14)^{2} - 4 (k) (8) = 0

196 - 32 k = 0

32 k = 196

k = \frac{196}{32} = \frac{49}{8}

For the quadratic equation

2 x^{2} + 4 x - 5 = 0

a = 2

b = 4

c = - 5

.Discriminant

D = b^{2} - 4 a c = (4)^{2} - 4 (2) (- 5) = 16 + 40 = 56

.Since

D = 56 > 0

, the equation has two distinct real roots.

Let the two consecutive positive integers be

x

and

x + 1

.According to the problem, the sum of their squares is 61.

x^{2} + (x + 1)^{2} = 61

x^{2} + x^{2} + 2 x + 1 = 61

2 x^{2} + 2 x + 1 - 61 = 0

2 x^{2} + 2 x - 60 = 0

x^{2} + x - 30 = 0

. This is the required quadratic equation.

Let the original speed of the train be

v

km/h.The distance is 300 km.Time taken at original speed

= \frac{300}{v}

hours.If the speed had been 5 km/h more, the new speed would be

(v + 5)

km/h.Time taken at increased speed

= \frac{300}{v + 5}

hours.According to the problem, the new time is 2 hours less than the original time.

\frac{300}{v + 5} = \frac{300}{v} - 2

Multiply by $v (v + 5)$ :

300 v = 300 (v + 5) - 2 v (v + 5)

300 v = 300 v + 1500 - 2 v^{2} - 10 v

0 = 1500 - 2 v^{2} - 10 v

2 v^{2} + 10 v - 1500 = 0

v^{2} + 5 v - 750 = 0

. This is the required quadratic equation.

The equation is

x^{2} - (a + b) x + a b = 0

.We can factorize this by splitting the middle term:

x^{2} - a x - b x + a b = 0

x (x - a) - b (x - a) = 0

(x - a) (x - b) = 0

So,

x - a = 0

x - b = 0

.The roots are

x = a

and

x = b

Let the breadth of the rectangular garden be

b

metres.The length is

l = 2 b + 2

metres.Area =

l \times b = (2 b + 2) b = 2 b^{2} + 2 b

.Given area is 120 sq m.

2 b^{2} + 2 b = 120

2 b^{2} + 2 b - 120 = 0

b^{2} + b - 60 = 0

. This is the quadratic equation for the breadth.Solving:

(b + 6) (b - 10) = 0

. Since breadth must be positive,

b = 10

m.Length

l = 2 (10) + 2 = 22

m.The dimensions are 10 m and 22 m.

\sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0

We need two numbers whose product is

(\sqrt{3}) (6 \sqrt{3}) = 18

and sum is

11

. These numbers are

9

and

2

\sqrt{3} x^{2} + 9 x + 2 x + 6 \sqrt{3} = 0

$\sqrt{3} x^{2} + 3 \sqrt{3} x + 6 x + 6 \sqrt{3} = 0$ (Rewriting $9 x$ as $3 \sqrt{3} x$ )
$\sqrt{3} x (x + 3) + 6 (x + \sqrt{3}) = 0$ This factorization is incorrect. Let's use correct splitting of terms.

\sqrt{3} x^{2} + 9 x + 2 x + 6 \sqrt{3} = 0

\sqrt{3} x (x + 3 \sqrt{3}) + 2 (x + 3 \sqrt{3}) = 0

(\sqrt{3} x + 2) (x + 3 \sqrt{3}) = 0

So,

\sqrt{3} x + 2 = 0

x + 3 \sqrt{3} = 0

x = - \frac{2}{\sqrt{3}} = - \frac{2 \sqrt{3}}{3}

x = - 3 \sqrt{3}

.The roots are

- \frac{2 \sqrt{3}}{3}

and

- 3 \sqrt{3}

For equal roots, the discriminant

b^{2} - 4 a c = 0

.In the equation

x^{2} - k x + 4 = 0

a = 1

b = - k

c = 4

(- k)^{2} - 4 (1) (4) = 0

k^{2} - 16 = 0

k^{2} = 16

k = \pm 4

Let the number of articles bought be

n

.The cost of each article was Rs. 600.So, the cost per article

= \frac{600}{n}

.If 10 more articles were bought for Rs. 600, the new number of articles is

(n + 10)

.The new cost per article

= \frac{600}{n + 10}

.According to the problem, the new cost is Rs. 2 less than the original cost.

\frac{600}{n + 10} = \frac{600}{n} - 2

Multiply by $n (n + 10)$ :

600 n = 600 (n + 10) - 2 n (n + 10)

600 n = 600 n + 6000 - 2 n^{2} - 20 n

0 = 6000 - 2 n^{2} - 20 n

2 n^{2} + 20 n - 6000 = 0

n^{2} + 10 n - 3000 = 0

. This is the required quadratic equation.

4 x^{2} + 4 \sqrt{3} x + 3 = 0

We can see this is a perfect square trinomial.

(2 x)^{2} + 2 (2 x) (\sqrt{3}) + (\sqrt{3})^{2} = 0

(2 x + \sqrt{3})^{2} = 0

So, $2 x + \sqrt{3} = 0$ .

2 x = - \sqrt{3}

x = - \frac{\sqrt{3}}{2}

.This is a repeated root.

If the discriminant

D = b^{2} - 4 a c < 0

, the quadratic equation

a x^{2} + b x + c = 0

has no real roots. The roots are complex conjugates.

Let the width of the path be

w

metres.The dimensions of the park are 20 m by 10 m.The dimensions of the park including the path will be

(20 + 2 w)

m by

(10 + 2 w)

m.Area of the park

= 20 \times 10 = 200

sq m.Area of the park including the path

= (20 + 2 w) (10 + 2 w) = 200 + 40 w + 20 w + 4 w^{2} = 4 w^{2} + 60 w + 200

.Area of the path = (Area of park + path) - (Area of park)

144 = (4 w^{2} + 60 w + 200) - 200

144 = 4 w^{2} + 60 w

4 w^{2} + 60 w - 144 = 0

$w^{2} + 15 w - 36 = 0$ .
Solving this quadratic equation for $w$ :
Using the quadratic formula: $w = \frac{- 15 \pm \sqrt{15^{2} - 4 (1) (- 36)}}{2 (1)} = \frac{- 15 \pm \sqrt{225 + 144}}{2} = \frac{- 15 \pm \sqrt{369}}{2}$ .
This does not give a simple integer or rational value for width. Let's recheck the problem statement or calculation.
Ah, the question does not state it has to be a simple value. However, typical CBSE questions lead to simpler answers. Let's assume there might be a slight error in the problem numbers as given, but we will proceed with the calculation.
$\sqrt{369} \approx 19.21$ . So $w \approx \frac{- 15 \pm 19.21}{2}$ .
$w \approx \frac{4.21}{2} \approx 2.105$ or $w \approx \frac{- 34.21}{2}$ (negative, ignore).
If we assume the numbers were intended to give simpler roots, for instance if the path area was 224 sq m:

4 w^{2} + 60 w = 224

4 w^{2} + 60 w - 224 = 0

$w^{2} + 15 w - 56 = 0$ .
This also doesn't factor easily.
Let's try factoring $w^{2} + 15 w - 36 = 0$ . We need two numbers whose product is -36 and sum is 15. No integers work.
Let's reconsider.
If area of path = 144, and park is 20x10.

(20 + 2 w) (10 + 2 w) - 200 = 144

200 + 40 w + 20 w + 4 w^{2} - 200 = 144

4 w^{2} + 60 w - 144 = 0

w^{2} + 15 w - 36 = 0

.Let's check the question formulation. If the path area leads to a cleaner number.Suppose width is 2m. Area of path =

4 (2^{2}) + 60 (2) = 16 + 120 = 136

. Close to 144.Suppose width is 2.1m. Area of path =

4 ({2.1}^{2}) + 60 (2.1) = 4 (4.41) + 126 = 17.64 + 126 = 143.64

. Very close.Let's check if there's a mistake in the NCERT example or typical question style. The sample problem is 4.4 Example 8. A pole. The equation was x^2+7x-60=0.Let's assume the question is correctly stated and requires the use of the quadratic formula.

w = \frac{- 15 \pm \sqrt{15^{2} - 4 (1) (- 36)}}{2} = \frac{- 15 \pm \sqrt{225 + 144}}{2} = \frac{- 15 \pm \sqrt{369}}{2}

.Since width must be positive,

w = \frac{- 15 + \sqrt{369}}{2}

metres.

\sqrt{369} = \sqrt{9 \times 41} = 3 \sqrt{41}

.So,

w = \frac{- 15 + 3 \sqrt{41}}{2}

metres.

For the equation

k x^{2} - 6 x - 1 = 0

to have real roots, the discriminant

D = b^{2} - 4 a c \geq 0

.Here,

a = k

b = - 6

c = - 1

(- 6)^{2} - 4 (k) (- 1) \geq 0

36 + 4 k \geq 0

4 k \geq - 36

k \geq - 9

.So, for real roots,

k

must be greater than or equal to

- 9

. If the question implied real and equal roots, then

k = - 9

. If it implies real roots (distinct or equal), then

k \geq - 9

Let the two positive integers be

x

and

y

.We are given

x - y = 3

(or

y - x = 3

). Let's assume

x > y

, so

x = y + 3

.The sum of their squares is 117:

x^{2} + y^{2} = 117

.Substitute

x = y + 3

(y + 3)^{2} + y^{2} = 117

y^{2} + 6 y + 9 + y^{2} = 117

2 y^{2} + 6 y + 9 - 117 = 0

2 y^{2} + 6 y - 108 = 0

y^{2} + 3 y - 54 = 0

. This is the required quadratic equation.Factorizing: We need two numbers whose product is -54 and sum is 3. These are 9 and -6.

(y + 9) (y - 6) = 0

.Since the integers are positive,

y = 6

.Then

x = y + 3 = 6 + 3 = 9

.The two integers are 6 and 9.

Let the width of the path be

w

metres.The dimensions of the lawn are 10 m by 8 m.The dimensions of the lawn including the path will be

(10 + 2 w)

m by

(8 + 2 w)

m.Area of the lawn

= 10 \times 8 = 80

sq m.Area of the lawn including the path

= (10 + 2 w) (8 + 2 w) = 80 + 20 w + 16 w + 4 w^{2} = 4 w^{2} + 36 w + 80

.Area of the path = (Area of lawn + path) - (Area of lawn)

120 = (4 w^{2} + 36 w + 80) - 80

120 = 4 w^{2} + 36 w

4 w^{2} + 36 w - 120 = 0

w^{2} + 9 w - 30 = 0

.Solving this quadratic equation for

w

using the quadratic formula:

w = \frac{- 9 \pm \sqrt{9^{2} - 4 (1) (- 30)}}{2 (1)} = \frac{- 9 \pm \sqrt{81 + 120}}{2} = \frac{- 9 \pm \sqrt{201}}{2}

.Since width must be positive,

w = \frac{- 9 + \sqrt{201}}{2}

metres.

\sqrt{201} \approx 14.18

w \approx \frac{- 9 + 14.18}{2} \approx \frac{5.18}{2} \approx 2.59

metres.

For a quadratic equation

a x^{2} + b x + c = 0

with roots

α

and

β

:Sum of roots

α + β = - \frac{b}{a}

Product of roots

α β = \frac{c}{a}

Given equation:

x^{2} - p x + q = 0

. Here

a = 1

b = - p

c = q

.Given roots are 2 and 3.Sum of roots

= 2 + 3 = 5

.So,

5 = - \frac{- p}{1} = p

. Thus,

p = 5

.Product of roots

= 2 \times 3 = 6

.So,

6 = \frac{q}{1} = q

. Thus,

q = 6

.The values are

p = 5

and

q = 6

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