# NCERT Solutions for Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.4

NCERT Solutions is the best guide for the students with detailed study material including the important topics. NCERT Solutions for Class 10 Chapter 4- Quadratic Equations is an important chapter and the students are advised to deal with it carefully. NCERT for Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.4 contains all the solutions to the exercise mentioned on page number 91 in the textbook.

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### Access other exercise solutions of Class 10 Maths Chapter 4- Quadratic Equations

Exercise 4.1 Solutionsâ€“ 2 Questions

Exercise 4.2 Solutionsâ€“ 6 Questions

Exercise 4.3 Solutionsâ€“ 11 Questions

### Access Answers of Maths NCERT Class 10 Chapter 4- Quadratic Equations Exercise 4.4

1. Find the nature of the roots of the following quadratic equations.Â If the real roots exist, find them;
(i) 2x2Â â€“ 3xÂ + 5 = 0
(ii) 3x2Â â€“Â 4âˆš3xÂ + 4 = 0
(iii) 2x2Â â€“Â 6xÂ + 3 = 0

Solutions:

(i) Given,

2x2Â â€“ 3xÂ + 5 = 0

Comparing the equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get

aÂ = 2,Â bÂ = -3 andÂ cÂ = 5

We know, Discriminant =Â b2Â â€“ 4ac

=Â ( â€“ 3)2Â â€“ 4 (2) (5) = 9 â€“ 40

= â€“ 31

As you can see, b2Â â€“ 4ac < 0

Therefore, no real root is possible for the given equation, 2x2Â â€“ 3xÂ + 5 = 0.

(ii) 3x2Â â€“ 4âˆš3xÂ + 4 = 0

Comparing the equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get

aÂ = 3,Â bÂ =Â -4âˆš3Â andÂ cÂ = 4

We know, Discriminant =Â b2Â â€“ 4ac

= (-4âˆš3)2Â â€“ 4(3)(4)

= 48 â€“ 48 = 0

AsÂ b2Â â€“ 4acÂ = 0,

Real roots exist for the given equation and they are equal to each other.

Hence the roots will be â€“b/2aÂ andÂ â€“b/2a.

â€“b/2aÂ = -(-4âˆš3)/2Ã—3 = 4âˆš3/6 = 2âˆš3/3 = 2/âˆš3

Therefore, the roots areÂ 2/âˆš3Â and 2/âˆš3.

(iii) 2x2Â â€“Â 6xÂ + 3 = 0

Comparing the equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get

aÂ = 2,Â bÂ = -6,Â cÂ = 3

As we know, Discriminant =Â b2Â â€“ 4ac

= (-6)2Â â€“ 4 (2) (3)

= 36 â€“ 24 = 12

AsÂ b2Â â€“ 4acÂ > 0,

Therefore, there are distinct real roots exist for this equation, 2x2Â â€“Â 6xÂ + 3 = 0.

= (-(-6) Â± âˆš(-62-4(2)(3)) )/ 2(2)

= (6Â±2âˆš3 )/4

= (3Â±âˆš3)/2

Therefore the roots for the given equation are (3+âˆš3)/2 and (3-âˆš3)/2

2. Find the values ofÂ kÂ for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2Â +Â kxÂ + 3 = 0
(ii)Â kxÂ (xÂ â€“ 2) + 6 = 0

Solutions:

(i) 2x2Â +Â kxÂ + 3 = 0

Comparing the given equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get,

aÂ = 2,Â bÂ = k andÂ cÂ = 3

As we know, Discriminant =Â b2Â â€“ 4ac

= (k)2Â â€“ 4(2) (3)

=Â k2Â â€“ 24

For equal roots, we know,

Discriminant = 0

k2Â â€“ 24 = 0

k2Â = 24

k = Â±âˆš24Â = Â±2âˆš6

(ii)Â kx(xÂ â€“ 2) + 6 = 0

orÂ kx2Â â€“ 2kxÂ + 6 = 0

Comparing the given equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get

aÂ =Â k,Â bÂ = â€“ 2kÂ andÂ cÂ = 6

We know, Discriminant =Â b2Â â€“ 4ac

= ( â€“ 2k)2 â€“ 4 (k) (6)

= 4k2 â€“ 24k

For equal roots, we know,

b2 â€“ 4acÂ = 0

4k2 â€“ 24kÂ = 0

4kÂ (kÂ â€“ 6) = 0

Either 4kÂ = 0 orÂ kÂ = 6 = 0

kÂ = 0 orÂ kÂ = 6

However, ifÂ kÂ = 0, then the equation will not have the terms â€˜x2â€˜ and â€˜xâ€˜.

Therefore, if this equation has two equal roots,Â kÂ should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2?Â If so, find its length and breadth.

Solution:

Let the breadth of mango grove beÂ l.

Length of mango grove will be 2l.

Area of mango grove = (2l) (l)= 2l2

2l2Â = 800

l2Â = 800/2 = 400

l2Â â€“ 400 =0

Comparing the given equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get

aÂ = 1,Â bÂ = 0,Â cÂ = 400

As we know, Discriminant =Â b2Â â€“ 4ac

=> (0)2Â â€“ 4 Ã— (1) Ã— ( â€“ 400) = 1600

Here,Â b2Â â€“ 4acÂ > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

lÂ = Â±20

As we know, the value of length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 Ã— 20 = 40 m

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution:

Letâ€™s say, the age of one friend be x years.

Then, the age of the other friend will be (20 â€“ x) years.

Four years ago,

Age of First friend = (xÂ â€“ 4) years

Age of Second friend = (20 â€“Â xÂ â€“ 4)Â = (16 â€“Â x) years

As per the given question, we can write,

(xÂ â€“ 4) (16 â€“Â x) = 48

16x â€“ x2Â â€“ 64 + 4xÂ = 48

Â â€“ x2Â +Â 20x â€“Â 112 = 0

x2Â â€“Â 20x +Â 112 = 0

Comparing the equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get

aÂ =Â 1,Â bÂ = -20Â andÂ cÂ = 112

Discriminant =Â b2Â â€“ 4ac

= (-20)2Â â€“ 4 Ã— 112

= 400 â€“ 448 = -48

b2Â â€“ 4acÂ < 0

Therefore, there will be no real solution possible for the equations. Hence, condition doesnâ€™t exist.

5.Â Is it possible to design a rectangular park of perimeter 80 and area 400 m2? 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 + bÂ = 40

Or,Â bÂ = 40 â€“Â l

Area of the rectangular park =Â lÃ—b = l(40 â€“ l) =Â 40lÂ â€“Â l2 = 400

l2Â â€“Â Â 40lÂ + 400Â = 0, which is a quadratic equation.

Comparing the equation withÂ ax2Â +Â bxÂ +Â cÂ = 0, we get

aÂ = 1,Â bÂ = -40,Â cÂ = 400

Since, Discriminant =Â b2Â â€“ 4ac

=(-40)2Â â€“ 4 Ã— 400

= 1600 â€“ 1600 = 0

Thus, b2Â â€“ 4acÂ = 0

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

Root of the equation,

lÂ = â€“b/2a

lÂ = -(-40)/2(1) = 40/2 = 20

Therefore, length of rectangular park,Â lÂ = 20 m

And breadth of the park,Â bÂ = 40 â€“Â lÂ = 40 â€“ 20 = 20 m.

### Key Features of NCERT Solutions Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.4

• The subject experts have provided the solutions after a lot of brainstorming.
• The answers are accurate.
• Each question in the NCERT Solutions Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.4 is explained in a stepwise manner.
• The solutions will help students score well in the second term examinations.

#### 1 Comment

1. Fazila Javaid

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Thanks