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

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

### Access Other Exercise Solutions of Class 10 Maths Chapter 4 – Quadratic Equations

Exercise 4.1 Solutions– 2 Questions

Exercise 4.2 Solutions– 6 Questions

Exercise 4.3 Solutions– 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. If the real roots exist, find them:
(i) 2x2Â – 3xÂ + 5 = 0
(ii) 3x2Â –Â 4âˆš3xÂ + 4 = 0
(iii) 2x2Â –Â 6xÂ + 3 = 0

Solutions:

(i) Given,

2x2Â – 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 = b2Â – 4ac

=Â ( – 3)2Â – 4 (2) (5) = 9 – 40

= – 31

As you can see, b2Â – 4ac < 0

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

(ii) 3x2Â – 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 – 4ac

= (-4âˆš3)2Â – 4(3)(4)

= 48 – 48 = 0

AsÂ b2Â – 4acÂ = 0,

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

Hence, the roots will be –b/2aÂ andÂ –b/2a.

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

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

(iii) 2x2Â –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 = b2Â – 4ac

= (-6)2Â – 4 (2) (3)

= 36 – 24 = 12

AsÂ b2Â – 4acÂ > 0,

Therefore, there are distinct real roots that exist for this equation, 2x2Â –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 k for each of the following quadratic equations so that they have two equal roots.
(i) 2x2Â +Â kxÂ + 3 = 0
(ii)Â kxÂ (xÂ – 2) + 6 = 0

Solutions:

(i) 2x2Â +Â kxÂ + 3 = 0

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

aÂ = 2,Â bÂ = k andÂ cÂ = 3

As we know, discriminant = b2Â – 4ac

= (k)2Â – 4(2) (3)

=Â k2Â – 24

For equal roots, we know,

Discriminant = 0

k2Â – 24 = 0

k2Â = 24

k = Â±âˆš24Â = Â±2âˆš6

(ii)Â kx(xÂ – 2) + 6 = 0

orÂ kx2Â – 2kxÂ + 6 = 0

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

aÂ =Â k,Â bÂ = – 2kÂ andÂ cÂ = 6

We know, Discriminant =Â b2Â – 4ac

= ( – 2k)2 – 4 (k) (6)

= 4k2 – 24k

For equal roots, we know,

b2 – 4acÂ = 0

4k2 – 24kÂ = 0

4kÂ (kÂ – 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 ‘x2‘ and ‘x‘.

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 the mango grove be l.

The length of the mango grove will be 2l.

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

2l2Â = 800

l2Â = 800/2 = 400

l2Â – 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 = b2Â – 4ac

=> (0)2Â – 4 Ã— (1) Ã— ( – 400) = 1600

Here,Â b2Â – 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, the breadth of the mango grove = 20 m.

Length of the 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 the two friends is 20 years. Four years ago, the product of their age in years was 48.

Solution:

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

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of first friend = (xÂ – 4) years

Age of second friend = (20 – xÂ – 4)Â = (16 –Â x) years

As per the given question, we can write,

(xÂ – 4) (16 –Â x) = 48

16x – x2Â – 64 + 4xÂ = 48

Â – x2Â +Â 20x –Â 112 = 0

x2Â –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 – 4ac

= (-20)2Â – 4 Ã— 112

= 400 – 448 = -48

b2Â – 4acÂ < 0

Therefore, there will be no real solution possible for the equations. Hence, the condition doesn’t exist.

5. Is it possible to design a rectangular park with a perimeter of 80 and an area of 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 l

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

l2Â 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 = b2Â – 4ac

=(-40)2Â – 4 Ã— 400

= 1600 – 1600 = 0

Thus, b2Â – 4acÂ = 0

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

The root of the equation,

lÂ = –b/2a

lÂ = -(-40)/2(1) = 40/2 = 20

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

And the breadth of the park, bÂ = 40 –Â lÂ = 40 – 20 = 20 m.

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

• 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 board examinations.