NCERT Solutions for Class 10 Maths Ex 4.3

# NCERT Solutions for Class 10 Maths Ex 4.3

In this post, you will find the NCERT solutions for class 10 maths ex 4.3. These solutions are based on the latest syllabus of NCERT Maths class 10.

NCERT Solutions for Class 10 Maths Ex 4.3

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x 3x + 5 = 0

(ii) 3x− 4√3x + 4 = 0

(iii) 2x 6x + 3 = 0

Solution: (i) 2x 3x + 5 = 0

Comparing the equation 2x 3x + 5 = 0 with the general form of a quadratic equation axbx + c = 0, we get

= 2, = −3 and = 5

Discriminant (D) = b− 4ac = (−3)– 4 × 2 × 5

= 9 – 40 = −31

Here, D < 0, which means that the equation has no real roots.

(ii) 3x− 4√3x + 4 = 0

Comparing the equation 3x− 4√3x + 4 = 0 with the general form of a quadratic equation axbx + c = 0, we get

= 3, = −4√3 and c = 4

Discriminant (D) = b− 4ac = (−4√3)− 4 × 3 × 4

= 48 – 48 = 0

Here, D = 0, which means that the equation has two equal real roots.

Using the quadratic formula, we have

⇒

Putting the values of ab and c, we get

⇒

Since the equation has two equal roots, it means

(iii) 2x 6x + 3 = 0

Comparing the equation 2x 6x + 3 = 0 with the general form of a quadratic equation axbx + c = 0, we get

= 2, = −6 and = 3

Discriminant (D) = b− 4ac = (−6)− 4 × 2 × 3

= 36 – 24 = 12

Here, D > 0, which means that the equation has two distinct real roots.

Using the quadratic formula, we have

⇒

Putting the values of ab and c, we get

⇒

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2xkx + 3 = 0

(ii) kx(− 2) + 6 = 0

Solution: (i) 2xkx + 3 = 0

We know that a quadratic equation has two equal real roots, if the value of discriminant (D) is equal to zero.

Comparing the equation 2xkx + 3 = 0 with the general form of a quadratic equation axbx + c = 0, we get

= 2, k and = 3

Discriminant (D) = b− 4ac = k– 4 × 2 × 3 = k− 24

Since the equation has two equal roots, we have

D = 0

k− 24 = 0

k2 = 24

k = ±√24 = ±2√6

k = 2√6, −2√6

(ii) kx(− 2) + 6 = 0

kx2 – 2kx + 6 = 0

Comparing the equation kx2 – 2kx + 6 = 0 with the general form of a quadratic equation axbx + c = 0, we get

k= −2k and = 6

Discriminant (D) = b− 4ac = (−2k)− 4 × × 6 = 4k− 24k

We know that a quadratic equation has two equal real roots, if the value of discriminant (D) is equal to zero.

Since the equation has two equal roots, we have

D = 0

4k− 24k = 0

4k(− 6) = 0

= 0, 6

According to the definition of a quadratic equation, we have

A quadratic equation is the equation of the form axbx + c = 0, where ≠ 0.

Therefore, in equation kx− 2kx + 6 = 0, we cannot have k = 0.

Therefore, we discard k = 0.

Hence, the answer is k = 6.

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

Solution:   Let the breadth of the rectangular mango grove be x m.

Then, the length of the rectangular mango grove = 2x m

Area of the rectangular mango grove = length × breadth= x × 2x = 2xm2

According to the question, we have

2x2 = 800

2x− 800 = 0

⇒ x− 400 = 0

Comparing the equation x− 400 = 0 with the general form of a quadratic equation axbx + c = 0, we get

= 1, = 0 and = −400

Discriminant (D) = b− 4ac = (0)− 4 × 1 × (−400) = 1600

Since D > 0, it means that the equation has two distinct real roots.

Therefore, it is possible to design a rectangular mango grove.

Using the quadratic formula, we have

⇒

Putting the values of ab and c, we get

⇒

= 20, −20

We discard the negative value of x because the breadth of a rectangle cannot be negative.

Therefore, x = breadth of the rectangle = 20 m

And, length of the rectangle = 2= 2 × 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: Let the age of first friend be x years.

Then, the age of second friend = (20 − x) years.

Four years ago, the age of first friend = (− 4) years

Four years ago, the age of second friend = (20 − x) − 4 = (16 − x) years

According to the question, we have

(− 4)(16 − x) = 48

16− x– 64 + 4x = 48

20x − x− 112 = 0

x− 20x + 112 = 0

Comparing the equation x− 20x + 112 = 0 with the general form of a quadratic equation axbx + c = 0, we get

= 1, = −20 and = 112

Discriminant (D) = b− 4ac = (−20)– 4 × 1 × 112 = 400 – 448 = −48

Here D < 0, which means we have no real roots for this equation.

Therefore, the give situation is not possible.

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

Solution: Let the length of the park be x m.

It is given that the area of the rectangular park = 400 m2

Length × breadth = 400 m2

Therefore, breadth of park = 400/x m

Perimeter of the rectangular park = 2(length breath) = 2(x + 400/x) m

It is given that the perimeter of the rectangular park = 80 m

According to the question, we have

⇒

2x+ 800 = 80x

2x− 80x + 800 = 0

x− 40x + 400 = 0

Comparing the equation x− 40x + 400 = 0 with the general form of a quadratic equation axbx + c = 0, we get

= 1, = −40 and = 400

Discriminant (D) = b− 4ac = (−40)− 4 × 1 × 400 = 1600 – 1600 = 0

Here D = 0.

Therefore, the two roots of the equation are real and equal, which means that it is possible to design a rectangular park of the perimeter 80 m and area 400 m2.

Using the quadratic formula, we have

⇒

Putting the values of ab and c, we get

Here, both the roots are equal to 20.

Therefore, the length of the rectangular park = 20 m

And, the breadth of the rectangular park = 400/x = 400/20 = 20 m