# 10th cbse maths solution for exercise 4.4 part 1

This page 10th cbse maths solution for exercise 4.4 part 1 is going to provide you solution for every problems that you find in the exercise no 4.4 part 1

## 10th CBSE maths solution for Exercise 4.4 part 1

(iii) 2 x² - 6 x + 3 = 0

Discriminant = b² - 4 a c

a = 2  b = - 6 and c = 3

= (-6)² - 4 (2) (3)

= 36 - 24

= 12 > 0

It has two distinct real roots.

We cannot factorize the given equation. To solve this we have to use the quadratic formula

x = (- b ± √ b² - 4 a c)/2a

x = [-(-6) ± √12]/2(2)

x = [6 ± √12]/4

x = [6 ± 2√3]/4

x = 2 [3 ± √3]/4

x = (3 ± √3)/2

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

(i) 2 x² + k x + 3 = 0

Since the equation has two equal roots

Discriminant = 0

b² - 4 a c = 0

a = 2  b = k and c = 3

k² - 4 (2) (3) = 0

k² - 24 = 0

k² = 24

k = √24

k = √2 x 2 x 2 x 3

k = 2√6

(ii) k x (x - 2) + 6 = 0

k x² - 2 k x + 6 = 0

Since the equation has two equal roots

Discriminant = 0

b² - 4 a c = 0

a = k  b = -2 k and c = 6

(-2 k)² - 4 (k) (6) = 0

4 k² - 24 k = 0

4 k (k - 6) = 0

4 k = 0            k - 6 = 0

k = 0                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 "x" be the breadth of the rectangular grove

length = 2 x

Area of rectangular grove = 800 m²

x (2x) = 800

2 x² = 800

x² = 800/2

x² = 400

x =  √400

x = 20

breadth of the rectangular grove = 20 m

length of rectangular grove = 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 "x" be the age of one friend

age of another friend = 20 - x

four years ago their ages was

x - 4 and 20 - x - 4

x - 4 and 16 - x

The product of their ages = 48

(x - 4) (16 - x) = 48

16 x - x² - 64 + 4 x = 48

- x²+ 20 x - 64 - 48 = 0

- x²+ 20 x - 112 = 0

multiplying the whole equation by negative

x²- 20 x + 112 = 0

by comparing this equation with general form of quadratic equation we get

a = 1  b = -20 and c = 112

Discriminant = b² - 4 a c

= (-2 0)² - 4 (1) (112)

= 400 - 448

= -48 < 0

Since discriminant is less than zero,there is no real roots for this equation. Therefore we can say the above 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:

perimeter of rectangular park = 80

2 (l + b) = 80

l + b =  40 -----(1)

Area of rectangular park = 400 m²

l x b = 400

b = 400/l ------(2)

Substitute (2) equation in the first equation

l + (400/l) = 40

(l² + 400)/l = 40

l² + 400 = 40 l

l² - 40 l + 400 = 0

l² - 20 l - 20 l + 400 =  0

l (l - 20) - 20 (l - 20) = 0

(l - 20) (l - 20) = 0

l - 20 = 0

l = 20 m

b = 400/20

b = 20 m

Therefore the length of the rectangular park = 20 m and breadth of the rectangular park = 20 m