159-NCERT-10-4-Quadratic Equations - Ex-4.4

NCERT

10th Mathematics

Exercise 4.4

Topic: 4 Quadratic Equations

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EXERCISE 4.4

Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² – 3x + 5 = 0 (ii) 3x² – 4√3 x + 4 = 0 (iii) 2x² – 6x + 3 = 0 

Explanation:

1) For the quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0, 

(b² - 4ac) is known as discriminant.

2) The quadratic equation ax² + bx + c = 0 has

(a) two distinct real roots, if b² – 4ac > 0,

(b) two equal real roots, if b² – 4ac = 0,

(c) no real roots, if b² – 4ac < 0.

Solution:

(i) 2x² – 3x + 5 = 0

1) The given equation is 2x² - 3x + 5 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 2x² - 3x + 5 = 0 with ax² + bx + c = 0, we have,

a = 2, b = - 3, c = 5.

3) First we will find:

b² - 4ac = (- 3)² - 4(2)(5)

b² - 4ac = 9 - 40

b² - 4ac = - 31 ------------------ equation 2. 

4) As b² - 4ac < 0, it has no real roots of the quadratic equation 2x² - 3x + 5 = 0.

(ii) 3x² – 4√3 x + 4 = 0

1) The given equation is 3x² – 4√3 x + 4 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 3x² – 4√3 x + 4 = 0 with ax² + bx + c = 0, 

we have,

a = 3, b = - 4√3, c = 4.

3) First we will find:

b² - 4ac = (- 4√3)² - 4(3)(4)

b² - 4ac = 48 - 48

b² - 4ac = 0 ------------------ equation 2. 

4) As, b² - 4ac = 0, it has two equal real roots, so from equation 2 and equation 3,

we have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (- 4√3) ± √0]/2(3)
x = (4√3 ± 0)/2(3)

x = 2(√3)/3

x = 2(√3)/(√3√3)

x = 2/√3

 5) So,  x = 2/√3 or x = 2/√3.

(iii) 2x² – 6x + 3 = 0

1) The given equation is 2x² – 6x + 3 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 2x² – 6x + 3 = 0 with ax² + bx + c = 0, 

we have,

a = 2, b = - 6, c = 3.

3) First we will find:

b² - 4ac = (- 6)² - 4(2)(3)

b² - 4ac = 36 - 24

b² - 4ac = 12 ------------------ equation 2. 

4) As, b² - 4ac ≥ 0, it has two distinct real roots, so from equation 2 and equation 3,

we have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (- 6) ± √12]/2(2)
x = (6 ± √12)/6

x = (6 ± 2√3)/6

x = (3 ± √3)/3

5) So,  x = (3 + √3)/3 or x = (3 - √3)/3.

Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x² + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

Explanation:

1) For the quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0, 

(b² - 4ac) is known as discriminant.

2) The quadratic equation ax² + bx + c = 0 has

(a) two equal real roots, if b2 – 4ac = 0,

Solution:

(i) 2x² + kx + 3 = 0

1) The given equation is 2x² + kx + 3 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 2x² + kx + 3 = 0 with ax² + bx + c = 0, 

we have,

a = 2, b = k, c = 3.

3) First we will find:

b² - 4ac = (k)² - 4(2)(3)

b² - 4ac = k² - 24 ------------------ equation 2. 

4) As the quadratic equation has two equal roots, 

b² - 4ac = 0

k² - 24 = 0

k² = 24 

k = ± √24 = ± 2√6

 5) So,  k = 2√6 or k = - 2√6.

(ii) kx (x – 2) + 6 = 0

1) The given equation is 

kx (x – 2) + 6 = 0

kx² – 2kx + 6 = 0 ------------------ equation 1.

2) Equate the coefficient of equation kx² – 2kx + 6 = 0 with ax² + bx + c = 0, 

we have,

a = k, b = - 2k, c = 6.

3) First we will find:

b² - 4ac = (- 2k)² - 4(k)(6)

b² - 4ac = 4k² - 24k ------------------ equation 2. 

4) As the quadratic equation has two equal roots, 

b² - 4ac = 0

4k² - 24k = 0

4k(k - 6) = 0
k(k - 6) = 0

 5) So,  k = 0 or k = 6.

Q3. 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.

1) Let the breadth of a rectangular mango grove be x m.

2) So, the length of a rectangular mango grove will be 2x m 

3) According to the problem, the area is 800, so

2x(x) = 800

2x² = 800

x² = 400

x² - 400 = 0 ------------------ equation 1.

4) Equate the coefficient of equation x² – 400 = 0 with ax² + bx + c = 0, 

we have,

a = 1, b = 0, c = - 400.

5) First we will find:

b² - 4ac = (0)² - 4(1)(- 400)

b² - 4ac = 1600 ------------------ equation 2.

6) As b² - 4ac = 1600 > 0, it has real roots, so from equation 1 and equation 2, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (0) ± √1600]/2

x = (0 ± 40)/2 

7) So,  x = (40)/2 or x = (- 40)/2, i.e. x = 20, or x = - 20.

8) As length is always positive, ignore x = - 20.

9) So, the breadth of the rectangular mango grove is 20 m and its length is 40 m.

Q4. 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 age in years was 48.

1) Let the present age of the first friend be x.

2) So, the present age of the second friend will be (20 - x).

3) 4 years ago, their ages will be (x - 4) and (20 - x - 4).

4) According to the problem,

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

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

16x - x² - 64 + 4x = 48

20x - x² - 64 - 48 = 0
20x - x² - 112 = 0
x² - 20x + 112 = 0 ------------------ equation 1.

5) Equate the coefficient of equation x² - 20x + 112 = 0 with ax² + bx + c = 0, so

a = 1, b = - 20, c = 112

6) First we will find:

b² - 4ac = (- 20)² - 4(1)(112)

b² - 4ac = 400 - 448

b² - 4ac = - 48 ------------------ equation 2.

7) As b² - 4ac = - 48 < 0, it has no real roots, so the given situation is impossible.

Q5. Is it possible to design a rectangular park of perimeter 80 m and an area

of 400 m²? If so, find its length and breadth.

1) Let the breadth of a rectangular park be x m.

2) As, the perimeter of a rectangular park = 80 m.

3) So, length = [(perimeter/2) - breadth]

length = [(80/2) - x]

length = (40 - x)

4) According to the problem,

x(40 - x) = 400

40x - x² = 400

x² - 40x + 400 = 0 ------------------ equation 1.

5) Equate the coefficient of equation x² - 40x + 400 = 0 with ax² + bx + c = 0, so

a = 1, b = - 40, c = 400

6) First we will find:

b² - 4ac = (- 40)² - 4(1)(400)

b² - 4ac = 1600 - 1600

b² - 4ac = 0 ------------------ equation 2.

7) As, b² - 4ac = 0, it has two equal real roots, so from equation 1 and equation 2,

we have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (- 40) ± √0]/2(1)
x = (40 ± 0)/2

x = (20 ± 0)

8) So,  x = 20, the breadth = 20 m and the length = 40 - 20 = 20 m.

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