In the previous section we completed a discussion on Probability. In this section we will see Second degree equations.
Consider a square. It's side is not known. We have to find the side using a given clue:
• When each side is increased by 1 metre, the perimeter becomes 36 metres
Solution:
1. Let the side of the original square be x
2. Then side of final square = (x+1)
3. So perimeter of final square = 4(x+1) = 36 m
4. 4x + 4 = 36 ⇒ 4x = 32 ⇒ x = 8 m
In this problem, in step (4), we formed an equation to obtain the value of x
Another problem:
Consider a square. It's side is not known. We have to find the side using a given clue:
• When each side is increased by 1 metre, the area becomes 36 square metres
Solution:
1. Let the side of the original square be x
2. Then side of final square = (x+1)×(x+1) = (x+1)2 = 36
⇒ x2 + 2x + 1 = 36
3. But in this problem, it is better to avoid expansion. So we will write:
(x+1)2 = 36 ⇒ (x+1) = √36 = 6
4. So x = 6-1 = 5 m
In this problem, in step (2), we formed an equation to obtain the value of x. Note that, in this problem, the exponent of x is 2.
Another problem:
See fig.29.1 below. We have a square sheet of thick paper. A box of height 5 cm and volume 500 cm3 has to be made from it. What should be the side of the square sheet?
Solution:
1. Consider fig.29.2 (a) below. From each corner of the original square, we have to remove a square.
2. The size of squares at all the four corners must be the same. Sides of these squares is shown in green colour.
3. When the squares are removed from the corners, a large square will be formed at the middle. This is shown in red colour.
4. When the paper is folded into the box, the green lines will become the height and the red lines will become the base. This is shown in fig (b).
5. Since the green line becomes the height, their length must be 5 cm.
6. Now consider fig (c). If we take the side of original square as x, the side of red square will be (x-10).
7. That means, square base of the final box will have a side of (x-10) cm.
So area of the square base = (x-10)2.
8. Thus volume of the final box = base area × height = (x-10)2×5.
9. It is given that, the volume should be 500 cm3.
10. So we can write: (x-10)2×5 = 500 cm3
⇒ (x2-20x+100)×5 = 500.
11. But in this problem, it is better to avoid expansion. So we write:
(x-10)2×5 = 500 ⇒ (x-10)2 = 500 ÷ 5 =100
⇒ (x-10) = √100 = 10 ⇒ x = 10 +10 = 20 cm
• So the side of the original square must be 20 cm
When each side of a square was reduced by 2 m, the area became 49 m2. What was the side of the original square?
Solution:
1. Let the side of the original square be x m
2. Then the final side will be (x-2) m
3. So final area = (x-2)2 = 49 m2
⇒ (x-2) = √49 = 7 ⇒ x = 7 + 2 = 9 m
4. So side of the original square is 9 m
Solved example 29.2
A square ground has a 2 m wide path all around it. The total area of the ground and path is 1225 m2. What is the area of the ground alone?
Solution:
1. The ground and path is shown in the fig.29.3 below:
2. Let the total width of the ground alone be x. Then, from the fig., it is clear that, total width of the square will be (x+4) m.
3. Then the total area will be (x+4)2.
5. This total area is given as 1225 m2. So we can write:
(x+4)2 = 1225 ⇒ (x+4) = √1225 ⇒ (x+4) = 35 ⇒ x = 31
6. So the area of the ground alone = x2 = 312 = 961 m2
Solved example 29.3
The square of a term in the arithmetic sequence 2, 5, 8, . . . is 2500. What is it's position?
Solution:
1. We have seen the formula for the nth term of an arithmetic sequence. Details here.
So we have: nth term = a+(n-1)d
2. In our present problem, a = 2 and d = 3
3. Also we have, the actual nth term = √2500 = 50
4. So we can write: 50 = 2+(n-1)×3 ⇒ 50 = 2+3n-3 ⇒ 50 = 3n-1 ⇒ 3n = 51 ⇒ n = 17
5. So the position of the term whose square is 2500 is 17
Solved example 29.4
Rs 2000 was deposited in a scheme in which interest is compounded annually. After two years the amount in the account was Rs 2205. What is the rate of interest?
Solution:
1. The formula to calculate the amount (A) after n years is:
A = P(1+ r⁄100)n. Where P is the principal amount, r is the rate of interest and n is the number of years for which the amount is deposited
2. In this problem we have to calculate r. Substituting the known values, we get:
2205 = 2000 × (1+ r⁄100)2
⇒ 2205⁄2000 = [(100+r)⁄100]2 ⇒ 1.1025 = [(100+r)⁄100]2
⇒ 1.1025 × 1002 = (100+r)2 ⇒ √[1.1025 × 1002] = √[(100+r)2]
⇒ 1.05 × 100 = (100 +r) ⇒ 105 = 100 + r ⇒ r = 5
3. So the rate of interest is 5%
Consider a square. It's side is not known. We have to find the side using a given clue:
• When each side is increased by 1 metre, the perimeter becomes 36 metres
Solution:
1. Let the side of the original square be x
2. Then side of final square = (x+1)
3. So perimeter of final square = 4(x+1) = 36 m
4. 4x + 4 = 36 ⇒ 4x = 32 ⇒ x = 8 m
In this problem, in step (4), we formed an equation to obtain the value of x
Another problem:
Consider a square. It's side is not known. We have to find the side using a given clue:
• When each side is increased by 1 metre, the area becomes 36 square metres
Solution:
1. Let the side of the original square be x
2. Then side of final square = (x+1)×(x+1) = (x+1)2 = 36
⇒ x2 + 2x + 1 = 36
3. But in this problem, it is better to avoid expansion. So we will write:
(x+1)2 = 36 ⇒ (x+1) = √36 = 6
4. So x = 6-1 = 5 m
In this problem, in step (2), we formed an equation to obtain the value of x. Note that, in this problem, the exponent of x is 2.
Another problem:
See fig.29.1 below. We have a square sheet of thick paper. A box of height 5 cm and volume 500 cm3 has to be made from it. What should be the side of the square sheet?
 |
| Fig.29.1 |
1. Consider fig.29.2 (a) below. From each corner of the original square, we have to remove a square.
 |
| Fig.29.2 |
3. When the squares are removed from the corners, a large square will be formed at the middle. This is shown in red colour.
4. When the paper is folded into the box, the green lines will become the height and the red lines will become the base. This is shown in fig (b).
5. Since the green line becomes the height, their length must be 5 cm.
6. Now consider fig (c). If we take the side of original square as x, the side of red square will be (x-10).
7. That means, square base of the final box will have a side of (x-10) cm.
So area of the square base = (x-10)2.
8. Thus volume of the final box = base area × height = (x-10)2×5.
9. It is given that, the volume should be 500 cm3.
10. So we can write: (x-10)2×5 = 500 cm3
⇒ (x2-20x+100)×5 = 500.
11. But in this problem, it is better to avoid expansion. So we write:
(x-10)2×5 = 500 ⇒ (x-10)2 = 500 ÷ 5 =100
⇒ (x-10) = √100 = 10 ⇒ x = 10 +10 = 20 cm
• So the side of the original square must be 20 cm
Now we will see some solved examples:
Solved example 29.1When each side of a square was reduced by 2 m, the area became 49 m2. What was the side of the original square?
Solution:
1. Let the side of the original square be x m
2. Then the final side will be (x-2) m
3. So final area = (x-2)2 = 49 m2
⇒ (x-2) = √49 = 7 ⇒ x = 7 + 2 = 9 m
4. So side of the original square is 9 m
Solved example 29.2
A square ground has a 2 m wide path all around it. The total area of the ground and path is 1225 m2. What is the area of the ground alone?
Solution:
1. The ground and path is shown in the fig.29.3 below:
 |
| Fig.29.3 |
3. Then the total area will be (x+4)2.
5. This total area is given as 1225 m2. So we can write:
(x+4)2 = 1225 ⇒ (x+4) = √1225 ⇒ (x+4) = 35 ⇒ x = 31
6. So the area of the ground alone = x2 = 312 = 961 m2
Solved example 29.3
The square of a term in the arithmetic sequence 2, 5, 8, . . . is 2500. What is it's position?
Solution:
1. We have seen the formula for the nth term of an arithmetic sequence. Details here.
So we have: nth term = a+(n-1)d
2. In our present problem, a = 2 and d = 3
3. Also we have, the actual nth term = √2500 = 50
4. So we can write: 50 = 2+(n-1)×3 ⇒ 50 = 2+3n-3 ⇒ 50 = 3n-1 ⇒ 3n = 51 ⇒ n = 17
5. So the position of the term whose square is 2500 is 17
Solved example 29.4
Rs 2000 was deposited in a scheme in which interest is compounded annually. After two years the amount in the account was Rs 2205. What is the rate of interest?
Solution:
1. The formula to calculate the amount (A) after n years is:
A = P(1+ r⁄100)n. Where P is the principal amount, r is the rate of interest and n is the number of years for which the amount is deposited
2. In this problem we have to calculate r. Substituting the known values, we get:
2205 = 2000 × (1+ r⁄100)2
⇒ 2205⁄2000 = [(100+r)⁄100]2 ⇒ 1.1025 = [(100+r)⁄100]2
⇒ 1.1025 × 1002 = (100+r)2 ⇒ √[1.1025 × 1002] = √[(100+r)2]
⇒ 1.05 × 100 = (100 +r) ⇒ 105 = 100 + r ⇒ r = 5
3. So the rate of interest is 5%
In the next section, we will see more details about second degree equations.
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