In the previous section we saw the formula to solve a second degree equation. We also saw a solved example. In this section we will see a few more solved examples.
Solved example 29.20
The perimeter of a rectangle is 42 m and it's diagonal is 15 m. What are the lengths of it's sides?
Solution:
1. Let the length and width of the rectangle be 'x' m and 'y' m respectively. See fig.29.20 below:
2. Given that perimeter = 42 m. So we can write:
2(x+y) = 42 ⟹ (x+y) = 21 ⟹ y = (21-x) m
3. Given that diagonal is 15 m. Applying Pythagoras theorem, we can write:
x2 + y2 = 152 ⟹ x2 + (21-x)2 = 152 ⟹ x2 + 441 -42x + x2 = 225
⟹ 2x2 -42x + 216 = 0 ⟹ x2 -21x + 108 = 0
Solution:
• Consecutive natural numbers starting from 1 will form an arithmetic progression with
♦ first term a = 1
♦ common difference d = 1
1. Let first n terms be added.
We have: Sum of first n terms of an arithmetic sequence = n⁄2[2a+(n-1)d]. See details here.
2. In this problem, a = 1 and d = 1. So we can write:
Sum of first n terms = n⁄2[2×1+(n-1)×1] = n⁄2[2+n-1] = n⁄2[1+n]
3. So we can write:
n⁄2[1+n] = 300 ⇒ n[1+n] = 600 ⇒ n+n2 = 600 ⇒ n2 + n - 600 = 0
Solution:
1. Let the number be 'x'. Then the reciprocal is '1⁄x'
2. So we can write: x - 1⁄x = 11⁄2 ⟹ (x2-1)⁄x = 3⁄2 ⟹ 2(x2 -1) = 3x
⟹ 2x2 - 2 = 3x ⟹ 2x2 -3x -2 = 0
Solution:
1. Let the number be 'x'. Then the reciprocal is '1⁄x'. For the time being, let us assume that their sum can be 11⁄2
2. So we can write: x + 1⁄x = 11⁄2 ⟹ (x2+1)⁄x = 3⁄2 ⟹ 2(x2 +1) = 3x
⟹ 2x2 + 2 = 3x ⟹ 2x2 -3x +2 = 0
Solution:
1. The actual problem was this:
A certain value of Area was given
A certain value (42 m) of perimeter was given. But it was wrongly noted down as '24 m'
It was required to find the length and width of the rectangle using the Area and perimeter.
2. We now have to find the area and the actual length and width using the computed length which is 10 m
3. Let us denote the correct perimeter as Pc and wrong perimeter as Pw
So we can write Pc = 42 and Pw = 24
4. We will denote the area as A itself. Because it was noted down correctly
5. Let us denote the correct length and width as lc and wc respectively
Also, the wrong length and width as lw and ww respectively
6. Then we can write: lw = 10 m
7. Using the wrong perimeter, we can write:
2 × (lw+ ww) = Pw = 24 m ⟹ 2 × (10 + ww) = 24 ⟹ (10 + ww) = 12 ⟹ ww = 2 m
8. The area was correctly noted down. (lw × ww) would give the correct area.
So we can write: A = lw × ww = 10 × 2 = 20
9. Now we have the correct area and correct perimeter. We can write:
(i) Pc= 42 = 2(lc+wc) ⟹ 21 = (lc+wc)
(ii) A = 20 = (lc × wc)
10. From the second equation above, we get: wc= 20⁄lc
11. Substituting this in (i) we get:
21 = (lc+20⁄lc)
12. For convenience, we will denote lc as 'x'. So we get:
21 = (x + 20⁄x) ⟹ 21 = (x2+20)⁄x ⟹ 21x = x2 + 20 ⟹ x2 - 21x + 20 = 0
Solution:
1. Let the second degree equation be ax2 + bx + c = 0
2. The number without x is 'c'. So the wrongly written equation is ax2 + bx + 24 = 0
3. Even when it was wrongly written, the solutions were obtained. They are: 4 and 6
4. So if we put x = 4 in (2), the equation will be satisfied. So we can write:
a×42 + b×4 + 24 = 0 ⟹ 16a + 4b +24 = 0
Dividing both sides by 4 we get: 4a + b + 6 = 0 ⟹ 4a + b = -6
5. Similarly, if we put x = 6 in (2), the equation will be satisfied. So we can write:
a×62 + b×6 + 24 = 0 ⟹ 36a + 6b +24 = 0
Dividing both sides by 6 we get: 6a + b + 4 = 0 ⟹ 6a + b = -4
6. From (5) we get b = -4 - 6a
7. Substituting this in (4) we get: 4a + (-4-6a) = -6 ⟹ 4a -4 -6a = -6
⟹ -2a = -2 ⟹ a = 1
8. Substituting this value of a in (5) we get: 6 × 1 + b = -4 ⟹ 6 +b = -4 ⟹ b = -10
• So the correct equation is x2 - 10x + -24 = 0
Solved example 29.20
The perimeter of a rectangle is 42 m and it's diagonal is 15 m. What are the lengths of it's sides?
Solution:
1. Let the length and width of the rectangle be 'x' m and 'y' m respectively. See fig.29.20 below:
 |
| Fig.29.20 |
2(x+y) = 42 ⟹ (x+y) = 21 ⟹ y = (21-x) m
3. Given that diagonal is 15 m. Applying Pythagoras theorem, we can write:
x2 + y2 = 152 ⟹ x2 + (21-x)2 = 152 ⟹ x2 + 441 -42x + x2 = 225
⟹ 2x2 -42x + 216 = 0 ⟹ x2 -21x + 108 = 0
4. This is of the form ax2 + bx + c = 0
Where: a = 1, b = (-21) and c = 108
5. So we can use the general formula to solve the equation
6. b2-4ac = (-21)2-4×1×108 = 441 - 432 = 9
• So √[b2-4ac] = √9 = ±3
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-21)±3 = 21 ±3 = (21+ 3) or (21 -3)
⟹ numerator = 24 or 18
• The denominator = 2a = 2×1 = 2
• Thus x = 24⁄2 or 18⁄2
⟹ x = 12 or 9
7. When x = 12, we get (from step 2): y = 21 - x = 21 -12 = 9
That is., length = x = 12 m and width = y = 9 m
Check: Perimeter = 2 × (12 + 9) = 2 × 21 = 42 m
8. When x = 9, we get (from step 2): y = 21 - x = 21 -9 = 12
That is., length = x = 9 m and width = y = 12 m
Check: Perimeter = 2 × (9 + 12) = 2 × 21 = 42 m
9. So both roots of 9 are acceptable in this problem
Solved example 29.21
How many consecutive natural numbers starting from 1 should be added to get 300?• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-21)±3 = 21 ±3 = (21+ 3) or (21 -3)
⟹ numerator = 24 or 18
• The denominator = 2a = 2×1 = 2
• Thus x = 24⁄2 or 18⁄2
⟹ x = 12 or 9
7. When x = 12, we get (from step 2): y = 21 - x = 21 -12 = 9
That is., length = x = 12 m and width = y = 9 m
Check: Perimeter = 2 × (12 + 9) = 2 × 21 = 42 m
8. When x = 9, we get (from step 2): y = 21 - x = 21 -9 = 12
That is., length = x = 9 m and width = y = 12 m
Check: Perimeter = 2 × (9 + 12) = 2 × 21 = 42 m
9. So both roots of 9 are acceptable in this problem
Solved example 29.21
Solution:
• Consecutive natural numbers starting from 1 will form an arithmetic progression with
♦ first term a = 1
♦ common difference d = 1
1. Let first n terms be added.
We have: Sum of first n terms of an arithmetic sequence = n⁄2[2a+(n-1)d]. See details here.
2. In this problem, a = 1 and d = 1. So we can write:
Sum of first n terms = n⁄2[2×1+(n-1)×1] = n⁄2[2+n-1] = n⁄2[1+n]
3. So we can write:
n⁄2[1+n] = 300 ⇒ n[1+n] = 600 ⇒ n+n2 = 600 ⇒ n2 + n - 600 = 0
4. This is of the form ax2 + bx + c = 0
Where: a = 1, b = 1 and c = -600
5. So we can use the general formula to solve the equation
6. b2-4ac = (1)2-4×1×-600 = 1+2400 = 2401
• So √[b2-4ac] = √2401 = ±49
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(1)±49 = -1 ±49 = (-1+ 49) or (-1 -49)
⟹ numerator = 48 or -50
• The denominator = 2a = 2×1 = 2
• Thus n = 48⁄2 or -50⁄2
⟹ n = 24 or -25
7. But the number of terms cannot be negative. So the negative root of 2401 is not acceptable in this problem. We get n = 24
Check: Sum = n⁄2[2a+(n-1)d] = 24⁄2[2×1+(24-1)×1] = 12 × (2 + 23) = 12 × 25 = 300
Solved example 29.22
The reciprocal of a positive number, subtracted from the number itself gives 11⁄2. What is the number?• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(1)±49 = -1 ±49 = (-1+ 49) or (-1 -49)
⟹ numerator = 48 or -50
• The denominator = 2a = 2×1 = 2
• Thus n = 48⁄2 or -50⁄2
⟹ n = 24 or -25
7. But the number of terms cannot be negative. So the negative root of 2401 is not acceptable in this problem. We get n = 24
Check: Sum = n⁄2[2a+(n-1)d] = 24⁄2[2×1+(24-1)×1] = 12 × (2 + 23) = 12 × 25 = 300
Solved example 29.22
Solution:
1. Let the number be 'x'. Then the reciprocal is '1⁄x'
2. So we can write: x - 1⁄x = 11⁄2 ⟹ (x2-1)⁄x = 3⁄2 ⟹ 2(x2 -1) = 3x
⟹ 2x2 - 2 = 3x ⟹ 2x2 -3x -2 = 0
3. This is of the form ax2 + bx + c = 0
Where: a = 2, b = -3 and c = -2
4. So we can use the general formula to solve the equation
5. b2-4ac = (-3)2-4×2×(-2) = 9 + 16 = 25
• So √[b2-4ac] = √25 = ±5
• So √[b2-4ac] = √25 = ±5
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-3)±5 = 3 ±5 = (3 + 5) or (3 - 5)
⟹ numerator = 8 or -2
• The denominator = 2a = 2×2 = 4
• Thus x = 8⁄4 or -2⁄4
• The numerator is: -b±√[b2-4ac] = -(-3)±5 = 3 ±5 = (3 + 5) or (3 - 5)
⟹ numerator = 8 or -2
• The denominator = 2a = 2×2 = 4
• Thus x = 8⁄4 or -2⁄4
⟹ x = 2 or -1⁄2
6. But it is given that the number is positive. So we must give the value of 2 for x. The negative root of 25 is not acceptable
Check: 2 - 1⁄2 = 3⁄2
Solved example 29.23
Can the sum of a number and it's reciprocal be 11⁄2? why?6. But it is given that the number is positive. So we must give the value of 2 for x. The negative root of 25 is not acceptable
Check: 2 - 1⁄2 = 3⁄2
Solved example 29.23
Solution:
1. Let the number be 'x'. Then the reciprocal is '1⁄x'. For the time being, let us assume that their sum can be 11⁄2
2. So we can write: x + 1⁄x = 11⁄2 ⟹ (x2+1)⁄x = 3⁄2 ⟹ 2(x2 +1) = 3x
⟹ 2x2 + 2 = 3x ⟹ 2x2 -3x +2 = 0
3. This is of the form ax2 + bx + c = 0
Where: a = 2, b = -3 and c = 2
4. So we can use the general formula to solve the equation
5. b2-4ac = (-3)2-4×2×(2) = 9 - 16 = -7
• So √[b2-4ac] = √(-7)
In writing the equation to construct a rectangle of specified perimeter and area, the perimeter was wrongly written as 24 instead of 42. The length of a side was then computed as 10 m. What is the area in the problem? What are the lengths of the rectangle in the correct problem?• So √[b2-4ac] = √(-7)
• But (-7) does not have any root.
• So (x + 1⁄x) can never be equal to 11⁄2
Solved example 29.24
• So (x + 1⁄x) can never be equal to 11⁄2
Solved example 29.24
Solution:
1. The actual problem was this:
A certain value of Area was given
A certain value (42 m) of perimeter was given. But it was wrongly noted down as '24 m'
It was required to find the length and width of the rectangle using the Area and perimeter.
2. We now have to find the area and the actual length and width using the computed length which is 10 m
3. Let us denote the correct perimeter as Pc and wrong perimeter as Pw
So we can write Pc = 42 and Pw = 24
4. We will denote the area as A itself. Because it was noted down correctly
5. Let us denote the correct length and width as lc and wc respectively
Also, the wrong length and width as lw and ww respectively
6. Then we can write: lw = 10 m
7. Using the wrong perimeter, we can write:
2 × (lw+ ww) = Pw = 24 m ⟹ 2 × (10 + ww) = 24 ⟹ (10 + ww) = 12 ⟹ ww = 2 m
8. The area was correctly noted down. (lw × ww) would give the correct area.
So we can write: A = lw × ww = 10 × 2 = 20
9. Now we have the correct area and correct perimeter. We can write:
(i) Pc= 42 = 2(lc+wc) ⟹ 21 = (lc+wc)
(ii) A = 20 = (lc × wc)
10. From the second equation above, we get: wc= 20⁄lc
11. Substituting this in (i) we get:
21 = (lc+20⁄lc)
12. For convenience, we will denote lc as 'x'. So we get:
21 = (x + 20⁄x) ⟹ 21 = (x2+20)⁄x ⟹ 21x = x2 + 20 ⟹ x2 - 21x + 20 = 0
22. This is of the form ax2 + bx + c = 0
Where: a = 1, b = -21 and c = 20
23. So we can use the general formula to solve the equation
24. b2-4ac = (-21)2-4×1×20 = 441 - 80 = 361
• So √[b2-4ac] = √361 = ±19
⟹ x = 20 or 1• So √[b2-4ac] = √361 = ±19
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-21)±19 = 21 ±19 = (21 + 19) or (21 - 19)
⟹ numerator = 40 or 2
• The denominator = 2a = 2×1 = 2
• Thus x = 40⁄2 or 2⁄2
• The numerator is: -b±√[b2-4ac] = -(-21)±19 = 21 ±19 = (21 + 19) or (21 - 19)
⟹ numerator = 40 or 2
• The denominator = 2a = 2×1 = 2
• Thus x = 40⁄2 or 2⁄2
25. So we get lc = 20 or 1
Substituting this in 9(i) we get: 21 = (20+wc) or 21 = (1+wc)
⟹ wc = 1 or 20
26. So, when lc = 20, wc = 1
• Then area = 20 × 1 = 20 m2
• This area is same as the area we calculated in (8).
27. when lc = 1, wc = 20
• Then area = 1 × 20 = 20 m2
• This area is same as the area we calculated in (8).
• So both roots of 361 are acceptable in this problem
Solved example 29.25
In copying a second degree equation, the number without x was written as 24 instead of -24. The answers found were 4 and 6. What are the answers of the correct problem?Substituting this in 9(i) we get: 21 = (20+wc) or 21 = (1+wc)
⟹ wc = 1 or 20
26. So, when lc = 20, wc = 1
• Then area = 20 × 1 = 20 m2
• This area is same as the area we calculated in (8).
27. when lc = 1, wc = 20
• Then area = 1 × 20 = 20 m2
• This area is same as the area we calculated in (8).
• So both roots of 361 are acceptable in this problem
Solved example 29.25
Solution:
1. Let the second degree equation be ax2 + bx + c = 0
2. The number without x is 'c'. So the wrongly written equation is ax2 + bx + 24 = 0
3. Even when it was wrongly written, the solutions were obtained. They are: 4 and 6
4. So if we put x = 4 in (2), the equation will be satisfied. So we can write:
a×42 + b×4 + 24 = 0 ⟹ 16a + 4b +24 = 0
Dividing both sides by 4 we get: 4a + b + 6 = 0 ⟹ 4a + b = -6
5. Similarly, if we put x = 6 in (2), the equation will be satisfied. So we can write:
a×62 + b×6 + 24 = 0 ⟹ 36a + 6b +24 = 0
Dividing both sides by 6 we get: 6a + b + 4 = 0 ⟹ 6a + b = -4
6. From (5) we get b = -4 - 6a
7. Substituting this in (4) we get: 4a + (-4-6a) = -6 ⟹ 4a -4 -6a = -6
⟹ -2a = -2 ⟹ a = 1
8. Substituting this value of a in (5) we get: 6 × 1 + b = -4 ⟹ 6 +b = -4 ⟹ b = -10
• So the correct equation is x2 - 10x + -24 = 0
9. This is of the form ax2 + bx + c = 0
Where: a = 1, b = -10 and c = -24
23. So we can use the general formula to solve the equation
24. b2-4ac = (-10)2-4×1×-24 = 100 + 96 = 196
• So √[b2-4ac] = √196 = ±14
⟹ x = 12 or -2• So √[b2-4ac] = √196 = ±14
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-10)±14 = 10 ±14 = (10 + 14) or (10 - 14)
⟹ numerator = 24 or -4
• The denominator = 2a = 2×1 = 2
• Thus x = 24⁄2 or -4⁄2
• The numerator is: -b±√[b2-4ac] = -(-10)±14 = 10 ±14 = (10 + 14) or (10 - 14)
⟹ numerator = 24 or -4
• The denominator = 2a = 2×1 = 2
• Thus x = 24⁄2 or -4⁄2
We have completed the present discussion on Second degree Equations. In the next chapter, we will see Trigonometry.
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