In the previous section we completed the discussion on the 'method of square completion' for solving second degree equations. In this section we will see the 'case of two answers' while solving second degree equations.
• We know that 4 × 4 = 16. So square root of 16 is 4
• We know that -4 × -4 = 16. So square root of 16 is -4
• If we have x2 = 16, x will be √16. So x can be 4 or -4.
• In the problems that we have done so far, we may get two answers. For example, consider the last problem (solved example 29.12) in the previous section.
• In step (9) we have: (n-0.5) = √30.25
• √30.25 can be either 5.5 or -5.5.
• Let us take 30.25 = 5.5. Then we can write:
(n-0.5) = 5.5 ⇒ n = 5.5 + 0.5 = 6
• Let us take 30.25 = -5.5. Then we can write:
(n-0.5) = -5.5 ⇒ n = -5.5 + 0.5 = -5.0
■ But n is the 'number of kids'. It cannot be a negative number. So we choose 'n = 6' as the answer.
■ So in the solution of second degree equations, the following procedure is adopted:
(i) Do all the calculations and write down both the possible solutions
(ii) Choose the one which gives a practical solution to the problem
Another example:
The length of a rectangle is 2 m more than it's breadth. Find the length and breadth if the area of the rectangle is 224 m2
Solution:
1. Let the breadth of the rectangle be x. Then length = (x+2) m
2. Given that Area = x(x+2) = 224
⇒ x2 + 2x = 224
3. The coefficient of x is 2.
• Half of the coefficient is 2⁄2 = 1
• Square of this is 12 = 1
4. Add this square to both sides. We get:
x2+2x+1 = 224+1 ⇒ x2+2x+1 = 225
5. But x2+2x+1 is (x+1)2
So we can write:
6. (x+1)2 = 225 ⇒ (x+1) = √225
7. But 225 has two roots: 15 and -15
8. Let us consider 15 first. We will get:
x+1 = 15 ⇒ x = 14
9. Now let us take -15. We will get:
x+1 = -15 ⇒ x = -15-1 = -16
10. But the breadth of a rectangle can never be negative. So we will choose x = 14
11. So we get breadth = 14 m. Then length = (14+2) = 16 m
Check: Area = 14×16 = 224 m
The product of a number and 2 more than that is 168. What are the numbers?
Solution:
1. Let the number be x. Then 2 more than that is (x+2)
2. We have: product = x(x+2) = 168
⇒ x2 + 2x = 168
3. The coefficient of x is 2.
• Half of the coefficient is 2⁄2 = 1
• Square of this is 12 = 1
4. Add this square to both sides. We get:
x2+2x+1 = 168+1 ⇒ x2+2x+1 = 169
5. But x2+2x+1 is (x+1)2
So we can write:
6. (x+1)2 = 169 ⇒ (x+1) = √169
7. But 169 has two roots: 13 and -13
8. Let us consider 13 first. We will get:
x+1 = 13 ⇒ x = 12
9. Now let us take -13. We will get:
x+1 = -13 ⇒ x = -13-1 = -14
10. If x is 12, the other number is (x+2) = 14
Their product = 12×14 = 168
11. If x is -14, the other number is (x+2) = -14+2 = -12
Their product = -14×-12 = 168
12. Both roots of 169 are acceptable in this problem
Solved example 29.14
Find two numbers with sum 4 and product 2
Solution:
1. Let one number be x and the other number be y
2. Then we have (x+y) = 4
3. Also we have xy = 2
4. From (2) we get y = (4-x)
5. Substituting this in (3) we get: x(4-x) = 2
⇒ 4x - x2 = 2 ⇒ x2 - 4x = -2
6. The coefficient of x is -4
• Half of the coefficient is -4⁄2 = -2
• Square of this is (-2)2 = 4
7. Add this square to both sides. We get:
x2-4x+4 = -2+4 ⇒ x2-4x+4 = 2
8. But x2-4x+4 is (x-2)2. So we can write:
9. (x-2)2 = 2 ⇒ (x-2) = √2
10. But 2 has two roots: √2 and -√2
11. Let us consider √2 first. We will get:
x-2 = √2 ⇒ x = 2+√2
• Then the other number y = 4-x = 4-(2+√2) = 4-2-√2 = 2-√2
• Their product = xy = (2+√2)(2-√2) = (22- (√2)2) = 4-2 = 2
♦ Recall the identity: (a+b)(a-b) = a2-b2
• So √2 is acceptable
12. Now let us take -√2. We will get:
x-2 = -√2 ⇒ x = 2-√2
• Then the other number y = 4-x = 4-(2-√2) = 4-2+√2 = 2+√2
• Their product = xy = (2-√2)(2+√2) = (22- (√2)2) = 4-2 = 2
• So -√2 is acceptable
■ Thus in this problem, both √2 and -√2 are acceptable
Solved example 29.15
How many terms of the arithmetic sequence 99, 97, 95, . . . must be added to get 900?
Solution:
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 = 99 and d = -2. So we can write:
Sum of first n terms = n⁄2[2×99+(n-1)×-2] = n⁄2[198-2n+2] = n⁄2[200-2n]
3. So we can write:
n⁄2[200-2n] = 900 ⇒ n[200-2n] = 1800 ⇒ 200n-2n2 = 1800 ⇒ 2n2 - 200n = -1800
4. The coefficient of the second degree term must be 1. So we divide both sides by 2. We get:
n2 -100n = -900
5. The coefficient of n is -100.
• Half of the coefficient is -100⁄2 = -50
• Square of this is (-50)2 = 2500
6. Add this square to both sides. We get:n2 -100n +2500 = -900 +2500 ⇒ n2 -100n +2500 = 1600
7. But n2 -100n +2500 is (n-50)2. So we can write:
8. (n-50)2 = 1600 ⇒ (n-50) = √1600
9. But 1600 has two roots: 40 and -40
10. Let us consider 40 first. We will get:
n-50 = 40 ⇒ n = 90
11. Let us consider -40. We will get:
n-50 = -40 ⇒ n = 10
12. Check:
• From step (2), Sum of first 10 terms = n⁄2[200-2n] = 10⁄2×[200-2×10] = 5×[200-20] = 5×180 = 900
• From step (2), Sum of first 90 terms = n⁄2[200-2n] = 90⁄2×[200-2×90] = 45×[200-180] = 45×20 = 900
■ So in this problem, both roots of 1600 are acceptable. Let us now see the reason why both the numbers give the same sum:
1. The given arithmetic sequence has the common difference d as -2. So the terms are decreasing
2. The terms must decrease upto zero and then, become negative
3. We know the method to check whether any particular number is a term in a given series. Using that method, we can find that zero is not a term in the series.
4. But (-1) is. This (-1) is the 51st term. So, from -1 onward, the terms are negative. This is shown in the fig.29.16 below
5. Note that the 10th term is 81. From the 1st term upto 10th term we get a sum of 900
6. The 11th term is 79. The last term, that is the 90th term is (-79).
7. The terms that occur after the 10th term up to the 90th term, cancels each other
8. So, the sum of terms from 11th term to 90th term is zero
9. Thus we get the sum as 900 for both n = 10 and n = 90
Solved example 29.16
The sum of a number and it's reciprocal is 21⁄6 . What is the number?
Solution:
1. Let the number be x. Then it's reciprocal is 1⁄x
2. So we can write: x + 1⁄x = 21⁄6 ⇒ x+ 1⁄x = 13⁄6
⇒ (x2+1)⁄x= 13⁄6 ⇒ 6(x2 + 1) = 13x ⇒ 6x2 + 6 = 13x
⇒ 6x2 -13x = -6
3. But we want the coefficient of x2 to be 1. So we divide both sides by 6. We get:
x2 - 13⁄6 x = -1
4. The coefficient of x is -13⁄6
• Half of the coefficient is -13⁄6 × 1⁄2 = -13⁄12
• Square of this is (-13⁄12)2 = 169⁄144
5. Add this square to both sides. We get:
x2 - 13⁄6 x + 169⁄144 = -1 + 169⁄144 ⇒ x2 - 13⁄6 x + 169⁄144 = (-144 + 169)⁄144
⇒ x2 - 13⁄6 x + 169⁄144 = 25⁄144
6. But x2 - 13⁄6 x + 169⁄144 is (x- 13⁄12)2. So we can write:
7. (x-13⁄12)2 = 25⁄144 ⇒ (x-13⁄12) = √[25⁄144]
8. But 25⁄144 has two roots: 5⁄12 and -5⁄12.
9. We will consider 5⁄12 first. We get:
• (x-13⁄12) = 5⁄12 ⇒ x = 5⁄12 + 13⁄12 ⇒ x = 18⁄12 ⇒ x = 3⁄2
• So the sum of the number and it's reciprocal is: 3⁄2 + 2⁄3 = (9+4)⁄6 = 13⁄6
• So the root 5⁄12 is acceptable
10. We will now consider -5⁄12 . We get:
• (x-13⁄12) = -5⁄12 ⇒ x = -5⁄12 + 13⁄12 ⇒ x = 8⁄12 ⇒ x = 2⁄3
• So the sum of the number and it's reciprocal is: 2⁄3 + 3⁄2 = (4+9)⁄6 = 13⁄6
• So the root -5⁄12 is also acceptable
Solved example 29.17
Two taps open into a tank. If both are opened, the tank would be filled in 12 minutes. The time taken to fill the tank by the smaller tap alone is 10 minutes more than the time taken to fill the tank by the larger tap alone. If the smaller tap alone is opened, what would be the time taken to fill the tank?
Solution:
1. Let the volume output by the smaller tap per minute be Vs
[In this problem, we do not have to calculate Vs. But it is better to know how Vs is calculated. Consider fig.29.17 below:
• Water is flowing through a pipe. When time = 0 seconds, that is, at the moment when a stop watch is started, the point A is at the edge of the pipe.
• After 60 seconds, let point B, which was l cm away from A, reach the edge of the pipe
• Then we can say: water between A and B flowed out of the pipe in 60 seconds.
• The 'inside cross sectional area' multiplied by l will be the volume that flowed out in 60 seconds
• So this volume is Vs]
• Then in 12 minutes, the smaller tap would give an output volume of 12Vs
• Also, let the time required by the smaller tap to fill the tank alone be Ts minutes
• Then in Ts minutes, the output from the smaller tank will be VsTs. And this VsTs will be the total capacity of the tank V
• So we can write: V = VsTs ⇒ Vs = V⁄Ts
2. Let the volume output by the bigger tap per minute be Vb
• Then in 12 minutes, the bigger tap would give an output volume of 12Vb
• Also, let the time required by the bigger tap to fill the tank alone be Tb minutes
• Then in Tb minutes, the output from the smaller tank will be VbTb. And this VbTb will be the total capacity of the tank V
• So we can write: V = VbTb ⇒ Vb = V⁄Tb
3. The total output from the two taps per minute = Vs + Vb
4. So total output from the two taps in 12 minutes = 12(Vs+Vb)
5. This total output in 12 minutes is equal to the total capacity of the tank. If we take V as the total capacity of the tank, we can write:
V = 12(Vs+Vb)
6. Substituting for Vs and Vb from (1) and (2) we get:
V = 12(V⁄Ts + V⁄Tb) ⇒ V = 12V(1⁄Ts+ 1⁄Tb) ⇒ 1 = 12(1⁄Ts+ 1⁄Tb)
7. Given that Ts = Tb + 10. Substituting this in (6) we get:
1 = 12(1⁄(Tb+10) + 1⁄Tb) ⇒ 1 = 12×[(Tb+Tb+10)⁄Tb(Tb+10) ]
⇒ Tb(Tb+10) = 12×(2Tb+10) ⇒ Tb2 + 10Tb = 24Tb+120 ⇒ Tb2 -14Tb = 120
8. The coefficient of Tb is -14
• Half of the coefficient is -14⁄2 = -7
• Square of this is (-7)2 = 49
9. Add this square to both sides. We get: Tb2 -14Tb + 49 = 120 + 49 ⇒ Tb2 -14Tb + 49 = 169
10. But Tb2 -14Tb + 49 is (Tb-7)2. So we can write:
11. (Tb-7)2 = 169 ⇒ (Tb-7) = √169
12. But 169 has two roots: 13 and -13
13. Let us consider 13 first. We will get:
Tb-7 = 13 ⇒ Tb = 20
14. Let us consider -14. We will get:
Tb-7 = -13 ⇒ Tb = -13+7 = -6
15. Negative time cannot be accepted. So the root -13 cannot be accepted
16. Thus out of the two roots of 169, we take 13. And we get Tb as 20 minutes
17. When Tb = 20, Ts = Tb+10 = 30 minutes
18. So, if the smaller tap alone is opened, it will take 30 minutes to fill the tank
• We know that 4 × 4 = 16. So square root of 16 is 4
• We know that -4 × -4 = 16. So square root of 16 is -4
• If we have x2 = 16, x will be √16. So x can be 4 or -4.
• In the problems that we have done so far, we may get two answers. For example, consider the last problem (solved example 29.12) in the previous section.
• In step (9) we have: (n-0.5) = √30.25
• √30.25 can be either 5.5 or -5.5.
• Let us take 30.25 = 5.5. Then we can write:
(n-0.5) = 5.5 ⇒ n = 5.5 + 0.5 = 6
• Let us take 30.25 = -5.5. Then we can write:
(n-0.5) = -5.5 ⇒ n = -5.5 + 0.5 = -5.0
■ But n is the 'number of kids'. It cannot be a negative number. So we choose 'n = 6' as the answer.
■ So in the solution of second degree equations, the following procedure is adopted:
(i) Do all the calculations and write down both the possible solutions
(ii) Choose the one which gives a practical solution to the problem
Another example:
The length of a rectangle is 2 m more than it's breadth. Find the length and breadth if the area of the rectangle is 224 m2
Solution:
1. Let the breadth of the rectangle be x. Then length = (x+2) m
2. Given that Area = x(x+2) = 224
⇒ x2 + 2x = 224
3. The coefficient of x is 2.
• Half of the coefficient is 2⁄2 = 1
• Square of this is 12 = 1
4. Add this square to both sides. We get:
x2+2x+1 = 224+1 ⇒ x2+2x+1 = 225
5. But x2+2x+1 is (x+1)2
So we can write:
6. (x+1)2 = 225 ⇒ (x+1) = √225
7. But 225 has two roots: 15 and -15
8. Let us consider 15 first. We will get:
x+1 = 15 ⇒ x = 14
9. Now let us take -15. We will get:
x+1 = -15 ⇒ x = -15-1 = -16
10. But the breadth of a rectangle can never be negative. So we will choose x = 14
11. So we get breadth = 14 m. Then length = (14+2) = 16 m
Check: Area = 14×16 = 224 m
We will now see some solved examples
Solved example 29.13The product of a number and 2 more than that is 168. What are the numbers?
Solution:
1. Let the number be x. Then 2 more than that is (x+2)
2. We have: product = x(x+2) = 168
⇒ x2 + 2x = 168
3. The coefficient of x is 2.
• Half of the coefficient is 2⁄2 = 1
• Square of this is 12 = 1
4. Add this square to both sides. We get:
x2+2x+1 = 168+1 ⇒ x2+2x+1 = 169
5. But x2+2x+1 is (x+1)2
So we can write:
6. (x+1)2 = 169 ⇒ (x+1) = √169
7. But 169 has two roots: 13 and -13
8. Let us consider 13 first. We will get:
x+1 = 13 ⇒ x = 12
9. Now let us take -13. We will get:
x+1 = -13 ⇒ x = -13-1 = -14
10. If x is 12, the other number is (x+2) = 14
Their product = 12×14 = 168
11. If x is -14, the other number is (x+2) = -14+2 = -12
Their product = -14×-12 = 168
12. Both roots of 169 are acceptable in this problem
Solved example 29.14
Find two numbers with sum 4 and product 2
Solution:
1. Let one number be x and the other number be y
2. Then we have (x+y) = 4
3. Also we have xy = 2
4. From (2) we get y = (4-x)
5. Substituting this in (3) we get: x(4-x) = 2
⇒ 4x - x2 = 2 ⇒ x2 - 4x = -2
6. The coefficient of x is -4
• Half of the coefficient is -4⁄2 = -2
• Square of this is (-2)2 = 4
7. Add this square to both sides. We get:
x2-4x+4 = -2+4 ⇒ x2-4x+4 = 2
8. But x2-4x+4 is (x-2)2. So we can write:
9. (x-2)2 = 2 ⇒ (x-2) = √2
10. But 2 has two roots: √2 and -√2
11. Let us consider √2 first. We will get:
x-2 = √2 ⇒ x = 2+√2
• Then the other number y = 4-x = 4-(2+√2) = 4-2-√2 = 2-√2
• Their product = xy = (2+√2)(2-√2) = (22- (√2)2) = 4-2 = 2
♦ Recall the identity: (a+b)(a-b) = a2-b2
• So √2 is acceptable
12. Now let us take -√2. We will get:
x-2 = -√2 ⇒ x = 2-√2
• Then the other number y = 4-x = 4-(2-√2) = 4-2+√2 = 2+√2
• Their product = xy = (2-√2)(2+√2) = (22- (√2)2) = 4-2 = 2
• So -√2 is acceptable
■ Thus in this problem, both √2 and -√2 are acceptable
Solved example 29.15
How many terms of the arithmetic sequence 99, 97, 95, . . . must be added to get 900?
Solution:
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 = 99 and d = -2. So we can write:
Sum of first n terms = n⁄2[2×99+(n-1)×-2] = n⁄2[198-2n+2] = n⁄2[200-2n]
3. So we can write:
n⁄2[200-2n] = 900 ⇒ n[200-2n] = 1800 ⇒ 200n-2n2 = 1800 ⇒ 2n2 - 200n = -1800
4. The coefficient of the second degree term must be 1. So we divide both sides by 2. We get:
n2 -100n = -900
5. The coefficient of n is -100.
• Half of the coefficient is -100⁄2 = -50
• Square of this is (-50)2 = 2500
6. Add this square to both sides. We get:n2 -100n +2500 = -900 +2500 ⇒ n2 -100n +2500 = 1600
7. But n2 -100n +2500 is (n-50)2. So we can write:
8. (n-50)2 = 1600 ⇒ (n-50) = √1600
9. But 1600 has two roots: 40 and -40
10. Let us consider 40 first. We will get:
n-50 = 40 ⇒ n = 90
11. Let us consider -40. We will get:
n-50 = -40 ⇒ n = 10
12. Check:
• From step (2), Sum of first 10 terms = n⁄2[200-2n] = 10⁄2×[200-2×10] = 5×[200-20] = 5×180 = 900
• From step (2), Sum of first 90 terms = n⁄2[200-2n] = 90⁄2×[200-2×90] = 45×[200-180] = 45×20 = 900
■ So in this problem, both roots of 1600 are acceptable. Let us now see the reason why both the numbers give the same sum:
1. The given arithmetic sequence has the common difference d as -2. So the terms are decreasing
2. The terms must decrease upto zero and then, become negative
3. We know the method to check whether any particular number is a term in a given series. Using that method, we can find that zero is not a term in the series.
4. But (-1) is. This (-1) is the 51st term. So, from -1 onward, the terms are negative. This is shown in the fig.29.16 below
 |
| Fig.29.16 |
6. The 11th term is 79. The last term, that is the 90th term is (-79).
7. The terms that occur after the 10th term up to the 90th term, cancels each other
8. So, the sum of terms from 11th term to 90th term is zero
9. Thus we get the sum as 900 for both n = 10 and n = 90
Solved example 29.16
The sum of a number and it's reciprocal is 21⁄6 . What is the number?
Solution:
1. Let the number be x. Then it's reciprocal is 1⁄x
2. So we can write: x + 1⁄x = 21⁄6 ⇒ x
⇒ (x2+1)⁄x= 13⁄6 ⇒ 6(x2 + 1) = 13x ⇒ 6x2 + 6 = 13x
⇒ 6x2 -13x = -6
3. But we want the coefficient of x2 to be 1. So we divide both sides by 6. We get:
x2 - 13⁄6 x = -1
4. The coefficient of x is -13⁄6
• Half of the coefficient is -13⁄6 × 1⁄2 = -13⁄12
• Square of this is (-13⁄12)2 = 169⁄144
5. Add this square to both sides. We get:
x2 - 13⁄6 x + 169⁄144 = -1 + 169⁄144 ⇒ x2 - 13⁄6 x + 169⁄144 = (-144 + 169)⁄144
⇒ x2 - 13⁄6 x + 169⁄144 = 25⁄144
6. But x2 - 13⁄6 x + 169⁄144 is (x- 13⁄12)2. So we can write:
7. (x-13⁄12)2 = 25⁄144 ⇒ (x-13⁄12) = √[25⁄144]
8. But 25⁄144 has two roots: 5⁄12 and -5⁄12.
9. We will consider 5⁄12 first. We get:
• (x-13⁄12) = 5⁄12 ⇒ x = 5⁄12 + 13⁄12 ⇒ x = 18⁄12 ⇒ x = 3⁄2
• So the sum of the number and it's reciprocal is: 3⁄2 + 2⁄3 = (9+4)⁄6 = 13⁄6
• So the root 5⁄12 is acceptable
10. We will now consider -5⁄12 . We get:
• (x-13⁄12) = -5⁄12 ⇒ x = -5⁄12 + 13⁄12 ⇒ x = 8⁄12 ⇒ x = 2⁄3
• So the sum of the number and it's reciprocal is: 2⁄3 + 3⁄2 = (4+9)⁄6 = 13⁄6
• So the root -5⁄12 is also acceptable
Solved example 29.17
Two taps open into a tank. If both are opened, the tank would be filled in 12 minutes. The time taken to fill the tank by the smaller tap alone is 10 minutes more than the time taken to fill the tank by the larger tap alone. If the smaller tap alone is opened, what would be the time taken to fill the tank?
Solution:
1. Let the volume output by the smaller tap per minute be Vs
[In this problem, we do not have to calculate Vs. But it is better to know how Vs is calculated. Consider fig.29.17 below:
 |
| Fig.29.17 |
• After 60 seconds, let point B, which was l cm away from A, reach the edge of the pipe
• Then we can say: water between A and B flowed out of the pipe in 60 seconds.
• The 'inside cross sectional area' multiplied by l will be the volume that flowed out in 60 seconds
• So this volume is Vs]
• Then in 12 minutes, the smaller tap would give an output volume of 12Vs
• Also, let the time required by the smaller tap to fill the tank alone be Ts minutes
• Then in Ts minutes, the output from the smaller tank will be VsTs. And this VsTs will be the total capacity of the tank V
• So we can write: V = VsTs ⇒ Vs = V⁄Ts
2. Let the volume output by the bigger tap per minute be Vb
• Then in 12 minutes, the bigger tap would give an output volume of 12Vb
• Also, let the time required by the bigger tap to fill the tank alone be Tb minutes
• Then in Tb minutes, the output from the smaller tank will be VbTb. And this VbTb will be the total capacity of the tank V
• So we can write: V = VbTb ⇒ Vb = V⁄Tb
3. The total output from the two taps per minute = Vs + Vb
4. So total output from the two taps in 12 minutes = 12(Vs+Vb)
5. This total output in 12 minutes is equal to the total capacity of the tank. If we take V as the total capacity of the tank, we can write:
V = 12(Vs+Vb)
6. Substituting for Vs and Vb from (1) and (2) we get:
V = 12(V⁄Ts + V⁄Tb) ⇒ V = 12V(1⁄Ts+ 1⁄Tb) ⇒ 1 = 12(1⁄Ts+ 1⁄Tb)
7. Given that Ts = Tb + 10. Substituting this in (6) we get:
1 = 12(1⁄(Tb+10) + 1⁄Tb) ⇒ 1 = 12×[(Tb+Tb+10)⁄Tb(Tb+10) ]
⇒ Tb(Tb+10) = 12×(2Tb+10) ⇒ Tb2 + 10Tb = 24Tb+120 ⇒ Tb2 -14Tb = 120
8. The coefficient of Tb is -14
• Half of the coefficient is -14⁄2 = -7
• Square of this is (-7)2 = 49
9. Add this square to both sides. We get: Tb2 -14Tb + 49 = 120 + 49 ⇒ Tb2 -14Tb + 49 = 169
10. But Tb2 -14Tb + 49 is (Tb-7)2. So we can write:
11. (Tb-7)2 = 169 ⇒ (Tb-7) = √169
12. But 169 has two roots: 13 and -13
13. Let us consider 13 first. We will get:
Tb-7 = 13 ⇒ Tb = 20
14. Let us consider -14. We will get:
Tb-7 = -13 ⇒ Tb = -13+7 = -6
15. Negative time cannot be accepted. So the root -13 cannot be accepted
16. Thus out of the two roots of 169, we take 13. And we get Tb as 20 minutes
17. When Tb = 20, Ts = Tb+10 = 30 minutes
18. So, if the smaller tap alone is opened, it will take 30 minutes to fill the tank
In the next section, we will see 'two answers' for any problem on second degree equations.
No comments:
Post a Comment