Chapter 29.6 - Solved examples on Second degree equations using Formula

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:

Fig.29.20

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

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?
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?
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

• 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

⟹ 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?
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)

• But (-7) does not have any root. 
• So (x + ¹⁄_x) can never be equal to 1¹⁄₂

Solved example 29.24

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?
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 P_c  and wrong perimeter as P_w
So we can write P_c = 42 and P_w = 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 l_c and w_c respectively
Also, the wrong length and width as l_w and w_w respectively
6. Then we can write: l_w = 10 m  
7. Using the wrong perimeter, we can write:
2 × (l_w+ w_w) = P_w = 24 m ⟹ 2 × (10 + w_w) = 24 ⟹ (10 + w_w) = 12 ⟹ w_w = 2 m
8. The area was correctly noted down. (l_w × w_w) would give the correct area.
So we can write: A = l_w × w_w = 10 × 2 = 20
9. Now we have the correct area and correct perimeter. We can write:
(i) P_c= 42 = 2(l_c+w_c) ⟹ 21 = (l_c+w_c)
(ii) A = 20 = (l_c × w_c)
10. From the second equation above, we get: w_c= ²⁰⁄_lc
11. Substituting this in (i) we get:
21 = (l_c+²⁰⁄_lc)
12. For convenience, we will denote l_c 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

• 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

⟹ x = 20 or 1

25. So we get l_c = 20 or 1
Substituting this in 9(i) we get: 21 = (20+w_c) or 21 = (1+w_c)
⟹ w_c = 1 or 20
26. So, when l_c = 20, w_c = 1
• Then area = 20 × 1 = 20 m²
• This area is same as the area we calculated in (8).
27. when l_c = 1, w_c = 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?
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

• 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

⟹ x = 12 or -2

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We have completed the present discussion on Second degree Equations. In the next chapter, we will see Trigonometry.

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Chapter 29.5 - Formula to solve Second degree Equations

In the previous section we completed the discussion on 'two answers'. In this section we will see the use of formula to solve second degree equations.

Consider some of the second degree equations that we saw in our discussions so far in this chapter:
x² + 2x = 224
x² - 4x = -2
n² -100n = -900
All those equations satisfy two conditions:
• There is only addition or subtraction between the terms
• The exponents in all terms are whole numbers
■ So they are all polynomials  (We saw details in an earlier chapter here)
• Consider the first equation above. We can write:
p(x) = x² + 2x - 224
■ If we put different values for x, will will get corresponding unique values for p(x)
For example:
• p(1) = 1² + 2×1 - 224 = 1+2-224 = -221
• p(2) = 2² + 2×2 - 224 = 4+4-224 = -216
• p(-¹⁄₂) = (-¹⁄₂)² + 2×(-¹⁄₂) - 224 = ¹⁄₄  -1  -224 = -224 ³⁄₄
■ We can ask a reverse question:
• We want p(x) to be a particular value. For that, what 'value of x' should we put?
An example:
• We have p(x) = x² + 2x - 224
• We want p(x) to be equal to zero. For that, what values of x need to be used?
Solution:
1. We want p(x) = 0
2. So we can write: x² + 2x - 224 = 0
⇒ x² + 2x = 224   
3. Using the method of square completion, we will get:
x = 14 or -16
4. So we can put x equal to either 14 or -16 to get p(x) equal to zero

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Suppose in the above example, we want p(x) to be equal to 1. How will we proceed?

Solution:

1. We want p(x) = 1

2. This is same as p(x) - 1 = 0

3. But p(x) = x² + 2x - 224

So we can write: x² + 2x - 224 -1 = 0 ⟹ x² + 2x - 225 = 0

4. This is a new equation. So let us put it as q(x). So we write:

q(x) = x² + 2x - 225

5. So now the problem becomes this:

• We have q(x) = x² + 2x - 225

• We want q(x) to be equal to zero. For that, what values of x need to be used?  

6. We want q(x) to equal to zero. So we can write: x² + 2x - 225 = 0

⟹ x² + 2x = 225

7. The coefficient of x is 2. 

• Half of the coefficient is ²⁄₂ = 1

• Square of this is 1² = 1

8. Add this square to both sides. We get:
x²+2x+1 = 225+1 ⟹ x²+2x+1 = 226

9. But x²+2x+1 is (x+1)²

 So we can write:

10. (x+1)² = 226 ⟹ (x+1) = √226 or -√226

11. First take √226. We get: x+1 = √226 ⟹ x = -1+√226

12. Now take -√226. We get: x+1 = -√226 ⟹ x = -1-√226

11. So we can put x equal to either (-1+√226) or (-1-√226) to get q(x) equal to zero

12. That means, we can put x equal to either (-1+√226) or (-1-√226) to get p(x) equal to 1

Check:

■ p(x) = x² + 2x - 224

• put x = (-1+√226). We get:

    ♦ p(-1+√226) = (-1+√226)² + 2(-1+√226) - 224

     ⟹ p(-1+√226) = 1 -2√226 + 226 -2 + 2√226 -224 = 1

• put x = (-1-√226). We get:

    ♦ p(-1-√226) = (-1-√226)² + 2(-1-√226) - 224

     ⟹ p(-1-√226) = 1 +2√226 + 226 -2 - 2√226 -224 = 1

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Now we will see the general method. That is:

• We have a second degree polynomial p(x). 

• We want the numbers that when put in place of x, will make p(x) equal to zero

• We want a general method (which can be used for any second degree polynomial) to find those numbers

■ The general method is to use a formula. 
• The first number is given by the formula:

• The second number is given by the formula:

• Note that the difference between the two is just the sign '+' and '-' between the two terms in the numerator. So we can combine the two formulae. This is shown below as Eq.29.1:

Eq.29.1:

The derivation of the above equation is done using 'square completion method'. It is given as a video presentation here.

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Now we will see some solved examples
Solved example 29.18
A rectangle is to be made on the ground using a 20 m long rope, with a wall as one side. See fig.29.18(a) below. 

Fig.29.18

The area enclosed must be 50 m². What should be the length of the sides?
Solution:
1. Let the width  of the rectangle be 'x' m
2. Then length will be equal to (20-2x) m. This is shown in fig (b)
3. So area of the rectangle = length × width = x(20-2x) = 20x-2x²
4. But the area must be 50 m². So we can write: 20x-2x² = 50
⟹ 2x² - 20x +50 = 0 

5. This is of the form ax² + bx + c = 0

Where: a = 2, b = (-20) and c = 50

6. So we can use the general formula to solve the equation

7. b²-4ac = (-20)²-4×2×50 = 400 - 400 = 0
• So √[b²-4ac] = √0 = ±0
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b²-4ac] = -(-20)±0 = 20 ±0 = 20
• The denominator = 2a = 2×2 = 4
• Thus x = ²⁰⁄₄ = 5
8. So width of the rectangle = x = 5 m
• Length of the rectangle = 20 - 2x = 20 - 10 = 10 cm
Check: Area = length × width = 10 × 5 = 50 m².
Note: In step (4), we could have divided both sides of the final equation by 2. Then further calculations would have been a little more easier. 

Solved example 29.19

• Consider any isosceles triangle. But it must satisfy one condition:

■ Each of the two base angles must be twice the apex angle. This is shown in the fig.29.19 below:

Fig.29.19

• We can easily find such an angle combination:

• We have: (2a + 2a + a) = 180^o ⟹ 5a = 180 ⟹ a = ¹⁸⁰⁄₅ = 36^o

• So each of the base angles must be 36 × 2 = 72^o and the apex angle must be 36^o

• So the angle combination is: (72, 72, 36). 

■ Triangles with this particular angle combination occurs naturally in all regular pentagons. 

• Consider the regular pentagon ABCDE in fig.29.20 below. Length of it's sides is 'k' cm

Fig.29.20

• AD and BD are two diagonals. Let their lengths be 'x' cm. These two diagonals and one side AB of the regular pentagon gives us the isosceles triangle.
• It's base angles are 72^o and apex angle is 36^o 

Let us now study one special feature of this triangle:

1. Draw the angle bisector at the vertex A. 

2. Let this angle bisector meet the side BD at F. This is shown in fig(b). 

3. So we get two angles: ∠DAF and ∠BAF. Each is 36^o. 

4. Now consider the new 𝛥ABF. Two of it's angles are known: ∠FAB = 36^o and ∠ABF = 72^o. So the third angle ∠AFB will be [180-(36+72)] = 72^o

5. So two angles are equal to 72^o. We can consider them as the base angles. Since they are equal, it is an isosceles triangle. The equal sides are AB and AF. So AF will be equal to 'k' cm
6. ΔAFD is  another such isosceles triangle. It's base angles are 36^o. So AF = DF = k cm
7. Since DF = k, we get BF = (x-k) cm
8. The angles in 𝛥AFB and the original triangle 𝛥ABD are the same. So they are similar.

9. Now we apply a special property that is applicable to any two similar triangles (Details here):

^{side opposite smallest angle in ΔAFB}⁄_{side opposite smallest angle in ΔABD} 

= ^{side opposite medium angle in ΔAFB}⁄_{side opposite smallest angle in ΔABD} 

= ^{side opposite largest angle in ΔAFB}⁄_{side opposite smallest angle in ΔABD} 

10. But angles in the two triangles are the same. That is.,

• Smallest angle in ΔAFB = Smallest angle in ΔABD

• Medium angle in ΔAFB = Medium angle in ΔABD

• Largest angle in ΔAFB = Largest angle in ΔABD

11. So we can write this:

• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,

^{side opposite 36 in ΔAFB}⁄_{side opposite 36 in ΔABD} 
= ^{side opposite 72 in ΔAFB}⁄_{side opposite 72 in ΔABD}

12. So we get: ^FB⁄_AB = ^AF⁄_AD ⟹ ^(x-k)⁄_k = ^k⁄_x

13. Cross multiplying, we get: x(x-k) = k² ⟹ x² - kx = k² ⟹ x² - kx - k² = 0

14. This is of the form ax² + bx + c = 0

Where: a = 1, b = (-k) and c = -(k²)

15. So we can use the general formula to solve the equation

16. b²-4ac = (-k)²-4×1×-(k²) = k² + 4k² = 5k² 
• So √[b²-4ac] = √[5k²] = ±k√5
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b²-4ac] = -(-k)±k√5 = k±k√5 = k(1±√5)
• √5 is greater than 1. So, if we use the '-' sign in '±', we will get a negative value for the numerator k(1±√5)
• Then the whole value of 'x' will become negative. Such a negative value for 'x' is not acceptable because 'x' is a length. So only '+√5' is acceptable. So the numerator is k(1+√5)
• The denominator = 2a = 2×1 = 2
• Thus x = ^[k(1+√5)]⁄₂.
17. So we get a general result:
• Given any regular pentagon of side 'k'
• Then it's diagonal will be equal to ^[(1+√5)]⁄₂ times 'k'

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In the next section, we will see a few more solved examples.

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Chapter 29.4 - Two possible solutions for Second degree Equations

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

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We will now see some solved examples

Solved example 29.13
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 51^st term. So, from -1 onward, the terms are negative. This is shown in the fig.29.16 below

Fig.29.16

5. Note that the 10^th term is 81. From the 1^st term upto 10^th term we get a sum of 900
6. The 11^th term is 79. The last term, that is the 90^th term is (-79). 
7. The terms that occur after the 10^th term up to the 90^th term, cancels each other
8. So, the sum of terms from 11^th term to 90^th 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:

Fig.29.17

• 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

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In the next section, we will see 'two answers' for any problem on second degree equations.

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