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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Chapter 29.3 - Another type of Square completion

In the previous section we saw the method of square completion for solving second degree equations. In this section we will see another type of square completion. In this case, we will adopt a different approach.
• A blue square and a green rectangle are given. See fig.29.9 below. 

Fig.29.9

• A certain area is removed from the blue square. 
• The area removed is exactly equal to that of the green rectangle
• This green rectangle has the same height as that of the square
• Width of the rectangle is 10 cm
• After removal, the net area becomes 600 cm²
• Then what is the area of the original blue square? 
Solution:
Step 1:
Let the side of the blue square be 'x'. We have to find 'x'. Split the rectangle length wise into two equal pieces. So width of each will be 5 cm. This is shown in fig.29.10 below. The two equal pieces are given two different colours. This is to see them distinctly from each other.

Fig.29.10

• If we put the two rectangles, over the blue square, the exposed blue portion will be 600 cm²
• So we can write: x² - 10x = 600 cm² ⇒ x² - [5x+5x] = 600 cm²
Step 2:
Align green rectangle on the top of the blue square in such a way that it covers the top portion of the square. This is shown in fig.29.11 below

Fig.29.11

• Now the area of the exposed blue colour = x² - 5x
• This exposed blue surface is not sufficient. We have to use the yellow rectangle also for covering. Then only we will get the net area as 600 cm²

Step 3:
Place the yellow rectangle on the right side of the square. But we are not able to align it correctly. A small portion of the yellow rectangle is left out. This is because, the green rectangle has already occupied the complete top portion. This situation is shown in fig.29.12 below:

Fig.29.12

• The exposed blue surface now is:
x² - [5x + 5(x-5)] = x² - [5x + 5x - 25] = x² - [10x -25]
• It is clear that, a 25 cm² is also to be covered
• Note that, the exposed blue surface is now a perfect square. Because 5 cm is equally removed from two perpendicular sides. 
• So area of the exposed blue surface = (x-5)(x-5) = (x-5)²
• A 25 cm² is to be removed from this
Step 4:
Cut out the 'left out portion of the yellow rectangle' and place it on the blue surface. This is shown in fig. 29.13 below. It is placed at the lower left corner.

Fig.29.13

• The exposed blue surface now is:
(x-5)² - 25 
• This quantity is 600 cm². So we can write:
(x-5)² - 25 = 600
• This is same as (x-5)² = 600 + 25
⇒ (x-5)² = 625 ⇒ (x-5) = √625
⇒ (x-5) = 25 ⇒ x = 30 cm
• So the side of the original blue square = 30 cm

Check: 
• Area of the blue square = x² = 30² = 900
• Area of the original green rectangle = 10x = 10×30 = 300
• Net area = 900 - 300 = 600 cm²

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Let us compare the two cases:

• In the first case, a 'term in x' was added to 'x²'
• In the present case, a 'term in x' is subtracted from 'x²'
■ In both the cases we are able to perform 'square completion'.
We need to do just two things:
(i) Take the square of the coefficient of x
(ii) Add it to both sides of the equation
This will become clear when we do more solved examples 

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Solved example 29.10 

Consider the isosceles triangle shown in the fig.29.14 below.

Fig.29.14

The height should be 2 m less than the base. The area of the triangle must be 12 m². What should be the length of the base? What are the lengths of the equal sides?
Solution:
1. Let the base = 'x' m
2. Then height = (x-2) m
3. Area = ¹⁄₂ × base × height = ¹⁄₂ × x × (x-2) = ¹⁄₂ × [x²-2x]
4. The area is given as 12 m². So we can write:
¹⁄₂ × [x²-2x] = 12 same as  [x²-2x]= 24
5. The coefficient of x is -2. 
• Half of the coefficient is ^-2⁄₂ = -1
• Square of this is (-1)² = 1
6. Add this square to both sides. We get:
x²-2x+1 = 24+1 ⇒ x²-2x+1 = 25
7. But x²-2x+1 is (x-1)² [Recall the identity: (A-B)² = A² -2AB + B²]
 So we can write:
8. (x-1)² = 25 ⇒ (x-1) = √25 = 5 ⇒ x = 6
• So the base is 6 m long

Check: Base = 6 m
• Height = 2 m less than base = 6-2 = 4 m
• Area =  ¹⁄₂ × base × height = ¹⁄₂ × 6 × 4 =  ²⁴⁄₂ = 12 m².
9. Now we can find the equal sides. Consider any on right triangle:
• Base of that right triangle = ⁶⁄₂ = 3 cm
• Altitude of that right triangle = height of the isosceles triangle = 4 cm
• So hypotenuse = √[3²+4²] = √[9+16] = √[25] = 5
• This hypotenuse is the length of equal sides.

Solved example 29.11
A 2.6 m long rod leans against a wall. It's foot is 1 m away from the wall. When the foot is moved a little away from the wall, it's upper end slides the same length down. How much further is the foot moved?
Solution:
1. Consider fig.29.15 below:

Fig.29.15

The rod, floor and the wall are shown in fig (a). Let us name the foot of the wall as A and the rod as BC. This is shown in fig (b).
• Given that length of the rod is 2.6 m. So we have BC = 2.6 m.
• Given that the foot of the rod is 1 m away from the wall. So we have AB = 1 m.
2. Let the foot of the rod be moved 'x' m further away from the wall. So the foot is now at B'
3. Let C' be the new position of the upper end of the rod. It is given that, the upper end slides the same distance down. So we get CC' = x m. This is shown in fig (c)
4. Consider the initial triangle ABC. It is right angled at A. Applying Pythagoras theorem, we get:
AC² = BC² - AB² ⇒ AC² = 2.6² - 1² ⇒ AC² = 6.76 - 1 = 5.76 ⇒ AC = √5.76 = 2.4
5. Consider the final triangle AB'C'. it is right angled at A. Applying Pythagoras theorem, we get:
AC'² = B'C'² - AB'²
⇒ (AC-x)² = 2.6² - (AB+x)²
⇒ 2.4² - 2 × 2.4 × x + x² = 2.6² - (1 + 2 × 1 × x + x²)
⇒ 2.4² - 4.8x + x² = 2.6² - 1 - 2x - x²
⇒ 2.4² - 2.6² + 1 - 2.8x + 2x² = 0
⇒ 2x² -2.8x + 5.76 - 6.76 +1 = 0
⇒ 2x² -2.8x = 0
⇒ x² -1.4x = 0 (Dividing both sides by 2)
6. The coefficient of x is -1.4. 
• Half of the coefficient is ^-1.4⁄₂ = -0.7
• Square of this is (-0.7)² = 0.49
7. Add this square to both sides. We get:
x²-1.4x+0.49 = 0+0.49 ⇒ x²-1.4x+0.49 = 0.49
8. But x²-1.4x+0.49 is (x-0.7)². So we can write:
9. (x-0.7)² = 0.49 ⇒ (x-0.7) = √0.49 = 0.7 ⇒ x = 0.7+0.7 = 1.4
• So the foot of the rod was moved a further distance of 1.4 m

Solved example 29.12
Thirty sweets were distributed equally among some kids. Sucking in the sweetness, a budding mathematician said: "Had we been one less, each would have got one more sweet." How many kids were there?
Solution:
1. Let the number of kids be n
2. Then the number of sweets each kid got = ³⁰⁄_n
3. If the number of kids were one less, new number number of kids would be (n-1)
4. Then the number of sweets that each kid get = ³⁰⁄_(n-1)
5. This number is one greater than the previous. So we can write:
³⁰⁄_(n-1) = (³⁰⁄_n) + 1 ⇒ ³⁰⁄_(n-1) = ⁽³⁰⁺ⁿ⁾⁄_n ⇒ 30n = (30+n)(n-1) 
⇒ 30n = 30n + n² -30 -n ⇒ 0 = n² -n -30 ⇒ n² -n = 30
6. The coefficient of x is -1 
• Half of the coefficient is ^-1⁄₂ = -0.5
• Square of this is (-0.5)² = 0.25
7. Add this square to both sides. We get:
n²-n+0.25 = 30+0.25 ⇒ n²-n+0.25 = 30.25
8. But n²-n+0.25 is (n-0.5)². So we can write:
9. (n-0.5)² = 30.25 ⇒ (n-0.5) = √30.25 = 5.5 ⇒ n = 5.5+0.5 = 6.0
• So there are 6 kids. 
Check:
• The number of sweets each get = ³⁰⁄₆ = 5
• If there are only 5 kids, each would get ³⁰⁄₅ = 6 sweets

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

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