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.
• 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.
• 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
• 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:
• 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.
• 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
Let us compare the two cases:
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:
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
• A blue square and a green rectangle are given. See fig.29.9 below.
 |
| Fig.29.9 |
• 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 |
• 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 |
• 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 |
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 |
(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
• 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
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.
• 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
Solved example 29.10
Consider the isosceles triangle shown in the fig.29.14 below. |
| Fig.29.14 |
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 |
• 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
In the next section, we will see why there are 'two answers' for any problem on second degree equations.
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