In the previous section we saw some basic details about second degree equations. In this section we will see some more details.
A square problem:
Consider the square shown in fig.29.4 below. It's side is not given.
We have to find the side. A clue is given:
• Area of the square is 225 cm2
Solution:
1. Let the side of the square be x.
2. Then area of the square will be x × x = x2
3. The area is given as 225. So we can write:
4. x2 = 225 ⇒ x = √225 = 15 cm
• The procedure is simple. In fact, once we understand the basics, we do not even have to write the steps.
Another square problem:
A blue square and a green rectangle are given. See fig.29.5 below:
We have to find the side of the blue square. We have to find it by using the following three clues:
• Height of the blue square and the green rectangle are the same
• Width of the green rectangle is 20 cm
• Total area of the square and rectangle is 224 cm2
Solution:
Step 1:
Split the rectangle length wise into two equal pieces. So width of each will be 10 cm. This is shown in fig.29.6 below
Note that, the fig.29.6 is split into two sides. A Left side and a right side. The quantities inside the two sides must always be equal
Step 2:
Align one green rectangle on the top of the blue square. Align the other green rectangle on the right side of the blue square. This is shown in fig.29.7 below
Note that, in this step also, the left side is equal to the right side. This is because, we have not added or subtracted any thing to any side
Step 3:
Place a magenta coloured square at the meeting point of the two green rectangles. This is shown in fig.29.8 below:
The side of the magenta square should be 10 cm. Then it will be a perfect fit. It is a perfect fit because, both the green rectangles have a width of 10 cm. Look at the resulting shape. It is a perfect square. So, if the side of the original blue square is taken as 'x', the side of the new square will be (x+10) cm. Area of the new square will be (x+10)2
• Now look at the right side in fig.29.8. An extra area of 100 cm2 is added. This is because, the area of the extra magenta square on the left side is 10 × 10 = 102 = 100 cm2
• Equating left side and right side, we can write:
(x+10)2 = (224+100) ⇒ (x+10)2 = 324 cm2
⇒ (x+10) = √324 ⇒ (x+10) = 18 ⇒ x = 8 cm
• Thus we found the side of the original blue square.
Check:
• Area of the blue square = x2 = 8 = 64
• Area of the original green rectangle = 20x = 20×8 = 160
• Total area = 64+160 = 224 cm2
♦ Only a term in x2 is present. Such problems can be easily solved.
• In the second problem, we have: x2 + 20x = 224
♦ So, in addition to the term in 'x2', there is a term in 'x' also.
♦ This term in 'x' is represented by the green rectangle (full green rectangle before splitting into two equal parts) in fig.29.5
Problems with an additional term in 'x' are often encountered in science and engineering. So we must learn how to solve them
Let us see the method:
• All such problems can be diagrammatically represented as in fig.29.5 that we saw at the beginning.
• So, the term in x2 will be the area of the blue square. Coefficient of x2 must be 1 only.
• The term in x will be the area of the green square.
• The third term (the constant term) will be the total area.
Let us see an example:
1. Consider the equation that we saw above: x2+8x =1209
2. Here x2 is the area of the blue square
3. 8x is the area of the green rectangle
4. Split the rectangle length wise into two equal parts. So each rectangle will have an area of 4x. 5. Align them on top and right side of the blue square
• Then we have: x2 + 4x + 4x = 1209
6. Now place the magenta square. it's side should be same as the width of the split rectangles. So area of magenta square = 4 × 4 = 16 cm2
7. When the magenta square is placed in position, we can write:
x2 + 4x + 4x + 16 = 1209 + 16
8. The quantity on the left side of the above equation is the area of a 'square with side (x+4)'.
• That area will be (x+4)2
9. The quantity on the right side is the total area including magenta square.
10. So we can write: (x+4)2 = 1209+16 = 1225
11. From this we get: (x+4) = √1225 = 35 same as x = 35-4 = 31
• So the value of x is 31.
Check:
Substituting this value of x in (1) we get:
x2+8x = 312+8×31 = 961 + 248 = 1209
♦ This is because, 8 cm, which is the width of the green rectangle, is split into two equal parts.
♦ If we can remember why this splitting is done, we do not have to write two '4x'. We can write it as '8x'.
• So step (7) becomes: x2+8x+16 = 1209+16
• In step (7) we have added 16 to both sides. From where did we get the 16?
Ans: It is the 'square of the half of the coefficient of x'
♦ Remember that, the magenta square will have a perfect fit only if it's side is equal to half of the green rectangle
• So there are in fact only two things to do:
(i) Take the square of the coefficient of x
(ii) Add it to both sides of the equation
• When we do the above two steps, we will have completed the larger square
• Once the larger square is completed, we can take the square root of the right side and then find x
Another example:
1. Consider the equation: x2+6x =720
2. Here x2 is the area of the blue square
3. 6x is the area of the green rectangle
4. Split the rectangle length wise into two equal parts. So each rectangle will have an area of 3x. 5. Align them on top and right side of the blue square
• Then we have: x2 + 3x + 3x = 720
6. Now place the magenta square. it's side should be same as the width of the split rectangles. So area of magenta square = 3 × 3 = 9 cm2
7. When the magenta square is placed in position, we can write:
x2 + 3x + 3x + 9 = 720 + 9
8. The quantity on the left side of the above equation is the area of a 'square with side (x+3)'.
• That area will be (x+3)2
9. The quantity on the right side is the total area including magenta square.
10. So we can write: (x+3)2 = 720+9 = 729
11. From this we get: (x+3) = √729 = 27 ⇒ x = 27-3 = 24
• So the value of x is 24.
Check:
Substituting this value of x in (1) we get:
x2+6x = 242+6×24 = 576 + 144 = 720
♦ This is because, 6 cm, which is the width of the green rectangle, is split into two equal parts.
♦ If we can remember why this splitting is done, we do not have to write two '3x'. We can write it as '6x'.
• So step (7) becomes: x2+6x+9 = 720+9
• In step (7) we have added 9 to both sides. From where did we get the 9?
Ans: It is the 'square of the half of the coefficient of x'
♦ Remember that, the magenta square will have a perfect fit only if it's side is equal to half of the green rectangle
• So there are in fact only two things to do:
(i) Take the square of the coefficient of x
(ii) Add it to both sides of the equation
• When we do the above two steps, we will have completed the larger square
• Once the larger square is completed, we can take the square root of the right side and then find x
From the above two examples, it is clear that, to complete the larger square, all we need to do is this:
• Add 'the square of half of the coefficient of x' to both sides
2. Now it is in the form: x2 + 2ax + a2 Where a = 3
3. But x2 + 2ax + a2 is in the form of an identity:
x2 + 2ax + a2 = (x+a)2
4. So we get: x2 + 6x + 9 = (x+3)2
• The left side is the area of the completed square
• The right side is the square of the side of the completed square
• Both are equal
Thus for solving problems, we do not need to draw the blue square and green rectangles. All we need to do is: Complete the square in the form of the identity: x2 + 2ax + a2 = (x+a)2
• For that, add 'the square of half of the coefficient of x' to both sides. Note that half of 2a is a
We will now see a solved example.
Solved example 29.5
1 added to the product of two consecutive even numbers gives 289. What are the numbers?
Solution:
1. Let x be the first of the two consecutive even numbers. Then the second will be (x+2)
2. Product of the two numbers is x(x+2)
3. 1 added to the product is [1+x(x+2)]
4. So we can write: [1+x(x+2)] = 289 ⇒ [1+x2+2x] = 289 ⇒ x2+2x = 288
5. The coefficient of x is 2.
• Half of the coefficient is 2⁄2 = 1
• Square of this is 12 = 1
6. Add this square to both sides. We get:
x2+2x+1 = 288+1 ⇒ x2+2x+1 = 289
7. But x2+2x+1 is (x+1)2. So we can write:
8. (x+1)2 = 289 ⇒ (x+1) = √289 = 17 ⇒ x = 16
9. So the first even number is 16. The next consecutive even number will be 18
Check: The product = 16 × 18 = 288
1 added to the product = 288+1 = 289
A square problem:
Consider the square shown in fig.29.4 below. It's side is not given.
 |
| Fig.29.4 |
• Area of the square is 225 cm2
Solution:
1. Let the side of the square be x.
2. Then area of the square will be x × x = x2
3. The area is given as 225. So we can write:
4. x2 = 225 ⇒ x = √225 = 15 cm
• The procedure is simple. In fact, once we understand the basics, we do not even have to write the steps.
Another square problem:
A blue square and a green rectangle are given. See fig.29.5 below:
 |
| Fig.29.5 |
• Height of the blue square and the green rectangle are the same
• Width of the green rectangle is 20 cm
• Total area of the square and rectangle is 224 cm2
Solution:
Step 1:
Split the rectangle length wise into two equal pieces. So width of each will be 10 cm. This is shown in fig.29.6 below
 |
| Fig.29.6 (Step 1) |
Step 2:
Align one green rectangle on the top of the blue square. Align the other green rectangle on the right side of the blue square. This is shown in fig.29.7 below
 |
| Fig.29.7 (Step 2) |
Step 3:
Place a magenta coloured square at the meeting point of the two green rectangles. This is shown in fig.29.8 below:
 |
| Fig.29.8 (Step 3) |
• Now look at the right side in fig.29.8. An extra area of 100 cm2 is added. This is because, the area of the extra magenta square on the left side is 10 × 10 = 102 = 100 cm2
• Equating left side and right side, we can write:
(x+10)2 = (224+100) ⇒ (x+10)2 = 324 cm2
⇒ (x+10) = √324 ⇒ (x+10) = 18 ⇒ x = 8 cm
• Thus we found the side of the original blue square.
Check:
• Area of the blue square = x2 = 8 = 64
• Area of the original green rectangle = 20x = 20×8 = 160
• Total area = 64+160 = 224 cm2
Let us compare the two problems:
• In the first problem, we have: x2 = 225♦ Only a term in x2 is present. Such problems can be easily solved.
• In the second problem, we have: x2 + 20x = 224
♦ So, in addition to the term in 'x2', there is a term in 'x' also.
♦ This term in 'x' is represented by the green rectangle (full green rectangle before splitting into two equal parts) in fig.29.5
Let us see the method:
• All such problems can be diagrammatically represented as in fig.29.5 that we saw at the beginning.
• So, the term in x2 will be the area of the blue square. Coefficient of x2 must be 1 only.
• The term in x will be the area of the green square.
• The third term (the constant term) will be the total area.
Let us see an example:
1. Consider the equation that we saw above: x2+8x =1209
2. Here x2 is the area of the blue square
3. 8x is the area of the green rectangle
4. Split the rectangle length wise into two equal parts. So each rectangle will have an area of 4x. 5. Align them on top and right side of the blue square
• Then we have: x2 + 4x + 4x = 1209
6. Now place the magenta square. it's side should be same as the width of the split rectangles. So area of magenta square = 4 × 4 = 16 cm2
7. When the magenta square is placed in position, we can write:
x2 + 4x + 4x + 16 = 1209 + 16
8. The quantity on the left side of the above equation is the area of a 'square with side (x+4)'.
• That area will be (x+4)2
9. The quantity on the right side is the total area including magenta square.
10. So we can write: (x+4)2 = 1209+16 = 1225
11. From this we get: (x+4) = √1225 = 35 same as x = 35-4 = 31
• So the value of x is 31.
Check:
Substituting this value of x in (1) we get:
x2+8x = 312+8×31 = 961 + 248 = 1209
Let us analyse the above steps.
• In step (5), there are two '4x'. ♦ This is because, 8 cm, which is the width of the green rectangle, is split into two equal parts.
♦ If we can remember why this splitting is done, we do not have to write two '4x'. We can write it as '8x'.
• So step (7) becomes: x2+8x+16 = 1209+16
• In step (7) we have added 16 to both sides. From where did we get the 16?
Ans: It is the 'square of the half of the coefficient of x'
♦ Remember that, the magenta square will have a perfect fit only if it's side is equal to half of the green rectangle
• So there are in fact only two things to do:
(i) Take the square of the coefficient of x
(ii) Add it to both sides of the equation
• When we do the above two steps, we will have completed the larger square
• Once the larger square is completed, we can take the square root of the right side and then find x
Another example:
1. Consider the equation: x2+6x =720
2. Here x2 is the area of the blue square
3. 6x is the area of the green rectangle
4. Split the rectangle length wise into two equal parts. So each rectangle will have an area of 3x. 5. Align them on top and right side of the blue square
• Then we have: x2 + 3x + 3x = 720
6. Now place the magenta square. it's side should be same as the width of the split rectangles. So area of magenta square = 3 × 3 = 9 cm2
7. When the magenta square is placed in position, we can write:
x2 + 3x + 3x + 9 = 720 + 9
8. The quantity on the left side of the above equation is the area of a 'square with side (x+3)'.
• That area will be (x+3)2
9. The quantity on the right side is the total area including magenta square.
10. So we can write: (x+3)2 = 720+9 = 729
11. From this we get: (x+3) = √729 = 27 ⇒ x = 27-3 = 24
• So the value of x is 24.
Check:
Substituting this value of x in (1) we get:
x2+6x = 242+6×24 = 576 + 144 = 720
Let us analyse the above steps.
• In step (5), there are two '3x'. ♦ This is because, 6 cm, which is the width of the green rectangle, is split into two equal parts.
♦ If we can remember why this splitting is done, we do not have to write two '3x'. We can write it as '6x'.
• So step (7) becomes: x2+6x+9 = 720+9
• In step (7) we have added 9 to both sides. From where did we get the 9?
Ans: It is the 'square of the half of the coefficient of x'
♦ Remember that, the magenta square will have a perfect fit only if it's side is equal to half of the green rectangle
• So there are in fact only two things to do:
(i) Take the square of the coefficient of x
(ii) Add it to both sides of the equation
• When we do the above two steps, we will have completed the larger square
• Once the larger square is completed, we can take the square root of the right side and then find x
• Add 'the square of half of the coefficient of x' to both sides
When we complete the square like this, we get a very interesting result:
1. Consider x2 + 3x + 3x + 9. Let us write it as: x2 + 2×3x + 92. Now it is in the form: x2 + 2ax + a2 Where a = 3
3. But x2 + 2ax + a2 is in the form of an identity:
x2 + 2ax + a2 = (x+a)2
4. So we get: x2 + 6x + 9 = (x+3)2
• The left side is the area of the completed square
• The right side is the square of the side of the completed square
• Both are equal
• For that, add 'the square of half of the coefficient of x' to both sides. Note that half of 2a is a
We will now see a solved example.
Solved example 29.5
1 added to the product of two consecutive even numbers gives 289. What are the numbers?
Solution:
1. Let x be the first of the two consecutive even numbers. Then the second will be (x+2)
2. Product of the two numbers is x(x+2)
3. 1 added to the product is [1+x(x+2)]
4. So we can write: [1+x(x+2)] = 289 ⇒ [1+x2+2x] = 289 ⇒ x2+2x = 288
5. The coefficient of x is 2.
• Half of the coefficient is 2⁄2 = 1
• Square of this is 12 = 1
6. Add this square to both sides. We get:
x2+2x+1 = 288+1 ⇒ x2+2x+1 = 289
7. But x2+2x+1 is (x+1)2. So we can write:
8. (x+1)2 = 289 ⇒ (x+1) = √289 = 17 ⇒ x = 16
9. So the first even number is 16. The next consecutive even number will be 18
Check: The product = 16 × 18 = 288
1 added to the product = 288+1 = 289
In the next section, we will see a few more solved examples.
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