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
• We can easily find such an angle combination:
6. ΔAFD is another such isosceles triangle. It's base angles are 36o. 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.
= side opposite 72 in ΔAFB⁄side opposite 72 in ΔABD
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
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
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
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 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.
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.
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
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 |
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
• 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 have: (2a + 2a + a) = 180o ⟹ 5a = 180 ⟹ a = 180⁄5 = 36o
• So each of the base angles must be 36 × 2 = 72o and the apex angle must be 36o
• 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
• 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 72o and apex angle is 36o
 |
| Fig.29.20 |
• It's base angles are 72o and apex angle is 36o
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 36o.
4. Now consider the new 𝛥ABF. Two of it's angles are known: ∠FAB = 36o and ∠ABF = 72o. So the third angle ∠AFB will be [180-(36+72)] = 72o
5. So two angles are equal to 72o. 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' cm6. ΔAFD is another such isosceles triangle. It's base angles are 36o. 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'
• 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'
In the next section, we will see a few more solved examples.
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