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If a= cos2π/7+ isin2π/7 , x= a+a2+a4 and y= a3+ a5+ a6. then x,y are the roots of the equation 1)p2+p+ 1=0 2)p2+p+ 2=0 3)p2+2p+2 4)p2+2p+ 3?
correct option is 2 . can u explain the answer?
correct option is 2 . can u explain the answer?
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If a= cos2π/7+ isin2π/7 , x= a+a2+a4 and y= a3+ a5+ a6. then x,y are t...
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If a= cos2π/7+ isin2π/7 , x= a+a2+a4 and y= a3+ a5+ a6. then x,y are t...
Given Information:
Let a = cos(2π/7) + isin(2π/7)
Let x = a * a^2 * a^4 = a^7
Let y = a^3 * a^5 * a^6 = a^14
To find:
The equation satisfied by x and y.
Solution:
Step 1: Find the value of a^7:
Since a = cos(2π/7) + isin(2π/7), we can write a^7 as (cos(2π/7) + isin(2π/7))^7.
Using De Moivre's theorem, we can expand (cosθ + isinθ)^n as cos(nθ) + isin(nθ). Applying this to a^7, we get:
a^7 = (cos(7(2π/7)) + isin(7(2π/7))) = (cos(2π) + isin(2π)) = 1 + i(0) = 1
Step 2: Find the value of a^14:
Since a = cos(2π/7) + isin(2π/7), we can write a^14 as (cos(2π/7) + isin(2π/7))^14.
Using De Moivre's theorem, we can expand (cosθ + isinθ)^n as cos(nθ) + isin(nθ). Applying this to a^14, we get:
a^14 = (cos(14(2π/7)) + isin(14(2π/7))) = (cos(4π) + isin(4π)) = 1 + i(0) = 1
Step 3: Find the equation satisfied by x and y:
We have x = a^7 = 1 and y = a^14 = 1.
Substituting these values into the equation p^2 + p - 2 = 0, we get:
1^2 + 1 - 2 = 0
Simplifying, we get:
1 + 1 - 2 = 0
2 - 2 = 0
0 = 0
Therefore, the equation satisfied by x and y is p^2 + p - 2 = 0.
Answer:
The correct option is 2) p^2 + p - 2 = 0.
Let a = cos(2π/7) + isin(2π/7)
Let x = a * a^2 * a^4 = a^7
Let y = a^3 * a^5 * a^6 = a^14
To find:
The equation satisfied by x and y.
Solution:
Step 1: Find the value of a^7:
Since a = cos(2π/7) + isin(2π/7), we can write a^7 as (cos(2π/7) + isin(2π/7))^7.
Using De Moivre's theorem, we can expand (cosθ + isinθ)^n as cos(nθ) + isin(nθ). Applying this to a^7, we get:
a^7 = (cos(7(2π/7)) + isin(7(2π/7))) = (cos(2π) + isin(2π)) = 1 + i(0) = 1
Step 2: Find the value of a^14:
Since a = cos(2π/7) + isin(2π/7), we can write a^14 as (cos(2π/7) + isin(2π/7))^14.
Using De Moivre's theorem, we can expand (cosθ + isinθ)^n as cos(nθ) + isin(nθ). Applying this to a^14, we get:
a^14 = (cos(14(2π/7)) + isin(14(2π/7))) = (cos(4π) + isin(4π)) = 1 + i(0) = 1
Step 3: Find the equation satisfied by x and y:
We have x = a^7 = 1 and y = a^14 = 1.
Substituting these values into the equation p^2 + p - 2 = 0, we get:
1^2 + 1 - 2 = 0
Simplifying, we get:
1 + 1 - 2 = 0
2 - 2 = 0
0 = 0
Therefore, the equation satisfied by x and y is p^2 + p - 2 = 0.
Answer:
The correct option is 2) p^2 + p - 2 = 0.
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If a= cos2π/7+ isin2π/7 , x= a+a2+a4 and y= a3+ a5+ a6. then x,y are the roots of the equation 1)p2+p+ 1=0 2)p2+p+ 2=0 3)p2+2p+2 4)p2+2p+ 3?correct option is 2 . can u explain the answer?
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