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12
If a ß are two zeroes of the polynomial x^2-3x-1
then find a quadratic
polynomial whose zeroes
are 1/a^2 and 1/b^2
Answers
Answer:
Heya !!!
P(X) = X²-3X+1
Here,
A = 1 , B = -3 and C = 1
Sum of zeroes = -B/A
Alpha + Beta = 3/1
And,
Product of zeroes = C/A
Alpha × Beta = 1/1
Sum of the zeroes of the Quadratic polynomial whose zeroes are Alpha/Beta , Beta/Alpha
Sum of zeroes = (Alpha/Beta + Beta/Alpha)
=> ( Alpha)² + (Beta)² / Alpha × Beta
=> ( Alpha + Beta)² - 2 Alpha × Beta / Alpha × Beta
=> (3)² - 2 × 1 / 1
=> 9 - 2
=> 7
And,
Product of zeroes = ( Alpha / Beta × Beta / Alpha)
=> ( Alpha × Beta ) / ( Alpha × Beta )
=> 1
Therefore,
Required Quadratic polynomial = X²-(Sum of zeroes)X + Product of zeroes
=> X² -7X + 1
HOPE IT WILL HELP YOU....... ;-)
Step-by-step explanation:
Answer:
Step-by-step explanation:
Heya !!!
P(X) = X²-3X+1
Here,
A = 1 , B = -3 and C = 1
Sum of zeroes = -B/A
Alpha + Beta = 3/1
And,
Product of zeroes = C/A
Alpha × Beta = 1/1
Sum of the zeroes of the Quadratic polynomial whose zeroes are Alpha/Beta , Beta/Alpha
Sum of zeroes = (Alpha/Beta + Beta/Alpha)
=> ( Alpha)² + (Beta)² / Alpha × Beta
=> ( Alpha + Beta)² - 2 Alpha × Beta / Alpha × Beta
=> (3)² - 2 × 1 / 1
=> 9 - 2
=> 7
And,
Product of zeroes = ( Alpha / Beta × Beta / Alpha)
=> ( Alpha × Beta ) / ( Alpha × Beta )
=> 1
Therefore,
Required Quadratic polynomial = X²-(Sum of zeroes)X + Product of zeroes
=> X² -7X + 1
HOPE IT WILL HELP YOU....... ;-)Heya !!!
P(X) = X²-3X+1
Here,
A = 1 , B = -3 and C = 1
Sum of zeroes = -B/A
Alpha + Beta = 3/1
And,
Product of zeroes = C/A
Alpha × Beta = 1/1
Sum of the zeroes of the Quadratic polynomial whose zeroes are Alpha/Beta , Beta/Alpha
Sum of zeroes = (Alpha/Beta + Beta/Alpha)
=> ( Alpha)² + (Beta)² / Alpha × Beta
=> ( Alpha + Beta)² - 2 Alpha × Beta / Alpha × Beta
=> (3)² - 2 × 1 / 1
=> 9 - 2
=> 7
And,
Product of zeroes = ( Alpha / Beta × Beta / Alpha)
=> ( Alpha × Beta ) / ( Alpha × Beta )
=> 1
Therefore,
Required Quadratic polynomial = X²-(Sum of zeroes)X + Product of zeroes
=> X² -7X + 1
HOPE IT WILL HELP YOU....... ;-)