find the quadratic polynomial with zeroes tan 60° and cot 30°
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Tan60°=√3,cot30°=√3
Let tan 60°be alpha,and cot 30°be beta
Then QP is x^2+2√3-3
Let tan 60°be alpha,and cot 30°be beta
Then QP is x^2+2√3-3
Answered by
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Given tan60° and cot30° are zeroes of an quadratic polynomial
Let alpha (a) be tan60°
Beta (ß) be cot30°
tan30°=3^(1/2)
cot60°=3^(1/2)
Sum of the zeroes = a+ß = 2×3^(1/2)
Product of the zeroes = aß = 3
The required quadratic equation
= k [ x^2 - (a+ß)x + aß ]
= k [ x^2 - 2*3^(1/2) + 3]
Put k=1
= x^2 -2*3^(1/2) + 3
Let alpha (a) be tan60°
Beta (ß) be cot30°
tan30°=3^(1/2)
cot60°=3^(1/2)
Sum of the zeroes = a+ß = 2×3^(1/2)
Product of the zeroes = aß = 3
The required quadratic equation
= k [ x^2 - (a+ß)x + aß ]
= k [ x^2 - 2*3^(1/2) + 3]
Put k=1
= x^2 -2*3^(1/2) + 3
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