Question

The quadractic polynomial whose zeroes are 3+2root2,3–2root2

Answer 1

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Step-by-step explanation:

Let the Zeros of the polynomial is α and β.

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α = 3+2√2 and,

β = 3–2√2

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X²–(α+β)x + α×β = 0

Putting the value of α and β

x²–(3+2√2+3–2√2)x + 3+2√2 × 3–2√2 = 0.

x²–6x+(3)²–(2√2)² = 0

x²–6x + 9–8 = 0

x²–6x + 1 = 0.

Required polynomial is x²–6x+1.

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