if the zeroes of the polynomial x³-3x²+x+1 are a-b,a,a+b,find a and b
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→ a = 1 and b = ±√2 .
Given polynomial is f(x) = x³ - 3x² + x + 1 .
Let α = ( a - b ) , β = a and γ = ( a + b ) .
→ α + β + γ = -b/a .
⇒ ( a - b ) + a + ( a - b ) = -(-3)/1 .
⇒ 3a = 3 .
⇒ a = 3/3 .
∴
And,
→
⇒ a( a - b ) + a( a + b ) + ( a + b )( a - b ) = 1/1 .
⇒
⇒ 3a² - b² = 1 .
⇒ ( 3 × 1² ) - b² = 1 . { ∵ a = 1 }
⇒ 3 - b² = 1 .
⇒ b² = 3 - 1 .
⇒ b² = 2 .
∴
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