guys give me answer of this...
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 .
Step-by-step explanation:
Given polynomial is f(x) = x³ - 3x² + x + 1 .
Here a = 1 , b = -3 , c = 1 , d = 1 .
Let α = ( a - b ) , β = a and γ = ( a + b ) .
As we know,
→ α + β + γ = -b/a .
⇒ ( a - b ) + a + ( a - b ) = -(-3)/1 .
⇒ 3a = 3 .
⇒ a = 3/3 .
∴ a = 1 .
And,
→ αβ + βγ + γα = c/a .
⇒ a( a - b ) + a( a + b ) + ( a + b )( a - b ) = 1/1 .
⇒ a² - ab + a² + ab + a² - b² = 1 .
⇒ 3a² - b² = 1 .
⇒ ( 3 × 1² ) - b² = 1 . { ∵ a = 1 }
⇒ 3 - b² = 1 .
⇒ b² = 3 - 1 .
⇒ b² = 2 .
∴ b = ±√2 .
Hence, it is solved .
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