Question

if the zeroes of the polynomial x³-3x²+x+1 are a-b,a,a+b,find a and b..​

Answer 1

Sum of zeroes = -( coefficient of x^2)/ coefficient of x^3

a- b + a + a+ b= -(-3) = 3

3a = 3

a= 1

Now as product of zeroes = - constant term /coefficient of x^3

( a-b)(a)( a+b) = -1

put a= 1

( 1- b) ( 1+ b) = -1

1 - b^2 = -1

b^2 = 2

b= √2

Answer 2

$<b>Answer:</b>$

→ a = 1  and b = ±√2 .

$<b>Step-by-step explanation:</b>$

Given polynomial is f(x) = x³ - 3x² + x + 1 .

$<b>Here  a = 1 , b = -3 , c = 1 , d = 1 .</b>$

Let α = ( a - b ) , β = a and γ = ( a + b ) .

$<b>As we know,</b>$

→ α + β + γ = -b/a .

⇒ ( a - b ) + a + ( a - b ) = -(-3)/1 .

⇒ 3a = 3 .

⇒ a = 3/3 .

∴ $\bf{\boxed{\boxed{a \:=\: 1}}}$

And,

→ $<b>αβ + βγ + γα = c/a .</b>$

⇒ a( a - b ) + a( a + b ) + ( a + b )( a - b ) = 1/1 .

⇒ $<b>a² - ab + a² + ab + a² - b² = 1 .</b>$

⇒ 3a² - b² = 1 .

⇒ ( 3 × 1² ) - b² = 1 .         { ∵ a = 1 }

⇒ 3 - b² = 1 .

⇒ b² = 3 - 1 .

⇒ b² = 2 .

∴ $<b>b = ±√2</b>$

$<b>Hence, it is solved </b>$