Question

If the zeroes of the polynomial x3-3x2+x+1 are in ap,then find its common difference

Answer 1

Sum of zeroes=-b/a

Product=-d/a

Now let us assume the terms are a-d,a and a+d.

With first term as a-d and common difference d

Now sum of zeroes=a-d+a+a+d=3a

3a=-b/a=3

a=1

Product=(a-d)a(a+d)

=a^4-a^2×d^2

=1-d^2

=-d/a

=-1

d^2=2

d=√2

Common difference is √2

Answer 2

$\huge{\boxed{\mathtt{\red{ANSWER}}}}$

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 (Ans)