Question

Q) If the zeroes of the polynomial
x³-3x²+x+1 are a-b, a, a+b, find a and b ??

Best answer required !!​

Answer 1

$\huge\bf\pink{Hello!!}$

Solution is given below⬇️⬇️⬇️⬇️

zeroes of a cubic polynomial are:-


α+β+℘=-b/a

substituting the values.

a-b+a+a+b=3
3a=3
a=1

αβ+β℘+℘α=c/a

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

a^2-ba+a^2+ab+a^2-b^2=1

3a^2-b^2=1

put value of a=1

-b^2=1-3=-2
b=√2


Tysm❣️✌️

Answer 2

Answer:

→ 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 .