Question

if the zeroes of the polynomial f(x)=x^3f(x)=x^3-3x^2+x+1 are a-b, a, a+b. find a and b ​

Answer 1

Answer:

Step-by-step explanation:

There are three zeroes given a-b, a, a+b

So,

     Alpha= a-b, beta = a, gaama = a+b

  Polynomial is given x^3-3x^2+x+1

                 Compare: ax^3+bx^2+cx+d

a=1, b=-3, c=1, d=1

Alpha + beta + gaama = -b/a

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

3a = 3/1

3a=3

a= 3/3

a= 1

We know that

Alpha*beta*gaama = -d/a

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

Put value of a

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

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

1-b^2 = -1

1 +1 = b^2

2 = b^2

Under root 2 = b

So , a= 1 b= under root 2

Note: jab square hatate h to under root lagta h

Answer 2

Answer :-

$\sf{Sum\ of\ zeroes\ =\ a - b + a + a + b}$

$\implies{\sf{3a}}$

$\sf{Sum of zeroes = \dfrac{-b}{a}}$

$\implies{\sf{3a = \dfrac{-(-3)}{1}}}$

$\implies{\sf{3a = 3}}$

$\implies{\sf{a = 1}}$

$\sf{Product\ of\ zeroes\ =\ (a - b) (a) (a + b)}$

$\implies{\sf{(1 - b) (1) (1 + b)}}$

$\implies{\sf{(1 - b) (1 + b)}}$

[Identity :- (a + b) (a - b) = a² - b²]

$\implies{\sf{{1}^{2} - {b}^{2}}}$

$\implies{\sf{1 - {b}^{2}}}$

$\sf{Product of zeroes = \dfrac{-d}{a}}$

$\implies{\sf{1 - {b}^{2} = -1}}$

$\implies{\sf{{- b}^{2} = - 1 - 1}}$

$\implies{\sf{{b}^{2} = 2}}$

$\implies{\sf{b = \pm \sqrt{2}}}$

So, the value of a & b are :-

$\boxed{\sf{a\ =\ 1}}$

$\boxed{\sf{b\ = \pm \sqrt{2}}}$