Question

f the zeros of the polynomial x3 – 3x2 + x + 1 are (a-b), a and (a+b). Find the value of a and b?

Answer 1

Answer:

$p(x) = {x}^{3} - 3 {x}^{2} + x + 1$

zeros =(a-b),a,(a+b)

if

$\alpha . \beta . \gamma \: are \: zeroes \: then \:$

$p(x) = {x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma$

so sum of roots =3

so a+b+a+a-b=3

3a=3

a=1.

Now product of roots =-1

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

substituting value of a

$(1 - {b}^{2} ) = - 1$

$b = \sqrt{2}$

Answer 2

Answer:

$a = 1 \: and \: b = + - \sqrt{2}$

Step-by-step explanation:

Given

$a - b = \alpha$

$a = \beta$

$a + b = \gamma$

and we know that.....

$\alpha + \beta + \gamma = - b \div a$

$\alpha + \beta + \gamma = - ( - 3) \div 1 = 3$

therefore

$a - b + a + a + b = 3$

$a = 1$

also we know that......

$\alpha \beta + \beta \gamma + \gamma \alpha = c \div a$

therefore....

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

$(1 - \: b )1 + 1(1 + b) + (1 + b)(1 - b) = 1$

$- b + b - 1 - {b}^{2} = 1$

${b}^{2} = 2$

$b = + - \sqrt{2}$