Question

If the zeroes of polynomial x3-3x2+x+1 are alpha-beta ,alpha and alpha +beta find alphan and beta

Answer 1

Given that cubic polynomial,
x³-3x²+x+1
then,
(a-b)+a+(a+b)=-b/a
3a=-(-3)/1
3a=3
a=3/3
a=1

Again,
(a-b)*a*(a+b)=-d/a
a(a²-b²)=-1/1
1(1²-b²=-1
1-b²=-1
-b²=-1-1
-b²=-2
b±2
b=±√2

that

a=1
b=√2

that's all Sujeet

Answer 2

Answer:

$\text{The value of }\alpha=1, \beta=\pm\sqrt2$

Step-by-step explanation:

Given α-β, α, α+β  are the zeroes of cubic polynomial  

$x^3-3x^2+x+1$  

we have to find the value of α and β

Polynomial: $x^3-3x^2+x+1$  

$\text{Sum of zeroes=}\frac{-b}{a}=\frac{-(-3)}{1}=3$

(α-β)+α+(α+β)=3

3α=3 ⇒ α=1

$\text{Product of zeroes=}\frac{-d}{a}=\frac{-1}{1}=-1$

$(\alpha-\beta)\alpha.(\alpha+\beta)=-1$

$\alpha((\alpha)^2-(\beta)^2)=-1$

Put α=1

$1-(\beta)^2=-1$

$(\beta)^2=2$

$\beta=\pm\sqrt2$

$\text{Hence, the value of }\alpha=1, \beta=\pm\sqrt2$