Question

if the zeroes of the polynomial x^3-3x^2+x+1 are alfa-beta,alfa,alfa+beta then alfa= and beta=​

Answer 1

$\huge\bf{\red{\overbrace{\underbrace{\purple{Given:}}}}}$

★$A\:cubic\:polynomial\:x^{3}-3x^{2}+x+1=p(x)$ (say)

★$\alpha, \alpha-\beta, \alpha+\beta\:are \:its\:zeroes$

$\huge\bf{\red{\overbrace{\underbrace{\purple{To\:\:Find:}}}}}$

★$The\: value \:of\:\alpha\:and\:\beta$

$\huge\bf{\red {\overbrace{\underbrace{\purple{Concept\:\:Used:}}}}}$

★We will use the formula for finding the sum of the zeroes of a, cubic polynomial, similar to $ax^{3}+bx^{2}+cx+d$ form.

$\huge\bf{\red {\overbrace{\underbrace{\purple{Answer:}}}}}$

We have,

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

With respect to standard form$ax^{3}+bx^{2}+cx+d$

a = 1

b = -3

c = 1

d = 1

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So, sum of zeroes is given by,

$\large\green{\boxed{\phi+\Delta+\gamma=\dfrac{-b}{a}=\dfrac{-Coefficient\:of\;x^{2}}{Coefficient\:of\:x^{3}}}}$

where,

$\phi,\Delta\:and\:\gamma$ are the zeroes of the cubic polynomial.

Now , using this,

$\implies \dfrac{-b}{a}=\alpha+\alpha-\beta \alpha+\beta$

$\implies \dfrac{-b}{a}=\alpha+\alpha-\cancel{\beta} +\alpha+\cancel{\beta}$

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

$\implies 3=3\alpha$

$\implies \alpha=\frac{3}{3}$

$\implies \alpha=\frac{\cancel{3}}{\cancel{3}}$

$\red{\underline{\boxed{\purple{.\degree.\:\alpha=1}}}}$

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Now,

$\alpha=1$ is a zero.

So, product of zeroes.

$\large\green{\boxed{\phi\Delta\gamma=\dfrac{-d}{a}=\dfrac{Constant\:\:term}{Coefficient\:of\:x^{3}}}}$

$\implies \alpha(\alpha+\beta) (\alpha-\beta) =\frac{-d}{a}$

$\implies \alpha[(\alpha)^{2}-(\beta) ^{2}]=-1$

$\implies 1(1-\beta^{2})=-1$

$\implies \beta^{2}=1+1$

$\implies \beta^{2}=2$

$\green{\underline{\boxed{\orange{.\degree.\beta=\sqrt{2}}}}}$

Therefore the value of $\alpha$ is 1 & $\beta$ is √2.

Answer 2

$\huge \bf \red{answer}$