[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\sf \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: polynomial \: p(x) = {x}^{2} + x + 1 \\ \\ [/tex]

We know,

[tex]\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]

[tex]\bf\implies \: \alpha + \beta = - \dfrac{1}{1} = - 1 \\ \\ [/tex]

Also,

[tex]\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]

[tex]\bf\implies \: \alpha \beta = \dfrac{1}{1} = 1 \\ \\ [/tex]

Now, Consider

[tex]\sf \: { \alpha }^{2} + { \beta }^{2} \\ \\ [/tex]

[tex]\sf \: = \: {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ [/tex]

[tex]\sf \: = \: {( - 1)}^{2} - 2(1) \\ \\ [/tex]

[tex]\sf \: = \: 1 - 2 \\ \\ [/tex]

[tex]\sf \: = \: - 1 \\ \\ [/tex]

Hence,

[tex]\sf \: \bf\implies \:{ \alpha }^{2} + { \beta }^{2} \: = \: - \: 1\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\sf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta (\alpha + \beta ) \\ \\ [/tex]

[tex]\sf \: {( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta \\ \\ [/tex]

Verified answer

Answer:

Question :-

• If alpha (α) and beta (β) are the zeroes of quadratic polynomial p(x) is x² + x + 1. Find the value of α² + β².

Given :-

• α and β are the zeroes of quadratic polynomial p(x) is x² + x + 1.

To Find :-

• What is the value of α² + β².

Solution :-

Given Equation :

[tex]\mapsto \bf p(x) =\: x^2 + x + 1[/tex]

By comparing with ax² + bx + c = 0, we get :

• a = 1
• b = 1
• c = 1

Now, we have to find the sum and product of zeroes :

❒ In case of sum of zeroes :

[tex]\implies \sf\boxed{\bold{Sum\: Of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}\\[/tex]

We Have :

• b = 1
• a = 1

So, by putting those values we get,

[tex]\implies \sf \alpha + \beta =\: \dfrac{- b}{a}[/tex]

[tex]\implies \sf \alpha + \beta =\: \dfrac{- 1}{1}[/tex]

[tex]\implies \sf\bold{\alpha + \beta =\: - 1}[/tex]

❒ In case of product of zeroes :

[tex]\implies \sf\boxed{\bold{Product\: Of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}\\[/tex]

We Have :

• c = 1
• a = 1

So, by putting those values we get,

[tex]\implies \sf \alpha\beta =\: \dfrac{c}{a}[/tex]

[tex]\implies \sf \alpha\beta =\: \dfrac{1}{1}[/tex]

[tex]\implies \sf\bold{\alpha\beta =\: 1}[/tex]

Now, we have to find the value of α² + β² :

As we know that :

[tex]\bigstar \: \: \sf\boxed{\bold{a^2 + b^2 =\: (a + b)^2 - 2ab}}\: \: \: \bigstar\\[/tex]

So, by using this identities we get,

[tex]\dashrightarrow \sf\bold{\alpha^2 + \beta^2 =\: (\alpha + \beta)^2 - 2\alpha\beta}\\[/tex]

We Have :

• α + β = - 1
• αβ = 1

So, by putting those values we get,

[tex]\dashrightarrow \sf \alpha^2 + \beta^2 =\: (- 1)^2 - 2(1)\\[/tex]

[tex]\dashrightarrow \sf \alpha^2 + \beta^2 =\: (- 1)(- 1) - (2 \times 1)\\[/tex]

[tex]\dashrightarrow \sf \alpha^2 + \beta^2 =\: 1 - 2\\[/tex]

[tex]\dashrightarrow \sf\bold{\underline{\alpha^2 + \beta^2 =\: - 1}}\\[/tex]

[tex]\small \sf\boxed{\bold{\therefore\: The\: value\: of\: \alpha^2 + \beta^2\: is\: - 1\: .}}\\[/tex]