Question

If alpha and beta are the roots of x^2+x+1=0, then find alpha^2+beta^2

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\alpha^{2}+\beta^{2}=-1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt:\implies {x}^{2} + x + 1 = 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: the \: roots \\ \\ \red{\underline \bold{To \: Find :}}\\ \tt: \implies \alpha^{2} + { \beta }^{2} = ?$

• According to given question :

$\tt: \implies {x}^{2} + x + 1 = 0 \\ \\ \tt \circ \: a = 1 \\ \\ \tt \circ \: b = 1 \\ \\ \tt \circ \: c = 1 \\ \\ \tt: \implies sum \: of \: zeroes = \frac{ - b}{a} \\ \\ \tt: \implies \alpha + \beta = \frac{ - 1}{1} \\ \\ \tt: \implies \alpha + \beta = - 1 \\ \\ \tt: \implies product \: of \: zeroes = \frac{c}{a} \\ \\ \tt: \implies \alpha \beta = \frac{1}{1} \\ \\ \tt: \implies \alpha \beta = 1 \\ \\ \bold{for \: finding \: value} \\ \tt: \implies \alpha^{2} + { \beta }^{2} \\ \\ \tt \circ \: {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy \\ \\ \tt: \implies {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ \tt: \implies ( - 1)^{2} - 2 \times 1 \\ \\ \tt: \implies 1 - 2 \\ \\ \tt: \implies - 1 \\ \\ \green{\tt \therefore { \alpha }^{2} + { \beta }^{2} = - 1}$

Answer 2

Answer:

-1

Step-by-step explanation:

Given that,

p(x) = x² + x + 1

This is a second degree polynomial in x. On comparing this polynomial with standard form of Polynomial of second degree,

ax² + bx + c,

We get,

a = 1

b = 1

c = 1

We know that,

Sun of Roots = -b/a

α + β = -1/1

α + β = -1

and,

Product of Roots = c/a

αβ = 1/1

αβ = 1

Now,

α² + β²

= α² + β² + 2αβ - 2αβ

= (α + β)² - 2αβ

= (-1)² - 2(1)

= 1 - 2

= -1