Question

If X square-x-2=0 then find alpha square+ bheta square​

Answer 1

Answer:

$\alpha + \beta = \frac{1}{1} = 1 \\ \alpha \beta = \frac{ - 2}{1} = - 2 \\ \\ \\ { \alpha }^{2} + { \beta }^{2} \\ = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ = {1}^{2} - 2 \times - 2 \\ = 1 + 4 \\ = 5$

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Answer 2

x² - x - 2 = 0

On comparing to the equation of ax²+bx+c = 0

here, a = 1, b= -1 and c = -2

As we know that :

$\alpha + \beta = \frac{ - b}{a} \: \: and \: \alpha . \beta = \frac{c}{a}$

Now, we have to find the value of

${ \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ = > { \alpha }^{2} + { \beta }^{2} = {(\frac{ -b }{a})}^{2} - \frac{2c}{a} = \frac{ {b}^{2} }{ {a}^{2} } - \frac{2c}{a} \\ \\ = > { \alpha }^{2} + { \beta }^{2} = \frac{ {b}^{2} - 2ac }{ {a}^{2} } \\ \\ = > { \alpha }^{2} + { \beta }^{2} = \frac{1 - 2 \times 1 \times ( - 2)}{1} = \frac{1 + 4}{1} = 5$

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