Question

If α and β are zeroes of the quadratic polynomial f(x)=x²+px+1 then find the values of (1) α²+β² (2) 1/α+1/β​

Answer 1

Given:

$\star \: \bold{ \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the \: polynomial} \: \\ \bold{ f \: (x) = {x}^{2} + px + 1}.$

To find :

•The values of :

$1) { \alpha }^{2} + { \beta }^{2} \\ \\ 2) \frac{1 }{ \alpha } + \frac{1}{ \beta }$

Solution :

• F(x) = x² +px +1 .

Compare with ax² + bx + c ,we get :

• a = 1 , b = p and c = 1

As we know that :

$\star \bold{\: sum \: of \: zeroes \: ( \alpha + \beta ) = \dfrac{ - b}{a} = \frac{ - p}{1} = - p} \\ \\ \star \bold{ product \: of \: zeroes \: ( \alpha \beta ) = \frac{c}{a} = \dfrac{1}{1} = 1}$

Now, find

$1) { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ \implies \: { \alpha }^{2} + { \beta }^{2} = {( - p)}^{2} - 2(1) \\ \\ \implies \: { \alpha }^{2} + { \beta }^{2} = {p}^{2} - 2$

$2) \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = \dfrac{ \beta + \alpha }{ \alpha \beta } = \dfrac{ - p}{1} = - p$

Answer 2

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