Question

If α and ß are the zeroes of the quadratic polynomial ax²+bx+c, then form a Quadratic polynomial whose zeroes are α+1/ß and ß+1/α.

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Answer 1

Question:

If α and ß are the zeroes of the quadratic polynomial ax²+bx+c, then form a Quadratic polynomial whose zeroes are α+1/ß and ß+1/α.

To find:

★ To find the equation.

Answer:

$\underline{ \: \sf \pink{ \bold{{x}^{2} + x + 1 = 0}} \: } \: is \: the \: equation.$

Given:

★ Quadratic equation,

$a {x}^{2} + bx + c = 0$

$\alpha + \frac{1}{ \beta } \: and \: \beta + \frac{1}{ \alpha } \: are \: roots.$

Step-by-step explanation:

$a {x}^{2} + bx + c = 0$

Where a = 1 ,b = 1 ,c = 1

$Sum \: of \: roots = \frac{ - b}{a}$

$\boxed{Sum \: of \: roots = - 1}$

$Product \: of \: roots = \frac{c}{a}$

$\boxed{Product \: of \: roots = 1}$

Sum of the roots,

$\big( \alpha + \frac{1}{ \beta } \big) + \big( \beta + \frac{1}{ \alpha } \big)$

$\implies \frac{ \alpha \beta + 1}{ \beta } + \frac{ \alpha \beta + 1}{ \alpha }$

$\implies \frac{ \alpha ( \alpha \beta + 1) + \beta ( \alpha \beta + 1)}{ \alpha \beta }$

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

$\implies \frac{ { \alpha }^{2} \beta + \alpha { \beta }^{2} + 2 }{ \alpha \beta }$

$\implies \frac{( \alpha \beta )( \alpha + \beta ) + 2}{ \alpha \beta }$

Now substituting the values,

$\implies \frac{1 ( - 1 )+ 2}{1}$

$\implies \frac{ - 1 + 2}{1}$

$\implies \frac{ - 1}{1}$

$\implies \boxed{ - 1}$

$\boxed{Sum \: of \: roots \: is \: - 1}$

Product of the roots,

$\big( \alpha + \frac{1}{ \beta } \big) \times \big( \beta + \frac{1}{ \alpha } \big)$

$\implies \frac{( \alpha + 1)( \beta + 1)}{ \alpha \beta }$

$\implies \frac{ (\alpha \beta) + ( \alpha + \beta ) + 1}{ \alpha \beta }$

Now substituting the values,

$\implies \frac{(1) + ( - 1) + 1}{1}$

$\implies \boxed{ 1}$

$\boxed{Product \: of \: roots = 1}$

The equation is,

${x}^{2} - \big( \alpha + \frac{1}{ \beta } \big) + \big( \beta + \frac{1}{ \alpha } \big) \times x + \big( \alpha + \frac{1}{ \beta } \big) \big( \beta + \frac{1}{ \alpha } \big) = 0$

$\implies {x}^{2} - ( - 1)x + 1 = 0$

$\implies \boxed{{x}^{2} + x + 1 = 0}$

$\therefore The \: equation \: is\: \bold{{x}^{2} + x + 1 = 0}.$