Question

if alpha and beta are the zeros of the quadritic polynomial such that alpha+beta=24 and alpha-beta=8. find the quadritic polynomial having alpha and beta as its zeroes​

Answer 1

Answer:

Required Polynomial $= > {x}^{2} - 24x + 128$

Step-by-step explanation:

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See The Attachment My Friend....

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Answer 2

Answer:

$\boxed{\sf \: \sf \: f(x) = k({x}^{2} - 24x + 128), \: where \: k \: \ne \: 0 \: }$

Step-by-step explanation:

Given that,

$\sf \: \alpha + \beta = 24 - - - (1)$

and

$\sf \: \alpha - \beta = 8 - - - (2)$

On adding equation (1) and (2), we get

$\sf \: 2 \alpha = 32 \: \: \implies\sf \: \alpha = 16$

On subtracting equation (2) from (1), we get

$\sf \: 2 \beta = 16 \: \: \implies\sf \: \beta = 8$

Now, the required quadratic polynomial f(x) whose zeroes are $\alpha$ and $\beta$ respectively is .

$\sf \: f(x) = k[ {x}^{2} - ( \alpha + \beta )x + \alpha \beta ], \: where \: k \: \ne \: 0$

$\sf \: f(x) = k[ {x}^{2} - ( 16 + 8 )x + 16 \times 8 ], \: where \: k \: \ne \: 0$

$\sf \: f(x) = k({x}^{2} - 24x + 128), \: where \: k \: \ne \: 0$

Hence, the required polynomial is

$\implies\sf \: \boxed{\sf \: \sf \: f(x) = k({x}^{2} - 24x + 128), \: where \: k \: \ne \: 0 \: }$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered} \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$