Question

HELLO FRIENDS✨✨✨✨✨✨

PLEASE answer this question
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if alpha and beta are the zeroes of the quadratic equation f(x)=x²-2x+3, find a polynomial whose roots are

(I)alpha+2 , beta+2.
(ii) alpha-1/alpha+1 , beta-1/beta + 1

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Answer 1

$\underline{\underline{\large{\mathfrak{Solution : }}}}$



$\underline{\textsf{ Given : }} \\ \\ \sf \implies f(x) \: = \: x^2 \: - \: 2x \: + \:3 \\ \\ \sf \alpha \: and \: \beta \: are \: its \: zeroes.$



$\textsf{Here ! } \\ \\ \sf \implies Coefficient \: of \: x^{2} ( a ) \: = \: 1 \\ \\ \sf \implies Coefficient \: of \: x(b) \: = \: -2 \\ \\ \sf \implies Constant \: term ( c) \: = \: 3$


$\sf Now :$



$\sf \implies Sum \: of \: zeroes \: = \: - \left ( \dfrac{b}{a} \right ) \\ \\ \sf \implies \alpha \: + \: \beta \: = \: - \left( \dfrac{-2}{1} \right) \\ \\ \sf \therefore \quad \alpha \: + \: \beta \: = \: 2 \quad...(1)$


$\sf And :$



$\sf \implies Product \: of \: zeroes \: = \: \dfrac{c}{a} \\ \\ \sf \implies \alpha \beta \: = \: \dfrac{3}{1} \\ \\ \sf \therefore \quad \alpha \beta \: = \: 3 \quad...(2)$




$\large{ \sf Question \: no. ( 1 )}$



$\underline{\textsf{We know that : }}$



$\sf P(x) \: = \: x^{2} \: - \: ( Sum \: of \: zeroes )x \: + \: (Product \: of \: zeroes ) \\ \\ \sf \qquad \: = \: {x}^{2} \: - \: ( \alpha \: + 2 \: + \: \beta \: + \: 2)x \: + \: ( \alpha \: + \: 2)( \beta \: + \: 2) \\ \\ \sf \: \qquad = {x}^{2} \: - \: ( \alpha \: + \: \beta \: + \: 4)x \: + \: (\alpha \beta \: + \: 2 \alpha \: + \: 2 \beta \: + \: 4) \\ \\ \sf \qquad \: = \: {x}^{2} \: - \: ( \alpha \: + \: \beta \: + \: 4)x \: + \: \{ \alpha \beta \: + \: 2( \alpha \: + \: \beta ) \: + \: 4 \}$



$\textsf{Plug the value of (1) and (2) :}$


$\sf \qquad \: = \: {x}^{2} \: - \: ( 2 \: + \: 4)x \: + \: \{ 3 \: + \: 2( 2) \: + \: 4 \} \\ \\ \sf \qquad \: = \: {x}^{2} \: - \: 6x \: + \: ( \: 3 \: + \: 4 \: + \: 4) \\ \\ \sf \qquad \: = \: {x}^{2} \: - \: 6x \: + \: 11$




$\large{ \sf Question \: no. ( 2 )}$



$\underline{\textsf{We know that : }}$


$\sf P(x) \: = \: x^{2} \: - \: ( Sum \: of \: zeroes )x \: + \: (Product \: of \: zeroes ) \\ \\ \sf\qquad \: = \: {x}^{2} \: - \: \left \{ \bigg( \dfrac{ \alpha \: - \: 1}{ \alpha \: + \: 1} \bigg) \: + \: \left( \dfrac{ \beta \: - \: 1}{ \beta \: + \: 1} \right)\right \}x \: + \\ \sf \qquad \: \: \: \: \left \{ \left( \dfrac{ \alpha \: - \: 1}{ \alpha \: + \: 1} \right) \left( \dfrac{ \beta \: - \: 1}{ \beta \: + \: 1 } \right ) \right \}$




$\sf = \: {x}^{2} \: - \: \left \{ \dfrac{( \alpha \: - \: 1)( \beta \: + \: 1) \: + \: ( \beta \: - \: 1)( \alpha \: + \: 1)}{( \alpha \: + \: 1)( \beta \: + \: 1)} \right \}x \: + \\ \sf \qquad \quad \: \: \left \{\dfrac{( \alpha \: - \: 1)( \beta \: - \: 1)}{( \alpha \: + \: 1)( \beta \: + \: 1)} \right \}$





$\sf = \: {x}^{2} \: - \: \left \{ \dfrac{ \alpha \beta \: + \: \cancel\alpha \: - \: \cancel\beta \: - \: 1\: + \: \alpha \beta \: - \: \cancel\alpha \: + \: \cancel\beta \: - \: 1} {\alpha \beta \: + \: \alpha \: + \: \beta \: + \: 1} \right \}x \: + \\ \sf \qquad \quad \: \: \left \{\dfrac{ \alpha \beta \: - \: \alpha \: - \: \beta \: + \: 1}{ \alpha \beta \: + \: \alpha \: + \: \beta \: + \: 1} \right \}$







$\sf = \: {x}^{2} \: - \: \left \{ \dfrac{ 2 \alpha \beta \: - \: 2} {\alpha \beta \: + \: ( \alpha \: + \: \beta ) \: + \: 1} \right \}x \: + \\ \sf \qquad \quad \: \: \left \{\dfrac{ \alpha \beta \: - \: ( \alpha \: + \: \beta ) \: + \: 1}{ \alpha \beta \: + \: ( \alpha \: + \: \beta) \: + \: 1} \right \}$





$\textsf{Plug the value of (1) and (2) :}$



$\sf = \: {x}^{2} \: - \: \left \{ \dfrac{ 2 \: \times \: 3 \: - \: 2} {3 \: + \: 2\: + \: 1} \right \}x \: + \: \left \{\dfrac{ 3 \: - \: 2\: + \: 1}{ 3 \: + \: 2\: + \: 1} \right \}$



$\sf = \: {x}^{2} \: - \: \left \{ \dfrac{ 6\: - \: 2} {6} \right \}x \: + \: \left \{\dfrac{ 4 \: - \: 2}{ 6} \right \}$



$\sf = \: {x}^{2} \: - \: \left \{ \dfrac{4} {6} \right \}x \: + \: \left \{\dfrac{ 2}{ 6} \right \}$



$\sf = \: {x}^{2} \: - \: \dfrac{2} {3} x \: + \: \dfrac{ 1}{ 3}$


$\underline{\textsf{Multiply by 3 ,}}$


$\sf = \: 3 \left( {x}^{2} \: - \: \dfrac{2} {3} x \: + \: \dfrac{ 1}{ 3} \right)$


$\sf = \: 3 {x}^{2} \: - \: 3 \left(\dfrac{2} {3} x \right)\: + \: 3\left(\dfrac{ 1}{ 3} \right) \\ \\ = \sf\: 3 {x}^{2} \: - \: 2x \: + \: 1$