Math, asked by kelrina10d, 11 months ago

If ‘α’ and ‘β’are the zeros of x 2 – x +2, form a quadratic polynomial whose zeros are (2α+1)
and (2β+1)

Answers

Answered by BrainlyConqueror0901
37

\blue{\bold{\underline{\underline{Answer:}}}}

\green{\tt{\therefore{Quadratic\:eqn=x^{2}-4x+11=0}}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

 \green{ \underline \bold{Given :}}  \\  \tt: \implies  {x}^{2}  - x + 2 = 0 \\ \\  \tt: \implies  \alpha  \: and \:  \beta  \: are \: zeroes  \\ \\  \red{ \underline \bold{To \: Find :}}  \\  \tt:  \implies Quadratic \: polynomial \: whose \: zeroes \: are \: (2 \alpha  + 1) \: and \: (2 \beta  + 1) = ?

• According to given question :

 \tt: \implies  {x}^{2}  - x + 2 \\  \\  \bold{For \: sum \: of \: zeroes} \\  \tt: \implies Sum \: of \: zeroes =   - \frac{b}{a}  \\  \\ \tt: \implies \alpha  +  \beta  =  \frac{ - ( - 1)}{1}   \\  \\  \tt: \implies \alpha  +  \beta  = 1 \\  \\  \bold{For \: product \: of \: zeroes} \\  \tt: \implies Product \: of \: zeroes=  \frac{c}{a} \\  \\ \tt: \implies \alpha\beta  = \frac{2}{1}  \\  \\   \tt: \implies \alpha\beta  =2  \\  \\ \bold{For \: sum \: of \: new \: zeroes} \\  \\  \tt: \implies(2 \alpha  + 1) + (2 \beta  + 1)  \\  \\ \tt: \implies 2 \alpha  + 2 \beta  + 2  \\  \\ \tt: \implies 2( \alpha  +  \beta  + 1)

\tt: \implies 2(1 + 1) \\  \\  \green{\tt: \implies 4} \\  \\  \bold{For \: product \: of \: new \: zeroes} \\ \tt: \implies  (2\alpha  + 1) (2\beta  + 1) \\  \\ \tt: \implies (4 \alpha  \beta  + 2 \alpha  + 2 \beta  +1) \\  \\ \tt: \implies 2(2 \alpha  \beta  +  \alpha  +  \beta ) + 1 \\  \\ \tt: \implies 2(2 \times 2 + 1) + 1 \\  \\ \tt: \implies 2(5) + 1 \\  \\  \green{\tt: \implies 11} \\  \\  \bold{As \: we \: know \: that} \\ \tt: \implies  {x}^{2}  - (Sum \: of \: zeroes)x + (Product \: of \: zeroes) = 0 \\  \\  \green{\tt: \implies  {x}^{2}  - 4x + 11 = 0 }

Answered by BrainlyPopularman
32

{ \bold{ \huge{ \green{ \underline{ANSWER} :  -  }}} } \\  \\  \\ { \bold{ \blue{ \:  \:  \: . \:  \:  \: given \:  \:  a \:  \: quadratic \:  \: polynomial \:  \: }}} \\ { \bold{ \blue{ \:  \:  \:  \:   \:  \:  \:  \: {x}^{2}  - x + 2.}}} \\  \\ { \bold{ \blue{ \:  \:  \:  \: . \:  \:  \: its \:  \: have \:  \: two \:  \: roots \:  \:  \alpha  \:  \: and \:  \:  \beta .}}} \\  \\ { \bold{ \red{ \implies \:  sum  \: \: of \:  \: root \:  \:  =   - \frac{b}{a} }}} \\  \\ { \bold{ \red{  \implies \:  \alpha  +  \beta  =   - (\frac{ - 1}{1}) = 1 }}} \\  \\  { \bold{ \orange{ \implies \: product \:  \: of \:  \: roots =  \frac{c}{a} }}} \\  \\{ \bold{ \orange{ \implies \:  \alpha  \beta  = 2}}} \\  \\  \\ { \bold{ \pink{ \implies \:sum \:  \: of \:  \: roots \: of \:  \: required \: equation  - }}} \\  \\ { \bold{ \pink{ \:  \:  \:  \:  = (2 \alpha  + 1)  + (2 \beta  + 1)}}} \\  \\ { \bold{ \pink{  \:  \:  \:  \:  =  \:  \: 2( \alpha  +  \beta ) + 2 }}} \\  \\ { \bold{ \pink{  \:  \:  \:  \:  =  \:  \: 2(1) + 2 = 4}}} \\  \\  \\  { \bold{ \green{ \implies \:product \:  \: of \:  \: roots \: of \:  \: required \: equation - }}} \\  \\ { \bold { \green{ \:  \:  \:  \:  = (2 \alpha  + 1)(2 \beta  + 1)}}} \\  \\ { \bold{ \green{ \:  \:  \:  \:  = 4 \alpha  \beta  + 2 \alpha  + 2 \beta  + 1}}} \\  \\ { \bold{ \green{ \:  \:  \:  \:  = 4(2) + 2(1) + 1 = 11}}} \\  \\ { \bold{ \blue{ \implies \: so \:  \: the \:  \: new \:  \: equation \:  \:  \: is - }}} \\ \\  \\ { \bold{ \blue{ \implies \:  {x}^{2}  - (sum \:  \: of \:  \: roots)x + product \:  \: of \:  \: roots = 0}}} \\  \\ \\  { \bold{ \blue{ \implies {\boxed { {x}^{2}  - 4x + 11 = 0}}}}}

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