if alpha and beta are the roots of the equation 3x^2-6x+1=0 then find the quadratic equation whose roots are 2alpha +beta and 2beta +alpha​

Question

if alpha and beta are the roots of the equation 3x^2-6x+1=0 then find the quadratic equation whose roots are 2alpha +beta and 2beta +alpha​

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Genesis 3 weeks 2021-09-09T04:41:03+00:00 1 Answer 0 views 0

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    2021-09-09T04:42:28+00:00

    Answer:

    From eq.,

    3 {x}^{2}  - 6x + 1 = 0

     \alpha  +  \beta  =  -  \frac{ - 6}{3}  = 2

     \alpha  \beta  =  \frac{1}{3}

    now, sum of the roots of new eq. is

    2 \alpha  +  \beta  + 2 \beta  +  \alpha  = 3( \alpha  +  \beta ) = 3 \times 2 = 6

    product of roots of new eq is

    (2 \alpha  +  \beta )(2 \beta  +  \alpha ) = 4 \alpha  \beta  + 2 { \beta }^{2}  + 2 { \alpha }^{2}  +  \alpha  \beta  =

    5 \alpha  \beta  + 2( { \alpha }^{2}  +  { \beta }^{2} ) = 5 \alpha  \beta  + 2(( \alpha  +  \beta ) {}^{2}  - 2 \alpha  \beta ) = 5 \times  \frac{1}{3}  + 2( {2}^{2}  - 2 \times  \frac{1}{3} ) =  \frac{5}{3}  + 2(4 -  \frac{2}{3} ) =

     \frac{5}{3}  +  \frac{20}{3}  =  \frac{25}{3}

    so quadratic equation is,

     {x}^{2}  - (sum \: of \: roots)x +  \: product \: of \: roots = 0

     {x}^{2}  - 6x +  \frac{25}{3}  = 0

    3 {x}^{2}  - 18x + 25 = 0

    Ans.

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