Posted by **Mihir Sriram** on Saturday, April 6, 2013 at 11:18am.

if alpha and beta are the zeroes of the polynomial f(x)=x^2-3x-2 , find a quadratic polynomial whose zeroes are 1/2(alpha)+beta and 1/2(beta)+alpha ?

Please... i have no idea !!

- Math -
**Bosnian**, Saturday, April 6, 2013 at 11:51am
The zeroes of the polynomial x ^ 2 - 3 x - 2 are

[ 3 - sqrt ( 17 ) ] / 2

[ 3 + sqrt ( 17 ) ] / 2

so:

alpha = [ 3 - sqrt ( 17 ) ] / 2

beta = [ 3 + sqrt ( 17 ) ] / 2

x1 = alpha / 2 + beta =

[ 3 - sqrt ( 17 ) ] / ( 2 * 2 ) + [ 3 + sqrt ( 17 ) ] / 2 =

[ 3 - sqrt ( 17 ) ] / 4 + 2 * [ 3 + sqrt ( 17 ) ] / ( 2 * 2 ) ] =

[ 3 - sqrt ( 17 ) ] / 4 + 2 * [ 3 + sqrt ( 17 ) ] / ( 2 * 2 ) =

[ 3 - sqrt ( 17 ) ] / 4 + [ 6 + 2 sqrt ( 17 ) ] / 4 =

[ 3 - sqrt ( 17 ) + 6 + 2 sqrt ( 17 ) ] / 4 =

[ sqrt ( 17 ) + 9 ] / 4

beta / 2 + alpha = alpha / 2 + beta =

[ 3 + sqrt ( 17 ) ] / ( 2 * 2 ) + [ 3 - sqrt ( 17 ) ] / 2 =

[ 3 + sqrt ( 17 ) ] / 4 + 2 * [ 3 - sqrt ( 17 ) ] / ( 2 * 2 ) ] =

[ 3 + sqrt ( 17 ) ] / 4 + 2 * [ 3 - sqrt ( 17 ) ] / ( 2 * 2 ) =

[ 3 + sqrt ( 17 ) ] / 4 + [ 6 - 2 sqrt ( 17 ) ] / 4 =

[ 3 + sqrt ( 17 ) + 6 - 2 sqrt ( 17 ) ] / 4 =

[ - sqrt ( 17 ) + 9 ] / 4

Now you must use Lagrange resolvents:

y = a x ^ 2 + b x + c = a ( x - x1 ) ( x - x2 )

in this case a = 1 so :

y = ( x - x1 ) ( x - x2 )

y = ( 1 / 4 )[ sqrt ( 17 ) + 9 ] * ( 1 / 4 )[ - sqrt ( 17 ) + 9 ]

y = [ x ^ 2 - 18 x + 64 ] / 16

y = x ^ 2 / 16 - 9 x / 8 + 4

- Math -
**Bosnian**, Saturday, April 6, 2013 at 12:29pm
x2 = beta / 2 + alpha =

[ 3 + sqrt ( 17 ) ] / ( 2 * 2 ) + [ 3 - sqrt ( 17 ) ] / 2 =

[ 3 + sqrt ( 17 ) ] / 4 + 2 * [ 3 - sqrt ( 17 ) ] / ( 2 * 2 ) ] =

[ 3 + sqrt ( 17 ) ] / 4 + 2 * [ 3 - sqrt ( 17 ) ] / ( 2 * 2 ) =

[ 3 + sqrt ( 17 ) ] / 4 + [ 6 - 2 sqrt ( 17 ) ] / 4 =

[ 3 + sqrt ( 17 ) + 6 - 2 sqrt ( 17 ) ] / 4 =

[ - sqrt ( 17 ) + 9 ] / 4

- Math -
**Mihir Sriram**, Sunday, April 7, 2013 at 2:56am
Thanks :) I understood (Y)

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