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
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 !!
3 answers
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
[ 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
Thanks :) I understood (Y)