if alpha and beta are the zeros of polynomial 2 x squared minus 5 x + 7 then find a polynomial whose zeros are 2 alpha plus 3 Beta and 3 alpha plus 2 Beta
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alpha and beta are the roots of
2x² -5x + 7
then,
alpha + beta = ( sum of roots )
= -(-5)/2 = 5/2
product of roots = alpha.beta = 7/2
now,
a new polynomial which roots are
(2alpha + 3beta ) and ( 3alpha + 2beta)
sum of roots = { (2alpha + 3beta )+ ( 3alpha + 2beta )} = 2(alpha + beta) +3(alpha + beta) = 5(alpha + beta)
= 5 × 5/2 = 25/2
product of roots = (2alpha +3beta)(3alpha+2beta)
=6alpha² + 4alpha.beta + 9alpha.beta + 6beta²
=6( alpha² + beta²) + 13.alpha.beta
=6{(alpha +beta)² -2alpha.beta}+13alpha.beta
=6(5/2)² + 7/2 = 150/4 + 7/2 = 164/4 = 41
now, polynomial is
x² -(sum of roots )x + product of roots
x² -(25/2)x +41
2x² -5x + 7
then,
alpha + beta = ( sum of roots )
= -(-5)/2 = 5/2
product of roots = alpha.beta = 7/2
now,
a new polynomial which roots are
(2alpha + 3beta ) and ( 3alpha + 2beta)
sum of roots = { (2alpha + 3beta )+ ( 3alpha + 2beta )} = 2(alpha + beta) +3(alpha + beta) = 5(alpha + beta)
= 5 × 5/2 = 25/2
product of roots = (2alpha +3beta)(3alpha+2beta)
=6alpha² + 4alpha.beta + 9alpha.beta + 6beta²
=6( alpha² + beta²) + 13.alpha.beta
=6{(alpha +beta)² -2alpha.beta}+13alpha.beta
=6(5/2)² + 7/2 = 150/4 + 7/2 = 164/4 = 41
now, polynomial is
x² -(sum of roots )x + product of roots
x² -(25/2)x +41
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