well, f(t) = (2t-3)(t-2)
so its roots are 3/2 and 2
your new roots are 4 and 5, so
g(t) = (t-4)(t-5) = t^2-9t+20
Find a quadratic polynomial whose zeroes are (2Alpha + 1) and (2Beta+1) if Alpha and Beta are the zeroes of the polynomial f(t)=2t^2-7t+6.
3 answers
using the property:
if ax^2 + bx + c = 0 has roots p and q
then p+q = -b/a
pq = c/a
so if using r for alpha and q for beta
then p+q = 7/2
pq = 6/2 = 3
new roots:
p+1 and q+1
p+1 + q+1 = 7/2 + 2 = 11/2
(p+1)(q+1)
= pq + p+q + 1
= 3 + 7/2 + 1
= 15/2
new function:
f(x) = 2x^2 - 11x + 15
if ax^2 + bx + c = 0 has roots p and q
then p+q = -b/a
pq = c/a
so if using r for alpha and q for beta
then p+q = 7/2
pq = 6/2 = 3
new roots:
p+1 and q+1
p+1 + q+1 = 7/2 + 2 = 11/2
(p+1)(q+1)
= pq + p+q + 1
= 3 + 7/2 + 1
= 15/2
new function:
f(x) = 2x^2 - 11x + 15
go with oobleck, didn't see the 2 in front of new roots
3 answers