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.
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
go with oobleck, didn't see the 2 in front of new roots
To find a quadratic polynomial with given zeroes, we need to use the fact that if α and β are the zeroes of a quadratic polynomial, then the polynomial can be expressed in the factored form as follows:
f(t) = a(t−α)(t−β)
where a is the leading coefficient. In this case, the quadratic polynomial f(t) = 2t^2 - 7t + 6 has zeroes α and β.
We are given that the zeroes are (2α + 1) and (2β + 1), so we can use this information to find α and β.
Given: 2α + 1 = 0
Subtracting 1 from both sides:
2α = -1
Dividing both sides by 2:
α = -1/2
Similarly, for the other zero:
2β + 1 = 0
Subtracting 1 from both sides:
2β = -1
Dividing both sides by 2:
β = -1/2
Now that we know the values of α and β, we can use them to find the quadratic polynomial.
Substituting α and β into the factored form:
f(t) = 2(t − α)(t − β)
f(t) = 2(t − (-1/2))(t − (-1/2))
f(t) = 2(t + 1/2)(t + 1/2)
Expanding the expression:
f(t) = 2(t^2 + t/2 + t/2 + 1/4)
f(t) = 2(t^2 + t + 1/4)
f(t) = 2t^2 + 2t + 1/2
Therefore, the quadratic polynomial whose zeros are (2α + 1) and (2β + 1) is f(t) = 2t^2 + 2t + 1/2.
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