I assume you meant x3*h(t). If so, just plug and chug. You need
x1(5+t)+x2(5-t)+x3(t^2) = 0
x3 t^2 + (x1-x2)t + 5(x1+x2) = 0
Now, (t-2)(t-3) = t^2 - 5t + 6
So, x3=1
x1-x2 = -5
5x1+5x2 = 6
x1 = -19/10
x2 = 31/10
Check:
(-19/10)(5+t)+(31/10)(5-t)+1(t^2) = t^2 - 5t + 6
Let f(t) = 5+t, g(t)=5-t and h(t) = t^2.
Find real numbers x1, x2, x3 so that x1*f(t) + x2*g(t) + x2*h(t) has roots t1 = 2 and t2 = 3
4 answers
nobody cares
@oobleck
Good bye, Kyra. Your ignorance and rudeness has just earned you a banning from Jiskha. We will not tolerate badmouthing anyone here.