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
Good bye, Kyra. Your ignorance and rudeness has just earned you a banning from Jiskha. We will not tolerate badmouthing anyone here.
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
nobody cares
To find the values of x1, x2, and x3, we need to set up a system of equations using the given information.
Substituting t1 = 2 and t2 = 3 into the expression x1*f(t) + x2*g(t) + x3*h(t), we get:
x1*f(2) + x2*g(2) + x3*h(2) = 0 (equation 1)
x1*f(3) + x2*g(3) + x3*h(3) = 0 (equation 2)
Now let's substitute the definitions of f(t), g(t), and h(t) into equations 1 and 2:
x1*(5+2) + x2*(5-2) + x3*(2^2) = 0 (equation 1)
x1*(5+3) + x2*(5-3) + x3*(3^2) = 0 (equation 2)
Simplifying these equations further:
7x1 + 3x2 + 4x3 = 0 (equation 1')
8x1 + 2x2 + 9x3 = 0 (equation 2')
Now we have a system of linear equations with variables x1, x2, and x3. We can use various methods to solve this system (e.g., substitution, elimination, or matrix methods). Let's solve it using the elimination method.
Multiplying equation 1' by 2 and equation 2' by 3 to eliminate the x1 terms:
14x1 + 6x2 + 8x3 = 0 (equation 1'')
24x1 + 6x2 + 27x3 = 0 (equation 2'')
Subtracting equation 1'' from equation 2'':
(24x1 + 6x2 + 27x3) - (14x1 + 6x2 + 8x3) = 0
Simplifying, we have:
10x1 + 19x3 = 0 (equation 3)
Now we have two linear equations with two variables: equation 3 and equation 1':
10x1 + 19x3 = 0 (equation 3')
7x1 + 3x2 + 4x3 = 0 (equation 1''')
We can solve this system by using substitution or elimination. Let's solve it by elimination.
Multiplying equation 3' by 3 and adding it to equation 1''':
(30x1 + 57x3) + (7x1 + 3x2 + 4x3) = 0
Simplifying, we have:
37x1 + 61x3 + 3x2 = 0 (equation 4)
From equation 4, we can now express x2 in terms of x1 and x3:
3x2 = -37x1 - 61x3
Dividing by 3:
x2 = (-37/3)x1 - (61/3)x3
Now we have expressions for x2 in terms of x1 and x3. We can assign a parameter, say "t", to either x1 or x3 to create a parametric solution. Let's choose t = x3:
x2 = (-37/3)x1 - (61/3)t (equation 5)
x3 = t (equation 6)
Now, let's express x1 and x2 solely in terms of the parameter t:
x1 = 2t (from equation 6)
x2 = (-37/3)(2t) - (61/3)t (from equation 5)
Simplifying further:
x1 = 2t
x2 = (-74/3)t - (61/3)t
x2 = (-135/3)t
So, the values of x1, x2, and x3 in terms of the parameter t are:
x1 = 2t
x2 = (-135/3)t
x3 = t
Therefore, the real numbers x1, x2, and x3 that satisfy the given conditions are:
x1 = 2t
x2 = (-135/3)t
x3 = t,
where t can be any real number.