What are solutions of the equation u(v(t))=0 if u(t)=t^2-16 and v(t)=t+2 ?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We need to determine the solutions of u(v(t)) = 0 given that u(t) = t^2 - 16 and v(t) = t + 2.
u(v(t)) = 0
=> u( t + 2) = 0
=> (t + 2)^2 - 16 = 0
=> (t + 2)^2 - 4^2 = 0
=> (t + 2 - 4)(t + 2 + 4) = 0
=> (t - 2)( t + 6) = 0
t - 2 = 0 => t = 2
t + 6 = 0 => t = -6
We get the solutions of u(v(t)) = 0 as t = 2 and t = -6
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll substitute v(t) by it's expression:
u(v(t)) = u((t+2))
u((t+2)) = (t+2)^2 - 16
We'll notice that we have a difference of squares:
u((t+2)) = (t+2-4)(t+2+4)
u((t+2)) = (t-2)(t+6)
We'll solve the equation:
u((t+2)) = 0 <=> (t-2)(t+6) = 0
We'll set each factor as zero:
t-2 = 0
t = 2
t+6=0
t=-6
The solutions of the equation u(v(t)) = 0 are: {-6 ; 2}.