# What are solutions of the equation u(v(t))=0 if u(t)=t^2-16 and v(t)=t+2 ?

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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**

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}.**