Answer
(a) $t=4$ or $t=8$
(b) $t=10.58$ or $t = 1.42$
(c) $t=-4$ or $t=16$
Work Step by Step
(a) We know that $V= 8$, and
V = $t^2-12t+40$
So, we can say that $t^2-12t+40 = 8$.
(Subtract 8 from each side to make it equal 0)
$t^2-12t+32 = 0$
We can solve this quadratic equation by factoring:
$t^2-8t-4t+32 = 0$
$t(t-8)-4(t-8) = 0$
$(t-4)(t-8) = 0$
We can find the values of $t$ by making each bracket equal to 0.
$(t-4)=0$ or $(t-8) = 0$
$t=4$ or $t=8$
(b) We know that $V= 25$, and
V = $t^2-12t+40$
So, we can say that $t^2-12t+40 = 25$.
(Subtract 25 from each side to make it equal to 0)
$t^2-12t+15 = 0$
We can solve this equation using the quadratic formula, as it is difficult to do by factoring:
$t = \frac{-b±\sqrt {b^2-4ac}}{2a}$
$t_{1}=\frac{-(-12)+\sqrt {(-12)^2-4(1)(15)}}{2(1)}$, $t_{2} = \frac{-(-12)-\sqrt {(-12)^2-4(1)(15)}}{2(1)}$
$t_{1}=\frac{12+\sqrt {84}}{2}$, $t_{2} = \frac{12-\sqrt {84}}{2}$
$t_{1}=10.58$, $t_{2} = 1.42$
(c) We know that $V= 104$, and
V = $t^2-12t+40$
So, we can say that $t^2-12t+40 = 104$.
(Subtract 104 from each side to make it equal to 0)
$t^2-12t-64 = 0$
We can solve this quadratic equation by factoring:
$t^2-16t+4t+-64 = 0$
$t(t-16)+4(t-16) = 0$
$(t+4)(t-16) = 0$
We can find the values of $t$ by making each bracket equal to 0.
$(t+4)=0$ or $(t-16) = 0$
$t=-4$ or $t=16$
