Answer
(a) $t=5$ or $t=2$
(b) $t=4$ or $t=3$
(c) $t= 5.56$ or $t= 1.44$
Work Step by Step
(a) We can make the equation $= 2$ to find the times at which the current, $i$ is $2A$.
$t^2-7t+12=2$ (Subtract 2 from each side to make the equation=0)
$t^2-7t+10=0$
We can solve this equation by factoring!
$t^2-7t+10=0$
$t^2-5t-2t+10=0$
$t(t-5)-2(t-5)=0$
$(t-5)(t-2)=0$
We can make both brackets $=0$ to find the solutions for $t$.
$(t-5)=0$ or $(t-2)=0$
$t=5$ or $t=2$
(b) We can make the equation $= 0$ to find the times at which the current, $i$ is $0A$.
$t^2-7t+12=0$
We can solve this equation by factoring!
$t^2-7t+12=0$
$t^2-4t-3t+12=0$
$t(t-4)-3(t-4)=0$
$(t-4)(t-3)=0$
We can make both brackets =0 to find the solutions for $t$.
$(t-4)=0$ or $(t-3)=0$
$t=4$ or $t=3$
(c) We can make the equation $= 4$ to find the times at which the current, $i$ is $4A$.
$t^2-7t+12=4$ (Subtract 4 from each side to make the equation=0)
$t^2-7t+8=0$
We can solve this equation using the quadratic formula, as it will be difficult to solve with factoring;
$t=\frac{-b±\sqrt {b^2-4ac}}{2a}$
$t_{1}= \frac{-(-7)+\sqrt {(-7)^2-4(1)(8)}}{2(1)}$, $t_{2}= \frac{-(-7)-\sqrt {(-7)^2-4(1)(8)}}{2(1)}$
$t_{1}= \frac{7+\sqrt {17}}{2}$, $t_{2}= \frac{7-\sqrt {17}}{2}$
$t_{1}= 5.56$, $t_{2}= 1.44$
