$$(t-4)^2-16=0$$

Which of the following is a value of $t$ that satisfies the equation above?

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Solve for $t$ by common algebraic methods:

$$(t-4)^2-16=0$$ $$(t-4)^2=16$$ $$t-4 = \pm \sqrt{16}$$ $$t-4= \pm 4$$ $$t-4 = 4 \text{ and } t-4=-4$$ $$t=8 \text{ and } t=0$$

Expand and factor:

$$(t-4)^2-16=0$$ $$t^2-8t+16-16=0$$ $$t^2-8t=0$$ $$t(t-8)=0$$ $$t=0 \text { and } t=8$$

Expand and use the quadratic formula:

$$(t-4)^2-16=0$$ $$t^2-8t+16-16=0$$ $$t^2-8t=0$$ $$t=\frac{-(-8) \pm \sqrt{(-8)^2-4(1)(0)}}{2(1)}$$ $$t=\frac{8 \pm \sqrt{64}}{2}$$ $$t = \frac{8+8}{2} \text { and } \frac{8-8}{2}$$ $$t=8 \text{ and } 0$$

Use the difference of squares to factor.

$$a^2-b^2=(a+b)(a-b)$$ $$(t-4)^2-16=0$$ $$(t-4)^2-4^2=0$$ $$((t-4)+4)((t-4)-4)=0$$ $$(t)(t-8)=0$$ $$t=0 \text{ and } 8$$

Plug in the answer choices:

Only the third choice $\boxed{8}$ is viable:

$$(t-4)^2-16=0$$ $$((8)-4)^2-16=0$$ $$4^2-16=0$$ $$16-16=0 \checkmark$$

As you can see, there are quite a lot of options to pick from. Use the one that your most comfortable with, but try some new methods from time to time. They allow you to double check your answer using a different method, in addition to improving your math mastery!