Plugging In: Basic Technique

What must be subtracted from $$(2a+\frac{b}{2})^{2}$$ to obtain $$2ab$$?

Incorrect. [[snippet]] __Plugging In__ $$\color{red}{a=3}$$ and $$\color{blue}{b=2}$$ for this answer choice gives > $$ \frac{\color{red}{a}^{2}+16\color{blue}{b}^{2}}{4} = \frac{\color{red}{3}^{2}+16\color{blue}{(2)}^{2}}{4} = \frac{9+16(4)}{4} = \frac{73}{4}$$. Since $$\frac{73}{4}$$ is less than 20, it is certainly too small. You can also see that it won't be an integer. That fact also eliminates this answer choice.

How did you solve this problem?

Incorrect. [[snippet]] __Plugging In__ $$\color{red}{a=3}$$ and $$\color{blue}{b=2}$$ for this answer choice gives > $$ 4(\color{red}{a}^{2}+\color{blue}{b}^{2}) = 4(\color{red}{3}^{2}+\color{blue}{2}^{2}) = $$ … Hey, wait! Before you calculate anything, notice that this will be a multiple of 4 since the values of $$a$$ and $$b$$ you used are integers. Since your **goal value** (37) is not a multiple of 4, you can automatically eliminate this answer.

Incorrect. [[snippet]] __Plugging In__ $$\color{red}{a=3}$$ and $$\color{blue}{b=2}$$ for this answer choice gives > $$ \left(\frac{\color{red}{a}}{2}+2\color{blue}{b}\right)^{2} = \left(\frac{\color{red}{3}}{2}+2\color{blue}{(2)}\right)^{2}$$. Since the inside of the parentheses will not be an integer, the final result will not be an integer. Calculate no further and eliminate E.

Variables in the answer choices is the surefire sign to __Plug In__. Don't miss an opportunity to do so. It will save you precious time and prevent careless errors. Bear in mind that you must master __Plugging In__ with easy problems before you move on to difficult ones.

Incorrect. Did you reach this answer choice by __Plugging In__ $$a=2$$ and $$b=4$$ (or $$a=1$$ and $$b=2$$) and seeing that C also works?

This is precisely why you need to plug in _all five answer choices_ when using __Plugging In__. For these particular numbers, both B and C will match your goal, but for other values only B will match your goal while C will be eliminated. The whole point of the Plugging In approach is that only one answer choice will match every goal, regardless of the numbers you use, while the other four answer choices may match the goal based on _some values_ but not for others. You are basically using the multiple choice format against itself. The idea is to continue Plugging In until four answer choices are eliminated by not matching a specific _goal_, so whatever is left must be the right answer choice because only one answer choice must be right. Choosing good numbers maximizes the chances of this happening in a single Plug-In with no need for further Plug-Ins to eliminate "persistent" answer choices.

[[snippet]] __Plugging In__ $$\color{red}{a=3}$$ and $$\color{blue}{b=2}$$ for this answer choice gives > $$\displaystyle \color{red}{a}^{2}+\color{blue}{b}^{2} = \color{red}{(3)}^{2}+\color{blue}{(2)}^{2} = 9+4 = 13$$. The number 13 doesn't match the **goal value** of 37, so eliminate it.

Correct. [[snippet]] __Plugging In__ $$\color{red}{a=3}$$ and $$\color{blue}{b=2}$$ for all the answer choices gives us the following: A. $$ \frac{\color{red}{a}^{2}+16{\color{blue}{b}}^{2}}{4} = \frac{\color{red}{3}^{2}+16\color{blue}{(2)}^{2}}{4} = \frac{9+16(4)}{4} = \frac{73}{4}$$ >Since $$\frac{73}{4}$$ is less than 20, it is certainly too small. You can also see that it won't be an integer. That fact also eliminates this answer choice. B. $$ \frac{16\color{red}{a}^{2}+\color{blue}{b}^{2}}{4} = \frac{16\color{red}{(3)}^{2}+\color{blue}{2}^{2}}{4} = \frac{16(9) + 4}{4} = 4(9) + 1 = 37$$ >This matches your **goal value**, but be sure to check all the others as well. C. $$ \color{red}{a}^{2}+\color{blue}{b}^{2} = \color{red}{3}^{2}+\color{blue}{2}^{2} = 9+4 = 13 \ne 37$$ >**POE**. D. $$ 4(\color{red}{a}^{2}+\color{blue}{b}^{2})$$ >This is simply 4 times answer choice C, or $$4\times13=52 \ne 37$$. **POE**. E. $$ \left(\frac{\color{red}{a}}{2}+2\color{blue}{b}\right)^{2} = \left(\frac{\color{red}{3}}{2}+2\color{blue}{(2)}\right)^{2}$$ >Since the inside of the parentheses will not be an integer, the final result will not be an integer. Calculate no further and eliminate E. Since B is the only answer choice that matches your **goal value** for these numbers, it has to be the right answer choice.

$$ \frac{{a}^{2}+16{b}^{2}}{4}$$

$$\frac{16{a}^{2}+{b}^{2}}{4}$$

$${a}^{2}+{b}^{2}$$

$$ 4({a}^{2}+{b}^{2})$$

$$\left(\frac{a}{2}+2b\right)^{2}$$

By __Plugging In__, how else?

Using algebra and variables.

Yes—what did I do wrong?

No—I must have made some other calculation error somewhere.

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