5.3: Solve Equations with Roots

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Aug 19, 2021
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- 5.2: Inequalities and Midpoints
- 5.4: Solving Linear Equations in One Variable

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Learning Outcomes

Solve equations that include square roots.

Square roots occur frequently in a statistics course, especially when dealing with standard deviations and sample sizes. In this section we will learn how to solve for a variable when that variable lies under the square root sign. The key thing to remember is that the square of a square root is what lies inside. In other words, squaring a square root cancels the square root.

Example $\PageIndex{1}$

Solve the following equation for $x$ .

$2+\sqrt{x-3}\:=\:6 \nonumber$

Solution

What makes this a challenge is the square root. The strategy for solving is to isolate the square root on the left side of the equation and then square both sides. First subtract 2 from both sides:

$\sqrt{x-3}=4 \nonumber$

Now that the square root is isolated, we can square both sides of the equation:

$\left(\sqrt{x-3}\right)^2=4^2 \nonumber$

Since the square and the square root cancel we get:

$x-3=16 \nonumber$

Finally add 3 to both sides to arrive at:

$x=19 \nonumber$

It's always a good idea to check your work. We do this by plugging the answer back in and seeing if it works. We plug in $x=19$ to get

$\begin{align*}2+\sqrt{19-3} &=2+\sqrt{16} \\[4pt] &=2+4 \\[4pt] &= 6 \end{align*}$

Yes, the solution is correct.

Example $\PageIndex{2}$

The standard deviation, $\sigma_\hat p$ , of the sampling distribution for a proportion follows the formula:

$\sigma_\hat p=\sqrt{\frac{p\left(1-p\right)}{n}} \nonumber$

Where $p$ is the population proportion and $n$ is the sample size. If the population proportion is 0.24 and you need the standard deviation of the sampling distribution to be 0.03, how large a sample do you need?

Solution

We are given that $p=0.24$ and $\sigma_{\hat p } = 0.03$

Plug in to get:

$0.03=\sqrt{\frac{0.24\left(1-0.24\right)}{n}} \nonumber$

We want to solve for $n$ , so we want $n$ on the left hand side of the equation. Just switch to get:

$\sqrt{\frac{0.24\left(1-0.24\right)}{n}}\:=\:0.03 \nonumber$

Next, we subtract:

$1-0.24\:=\:0.76 \nonumber$

And them multiply:

$0.24\left(0.76\right)=0.1824 \nonumber$

This gives us

$\sqrt{\frac{0.1824}{n}}\:=\:0.03 \nonumber$

To get rid of the square root, square both sides:

$\left(\sqrt{\frac{0.1824}{n}}\right)^2\:=\:0.03^2 \nonumber$

The square cancels the square root, and squaring the right hand side gives:

$\frac{0.1824}{n}\:=\:0.0009 \nonumber$

We can write:

$\frac{0.1824}{n}\:=\frac{\:0.0009}{1} \nonumber$

Cross multiply to get:

$0.0009\:n\:=\:0.1824 \nonumber$

Finally, divide both sides by 0.0009:

$n\:=\frac{\:0.1824}{0.0009}=202.66667 \nonumber$

Round up and we can conclude that we need a sample size of 203 to get a standard error that is 0.03. We can check to see if this is reasonable by plugging $n = 203$ back into the equation. We use a calculator to get:

$\sqrt{\frac{0.24\left(1-0.24\right)}{203}}\:=\:0.029975 \nonumber$

Since this is very close to 0.03, the answer is reasonable.

Exercise

The standard deviation, $\sigma_\bar x$ , of the sampling distribution for a mean follows the formula:

$\sigma_\bar x=\frac{\sigma}{\sqrt{n}} \nonumber$

Where $\sigma$ is the population standard deviation and $n$ is the sample size. If the population standard deviation is 3.8 and you need the standard deviation of the sampling distribution to be 0.5, how large a sample do you need?

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