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Statistics LibreTexts

4.3: Solve Equations with Roots

  • Page ID
    7038
  • 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?