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Mathematics LibreTexts

8.1: Introduction to Radical Notation

  • Page ID
    19898
  • ( \newcommand{\kernel}{\mathrm{null}\,}\)

    We know how to square a number. For example:

    • 52=25
    • (5)2=25

    Taking the square root of a number is the opposite of squaring.

    • The nonnegative square root of 25 is 5.
    • The negative square root of 25 is 5.

    Thus, when searching for a square root of a number, we are searching for number whose square is equal to our number.

    Example 8.1.1

    Find the square roots of 81.

    Solution

    We are looking for a number whose square is 81.

    • Because 92=81, the nonnegative square root of 81 is 9.
    • Because (9)2=81, the negative square root of 81 is 9.

    Hence, 81 has two square roots, 9 and 9.

    Exercise 8.1.1

    Find the square roots of 64.

    Answer

    8 and 8

    Example 8.1.2

    Find the square roots of 0.

    Solution

    We are looking for a number whose square is 0.

    • Because 02=0, the nonnegative square root of 0 is 0.

    No other number squared will equal zero. Hence, zero has exactly one square root, namely zero.

    Exercise 8.1.2

    Find the square roots of 100.

    Answer

    10 and 10

    Example 8.1.3

    Find the square roots of 36.

    Solution

    We are looking for a number whose square is 36. However, every time you square a real number, the result is never negative. Hence, 36 has no real square roots.1

    Exercise 8.1.3

    Find the square roots of 25.

    Answer

    no real square roots

    The introductions in Examples 8.1.1, 8.1.2, and 8.1.3 lead to the following definition.

    Defining the Square Roots of a Number

    The solutions of x2=a are called square roots of a.

    Case: a>0. The equation x2=a has two real solutions, namely x=±a.

    • The notationa calls for the nonegative square root.
    • The notation a calls for the negative square root.

    Case: a=0. The equation x2=0 has exactly one solution, namely x=0.

    Case: a<0. The equation x2=a has no real solutions.

    Example 8.1.4

    Solve x2=9 for x, then simplify your answers.

    Solution

    Because the right-hand side of x2=9 is positive, the equation has two solutions.

    x2=9Original equation.x=±9Two answers: 9 and 9

    To simplify these answers, we need to understand the following facts:

    • 9 calls for the nonnegative square root of 9. Because (3)2=9, the nonnegative square root of 9 is 3. Hence, 9=3.
    • 9 calls for the negative square root of 9. Because (3)2=9, the negative square root of 9 is 3. Hence, 9=3.

    Thus, the solutions of x2=9 are x=±3, which is equivalent to saying “x=3 or x=3.”

    Exercise 8.1.4

    Solve x2=16 for x, then simplify your answers.

    Answer

    4, 4

    Example 8.1.5

    Solve x2=0 for x, then simplify the answer.

    Solution

    There is only one number whose square equals 0, namely 0.

    x2=0Original equation.x=0One answer: (0)2=0.

    Thus, the only solution of x2=0 is x=0. Consequently, the nonnegative square root of zero is zero. Hence, 0=0.

    Exercise 8.1.5

    Solve x2=49 for x, then simplify the answers.

    Answer

    7, 7

    Example 8.1.6

    Solve x2=4 for x, then simplify the answer.

    Solution

    You cannot square a real number and get a negative result. Hence, x2=4 has no real solutions. Therefore, 4 is not a real number.

    Exercise 8.1.6

    Solve x2=9 for x, then simplify the answers.

    Answer

    no real solutions

    Example 8.1.7

    Simplify each of the following:

    1. 121
    2. -225
    3. 100
    4. -324

    Solution

    Remember, the notation a calls for the nonnegative square root of a, while the notation -a calls for the negative square root of a.

    1. Because 112=121, the nonnegative square root of 225 is 15. Thus:225=15
    2. Because (15)2=225, the nonnegative square root of 121 is 11. Thus:121=11
    3. You cannot square a real number and get 100. Therefore, 100 is not a real number.
    4. Because (18)2=324, the nonnegative square root of 324 is 18. Thus:324=18

    Exercise 8.1.7

    Simplify: 144

    Answer

    12

    Squaring “undoes” taking the square root.

    Squaring square roots

    If a>0, then both a and a are solutions of x2=a. Consequently, if we substitute each of them into the equation x2=a, we get:

    (a)2=a and (a)2=a

    Example 8.1.8

    Simplify each of the following expressions:

    1. (5)2
    2. (7)2
    3. (11)2

    Solution

    We’ll handle each case carefully.

    1. Because 5 is a solution of x2=5, if we square 5, we should get 5.(5)2=5
    2. Because 7 is a solution of x2=7, if we square 7, we should get 7.(7)2=7
    3. Because x2=11 has no real answers, 11 is not a real number. Advanced courses such as college algebra or trigonometry will introduce the complex number system and show how to handle this expression.

    Exercise 8.1.8

    Simplify: (21)2

    Answer

    21

    Using the Graphing Calculator

    Up to this point, the equation x2=a has involved perfect squares. For example, if we start with x2=25, then the solutions are x=±25. Because 25 is a perfect square, we can simplify further, arriving at x=±5.

    However, the right-hand side of x2=a does not have to be a perfect square. For example, the equation x2=7 has two real solutions, x=±7. Because 7 is not a perfect square, we cannot simplify further. In the next example, we’ll use the graphing calculator to compare this algebraic solution with a graphical solution and hopefully provide some assurance that 7 and 7 are perfectly valid solutions of x2=7.

    Example 8.1.9

    Use the graphing calculator to solve x2=7. Then solve the equation algebraically and compare answers.

    Solution

    Enter each side of the equation x2=7 in the Y= menu (see Figure 8.1.1), then select 6:ZStandard to produce the image in Figure 8.1.1.

    fig 8.1.1.png
    Figure 8.1.1: Sketch each side of x2=7.

    Use the 5:intersect utility on the CALC menu to find the points of intersection. Press ENTER in response to “First curve,” press ENTER in response to “Second curve,” then use the arrow keys to move the cursor closer to the point of intersection on the left than the one on the right. Press ENTER in response to “Guess.” This will produce the point of intersection shown in image on the left in Figure 8.1.2. Repeat the procedure to find the point of intersection in the image on the right in Figure 8.1.2.

    fig 8.1.2.png
    Figure 8.1.2: Finding points of intersection.

    The approximate solutions of x2=7 are x2.645751 and x2.6457513.

    Reporting the solution on your homework: Duplicate the image in your calculator’s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves.

    • Label the horizontal and vertical axes with x and y, respectively (see Figure 8.1.3).
    • Place your WINDOW parameters at the end of each axis (see Figure 8.1.3).
    • Label each graph with its equation (see Figure 8.1.3).
    • Drop dashed vertical lines through each point of intersection. Shade and label the x-values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x2=7 (see Figure 8.1.3).
    fig 8.1.3.png
    Figure 8.1.3: Reporting your graphical solution on your homework.

    Now we solve the equation algebraically.

    x2=7x=±7

    At this point, the question is: “Do these algebraic solutions match the graphical solutions in Figure 8.1.3?” Let’s use our calculator to compare results. Locate the square root symbol on the calculator case above the x2 key in the leftmost column on the calculator keyboard. Note that we will have to use the 2nd key to access this operator. Enter (7) and press ENTER. Then enter (7) and press ENTER. The results are shown in Figure 8.1.4.

    fig 8.1.4.png
    Figure 8.1.4: Computing 7 and 7.

    Thus, 72.645751311 and 72.645751311. Note how these closely match the graphical approximations in Figure 8.1.3.

    Exercise 8.1.9

    Solve the equation x2=5 both algebraically and graphically, then compare your answers.

    Answer

    5, 5

    Ex 8.1.9.png

    Example 8.1.10

    Use the graphing calculator to solve x2=5. Then solve the equation algebraically and compare answers.

    Solution

    Enter each side of the equation x2=5 in the Y= menu (see Figure 8.1.5), then select 6:ZStandard to produce the image in Figure 8.1.5.

    fig 8.1.5.png
    Figure 8.1.5: Sketch each side of x2=5.

    Reporting the solution on your homework: Duplicate the image in your calculator’s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves.

    • Label the horizontal and vertical axes with x and y, respectively (see Figure 8.1.6).
    • Place your WINDOW parameters at the end of each axis (see Figure 8.1.6).
    • Label each graph with its equation (see Figure 8.1.6).

    Because there are no points of intersection, the graph in Figure 8.1.6 informs us that the equation x2=5 has no real solutions. Now we solve the equation algebraically. x2=5 However, you cannot square a real number and get a negative answer. Hence, the equation x2=5 has no real solutions. This agrees completely with the graph in Figure 8.1.6.

    fig 8.1.6.png
    Figure 8.1.6: Reporting your graphical solution on your homework.

    Approximating Square Roots

    Table 8.1.1: List of Squares
    n n2
    0 0
    1 1
    2 4
    3 9
    4 16
    5 25
    6 36
    7 49
    8 64
    9 81
    10 100
    11 121
    12 144
    13 169
    14 196
    15 225
    16 256
    17 289
    18 324
    19 361
    20 400
    21 441
    22 484
    23 529
    24 576
    25 625

    The squares in the “List of Squares” shown in Table 8.1.1 are called perfect squares. Each is the square of a whole number. But not all numbers are perfect squares. For example, in the case of 24, there is no whole number whose square is equal to 24. However, this does not prevent 24 from being a perfectly good number.

    We can use the “List of Squares” to find decimal approximations when the radicand is not a perfect square.

    Example 8.1.11

    Estimate 24 by guessing. Use a calculator to find a more accurate result and compare this result with your guess.

    Solution

    From the “List of Squares,” note that 24 lies between 16 and 25, so 24 will lie between 4 and 5, with 24 much closer to 5 than it is to 4.

    Example 8.1.11.png

    Let’s guess 244.8. As a check, let’s square 4.8.(4.8)2=(4.8)(4.8)=23.04 Not quite 24! Clearly, 24 must be a little bit bigger than 4.8.

    Let’s use a scientific calculator to get a better approximation. From our calculator, using the square root button, we find 244.89897948557.

    Even though this is better than our estimate of 4.8, it is still only an approximation. Our calculator was only capable of providing 11 decimal places. However, the exact decimal representation of 24 is an infinite decimal that never terminates and never establishes a pattern of repetition.

    Just for fun, here is a decimal approximation of 24 that is accurate to 1000 places, courtesy of www.wolframalpha.com

    Example 8.1.11a.png

    If you were to multiply this number by itself (square the number), you would get a number that is extremely close to 24, but it would not be exactly 24. There would still be a little discrepancy.

    Exercise 8.1.11

    Estimate: 83

    Answer

    9.1

    Reference

    1 When we say that 36 has no real square roots, we mean there are no real numbers that are square roots of 36. The reason we emphasize the word real in this situation is the fact that 36 does have two square roots that are elements of the complex numbers, a set of numbers that are usually introduced in advanced courses such as college algebra or trigonometry.


    This page titled 8.1: Introduction to Radical Notation is shared under a CC BY-NC-ND 3.0 license and was authored, remixed, and/or curated by David Arnold via source content that was edited to the style and standards of the LibreTexts platform.