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3.1 Roots

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Revision as of 11:27, 13 July 2008

Theory

Exercises

Content:

Quadratic roots and n:th roots
Manipulating roots

Learning outcomes:

After this section, you will have learned:

How to calculate the square root of some simple integers.
That the square root of a negative number has not been defined.
That the square root of a number denotes the positive root.
How to manipulate roots in the simplification of expressions containing roo
To recognise when the methods of manipulating roots are valid.
How to simplify expressions containing quadratic roots in the denominator.
When the n:th root of a negative number is defined (n odd).

Quadratic roots

right The well-known symbol \displaystyle \sqrt{a}, the square root of \displaystyle a, is used to describe the number that when multiplied by itself gives \displaystyle a. However, one has to be a little more precise in defining this symbol.

The equation \displaystyle x^2 = 4 has two solutions \displaystyle x = 2 and \displaystyle x = -2, since both \displaystyle 2\cdot 2 = 4 and \displaystyle (-2)\cdot(-2) = 4. It would then be logical to suppose that \displaystyle \sqrt{4} can be either \displaystyle -2 or \displaystyle 2, dvs. \displaystyle \sqrt{4}= \pm 2, but \displaystyle \sqrt{4} only denotes the positive number \displaystyle 2.

Square root \displaystyle \sqrt{a} "means the non-negative number that multiplied by itself gives \displaystyle a, i.e. the non-negative solution to the equation \displaystyle x^2 = a.

Square root of \displaystyle a can also be written as \displaystyle a^{1/2}.

It is therefore wrong to state that \displaystyle \sqrt{4}= \pm 2, but correct to state that the equation \displaystyle x^2 = 4 has the solution \displaystyle x = \pm 2.

Example 1

\displaystyle \sqrt{0}=0 \quad because \displaystyle 0^2 = 0 \cdot 0 = 0 and \displaystyle 0 is not negative.
\displaystyle \sqrt{100}=10 \quad since \displaystyle 10^2 = 10 \cdot 10 = 100 and \displaystyle 10 is a positive number.
\displaystyle \sqrt{0{,}25}=0{,}5 \quad since \displaystyle 0{,}5^2 = 0{,}5 \cdot 0{,}5 = 0{,}25 and \displaystyle 0{,}5 is positive.
\displaystyle \sqrt{2} \approx 1{,}4142 \quad since \displaystyle 1{,}4142 \cdot 1{,}4142 \approx 2 and \displaystyle 1{,}4142 is positive.
The equation \displaystyle x^2=2 has the solutions \displaystyle x=\sqrt{2} \approx 1{,}414 and \displaystyle x = -\sqrt{2} \approx -1{,}414.
\displaystyle \sqrt{-4}\quad is not defined, since there is no real number \displaystyle x that satisfies \displaystyle x^2=-4.
\displaystyle \sqrt{(-7)^2} = 7 \quad because \displaystyle \sqrt{(-7)^2} = \sqrt{(-7) \cdot (-7)} = \sqrt{49} = \sqrt{ 7 \cdot 7} = 7.

When taking square roots, it is useful to know some methods of calculation. As \displaystyle \sqrt{a} = a^{1/2} we can use the laws of exponents as "laws of roots" For example, we have

\displaystyle \sqrt{9\cdot 4}

 = (9\cdot 4)^{1/2}
 = 9^{1/2}\cdot 4^{1/2}
 = \sqrt{9}\cdot \sqrt{4}\mbox{.}

In this way we obtain the following rules for quadratic roots, which applies to all real numbers \displaystyle a, b \ge 0:

\displaystyle \begin{align*}

   \sqrt{ab} &= \sqrt{\vphantom{b}a}\cdot \sqrt{b}\\[4pt]
   \sqrt{\frac{a}{b}} &= \frac{\sqrt{a}}{\sqrt{b}}\\[4pt]
   a\sqrt{b} &= \sqrt{a^2b}
 \end{align*}

( We must however, in the above division, assume as always that b is not 0.)

Example 2

\displaystyle \sqrt{64\cdot 81} = \sqrt{64}\cdot \sqrt{81} = 8\cdot 9 = 72
\displaystyle \sqrt{\frac{9}{25}} = \frac{\sqrt{9}}{\sqrt{25}} = \frac{3}{5}
\displaystyle \sqrt{18} \cdot \sqrt{2} = \sqrt{18 \cdot 2} = \sqrt{36} = 6
\displaystyle \frac{\sqrt{75}}{\sqrt{3}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5
\displaystyle \sqrt{12} = \sqrt{ 4 \cdot 3 } = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}

Note that the above calculations assume that \displaystyle a and \displaystyle b \ge 0. If \displaystyle a and \displaystyle b are negative (< 0) then \displaystyle \sqrt{a} and \displaystyle \sqrt{b} are not defined as real numbers. It is tempting to write , for example,

\displaystyle -1 = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{ (-1) \cdot (-1) } = \sqrt{1} = 1

but something here cannot be right. The explanation is that \displaystyle \sqrt{-1} is not a real number, which means the laws of roots discussed above may not be used.

Higher order roots

The cube root of a number l \displaystyle a is defined as the number that multiplied by itself three times gives \displaystyle a, and is denoted as \displaystyle \sqrt[\scriptstyle 3]{a}.

Example 3

\displaystyle \sqrt[\scriptstyle 3]{8} = 2 \quad as \displaystyle 2 \cdot 2 \cdot 2=8.
\displaystyle \sqrt[\scriptstyle 3]{0{,}027} = 0{,}3 \quad since \displaystyle 0{,}3 \cdot 0{,}3 \cdot 0{,}3 = 0{,}027.
\displaystyle \sqrt[\scriptstyle 3]{-8} = -2 \quad because \displaystyle (-2) \cdot (-2) \cdot (-2)= -8.

Note that, unlike quadratics roots, cube roots are also defined for negative numbers.

For any positive integers \displaystyle n one can define the the n:th root of a number\displaystyle a as

if \displaystyle n is even and \displaystyle a\ge0 then \displaystyle \sqrt[\scriptstyle n]{a} is the non-negative number that when multiplied by itself \displaystyle n times gives \displaystyle a,
if \displaystyle n is odd, \displaystyle \sqrt[\scriptstyle n]{a} is the number that when multiplied by itself \displaystyle n times gives \displaystyle a.

The root \displaystyle \sqrt[\scriptstyle n]{a} can also be written as \displaystyle a^{1/n}.

Example 4

\displaystyle \sqrt[\scriptstyle 4]{625} = 5\quad since \displaystyle 5 \cdot 5 \cdot 5 \cdot 5 = 625.
\displaystyle \sqrt[\scriptstyle 5]{-243} = -3\quad because \displaystyle (-3) \cdot (-3) \cdot (-3) \cdot (-3) \cdot (-3) = -243.
\displaystyle \sqrt[\scriptstyle 6]{-17}\quad is not defined as \displaystyle 6 is even and \displaystyle -17 is a negative number.

For \displaystyle n:th roots the same rules apply as for quadratic roots if \displaystyle a, \, b \ge 0. Note that if \displaystyle n is odd these methods apply even for negative \displaystyle a and \displaystyle b, that is, for all real numbers \displaystyle a and \displaystyle b.

\displaystyle \begin{align*}

   \sqrt[\scriptstyle n]{ab}
     &= \sqrt[\scriptstyle n]{\vphantom{b}a}\cdot
          \sqrt[\scriptstyle n]{b}\\[4pt]
   \sqrt[\scriptstyle n]{\frac{a}{b}}
     &= \frac{\sqrt[\scriptstyle n]{a}}{\sqrt[\scriptstyle n]{b}}\\[4pt]
   a\,\sqrt[\scriptstyle n]{b}
     &= \sqrt[\scriptstyle n]{a^nb}
 \end{align*}

Simplification of expressions containing roots

Often, one can significantly simplify expressions containing roots by using the usual methods for roots . As is also the case when using the laws of exponents, it is desirable to reduce expressions into as"small" roots as possible. For example, it is a good idea to do the following

\displaystyle \sqrt{8}

 = \sqrt{4\cdot2}
 = \sqrt{4} \cdot \sqrt{2}
 = 2\sqrt{2}

because it helps simplification as we see here

\displaystyle \frac{\sqrt{8}}{2}

 = \frac{2 \sqrt{2}}{2}
 = \sqrt{2}\mbox{.}

By rewriting expressions containing roots in terms of "small" roots one can also sum roots of "the same kind", e.g.

\displaystyle \sqrt{8} + \sqrt{2}

 = 2\sqrt{2} + \sqrt{2}
 = (2+1)\sqrt{2}
 = 3\sqrt{2}\mbox{.}

Example 5

\displaystyle \frac{\sqrt{8}}{\sqrt{18}} = \frac{\sqrt{2 \cdot 4}}{\sqrt{2 \cdot 9}} = \frac{\sqrt{2 \cdot 2 \cdot 2}}{\sqrt{2 \cdot 3 \cdot 3}} = \frac{\sqrt{2 \cdot 2^2}}{\sqrt{2 \cdot 3^2}} = \frac{2\sqrt{2}}{3\sqrt{2}} = \frac{2}{3}
\displaystyle \frac{\sqrt{72}}{6} = \frac{\sqrt{8 \cdot 9}}{ 2 \cdot 3} = \frac{\sqrt{2 \cdot 2 \cdot 2 \cdot 3 \cdot 3}}{ 2 \cdot 3} = \frac{\sqrt{2^2 \cdot 3^2 \cdot 2}}{ 2 \cdot 3} = \frac{2 \cdot 3\sqrt{2}}{2 \cdot 3} = \sqrt{2}
\displaystyle \sqrt{45} + \sqrt{20} = \sqrt{9\cdot5} + \sqrt{4\cdot5} = \sqrt{3^2\cdot5} + \sqrt{2^2\cdot5} = 3\sqrt{5} + 2\sqrt{5}\vphantom{\bigl(}
\displaystyle \phantom{\sqrt{45} + \sqrt{20}\vphantom{\bigl(}}{} = (3+2)\sqrt{5} = 5\sqrt{5}
\displaystyle \sqrt{50} + 2\sqrt{3} -\sqrt{32} + \sqrt{27}\vphantom{\Bigl(} = \sqrt{5 \cdot 10} + 2\sqrt{3} -\sqrt{2 \cdot 16} + \sqrt{3 \cdot 9}
\displaystyle \phantom{\sqrt{50} + 2\sqrt{3} -\sqrt{32} + \sqrt{27}\vphantom{\Bigl(}}{} = \sqrt{5 \cdot 2 \cdot 5} + 2\sqrt{3} -\sqrt{2 \cdot 4 \cdot 4} + \sqrt{3 \cdot 3 \cdot 3}\vphantom{a^{b^c}}
\displaystyle \phantom{\sqrt{50} + 2\sqrt{3} -\sqrt{32} + \sqrt{27}\vphantom{\Bigl(}}{} = \sqrt{5^2 \cdot 2 } + 2\sqrt{3} -\sqrt{2^2 \cdot 2^2 \cdot 2} + \sqrt{3 \cdot 3^2}\vphantom{a^{\textstyle b^{\textstyle c}}}
\displaystyle \phantom{\sqrt{50} + 2\sqrt{3} -\sqrt{32} + \sqrt{27}\vphantom{\Bigl(}}{} = 5\sqrt{2} +2\sqrt{3} - 2 \cdot 2\sqrt{2} + 3\sqrt{3}\vphantom{a^{\textstyle b^{\textstyle c}}}
\displaystyle \phantom{\sqrt{50} + 2\sqrt{3} -\sqrt{32} + \sqrt{27}\vphantom{\Bigl(}}{} = (5-4)\sqrt{2} + (2+3)\sqrt{3}\vphantom{a^{\textstyle b^{\textstyle c}}}
\displaystyle \phantom{\sqrt{50} + 2\sqrt{3} -\sqrt{32} + \sqrt{27}\vphantom{\Bigl(}}{} = \sqrt{2} + 5\sqrt{3}\vphantom{a^{\textstyle b^{\textstyle c}}}
\displaystyle \frac{ 2\cdot\sqrt[\scriptstyle3]{3} }{ \sqrt[\scriptstyle3]{12} } = \frac{ 2\cdot\sqrt[\scriptstyle3]{3} }{ \sqrt[\scriptstyle3]{3 \cdot 4} } = \frac{ 2\cdot\sqrt[\scriptstyle3]{3} }{ \sqrt[\scriptstyle3]{3} \cdot \sqrt[\scriptstyle3]{4} } = \frac{ 2 }{ \sqrt[\scriptstyle3]{4} } = \frac{ 2 }{ \sqrt[\scriptstyle3]{2 \cdot 2} } = \frac{ 2 }{ \sqrt[\scriptstyle3]{2} \cdot \sqrt[\scriptstyle3]{2} } \cdot \displaystyle \frac{\sqrt[\scriptstyle3]{2}}{ \sqrt[\scriptstyle3]{2}} = \frac{ 2\cdot\sqrt[\scriptstyle3]{2} }{ 2 } = \sqrt[\scriptstyle3]{2}
\displaystyle (\sqrt{3} + \sqrt{2}\,)(\sqrt{3} - \sqrt{2}\,) = (\sqrt{3}\,)^2-(\sqrt{2}\,)^2 = 3-2 = 1 where we have used the difference of two squares \displaystyle (a+b)(a-b) = a^2 - b^2 with \displaystyle a=\sqrt{3} and \displaystyle b=\sqrt{2}.

Rational root expressions

When roots appear in a rational expression one often want to avoid roots in the denominator (because it is difficult with hand calculations to divide by irrational numbers). By multiplying the numerator and denominator by \displaystyle \sqrt{2} for example, one obtains

\displaystyle \frac{1}{\sqrt{2}}

 = \frac{1\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}}
 = \frac{\sqrt{2}}{2}

which usually is preferable.

In other cases, you can take advantage of the method of difference of two squares, \displaystyle (a+b)(a-b) = a^2 - b^2, where one multiplies the numerator and denominator by the denominators “conjugate” and the root sign is eliminated from the denominator by squaring, as in the following,

\displaystyle \begin{align*}

   \frac{\sqrt{3}}{\sqrt{2}+1}
     &= \frac{\sqrt{3}}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1}
      = \frac{\sqrt{3}\,(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}\\[4pt]
     &= \frac{\sqrt{3}\cdot\sqrt{2} - \sqrt{3}\cdot1}{(\sqrt{2}\,)^2 - 1^2 }
      = \frac{\sqrt{3 \cdot 2} - \sqrt{3}}{ 2 - 1 }
      = \frac{\sqrt{6} - \sqrt{3}}{ 1 }
      = \sqrt{6} - \sqrt{3}\mbox{.}
 \end{align*}

Example 6

\displaystyle \frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}} = \frac{10\sqrt{15}}{5} = 2\sqrt{15}
\displaystyle \frac{1+\sqrt{3}}{\sqrt{2}} = \frac{(1+\sqrt{3})\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} = \frac{\sqrt{2}+\sqrt{6}}{2}
\displaystyle \frac{3}{\sqrt{2}-2} = \frac{3(\sqrt{2}+2)}{(\sqrt{2}-2)(\sqrt{2}+2)} = \frac{3\sqrt{2}+6}{(\sqrt{2}\,)^2-2^2} = \frac{3\sqrt{2}+6}{2-4} = -\frac{3\sqrt{2}+6}{2}
\displaystyle \frac{\sqrt{2}}{\sqrt{6}+\sqrt{3}} = \frac{\sqrt{2}\,(\sqrt{6}-\sqrt{3}\,)}{(\sqrt{6}+\sqrt{3}\,) (\sqrt{6}-\sqrt{3}\,)} = \frac{\sqrt{2}\,\sqrt{6}-\sqrt{2}\,\sqrt{3}}{(\sqrt{6}\,)^2 -(\sqrt{3}\,)^2}\vphantom{\Biggl(}
\displaystyle \phantom{\frac{\sqrt{2}}{\sqrt{6}+\sqrt{3}}\vphantom{\Biggl(}}{} = \frac{\sqrt{2}\,\sqrt{2\cdot 3}-\sqrt{2}\,\sqrt{3}}{6-3} = \frac{2\sqrt{3}-\sqrt{2}\,\sqrt{3}}{3} = \frac{(2-\sqrt{2}\,)\sqrt{3}}{3} \vphantom{\displaystyle\frac{a^{\textstyle b^{\textstyle c}}}{b}}

Exercises

Study advice

Basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.

Keep in mind that:

Square root of a number is always non-negative (that is positive or zero)!

Rules for roots are actually special case of laws of exponents .

For example: \displaystyle \sqrt{x}=x^{1/2}.

Reviews

For those of you who want to deepen your studies or need more detailed explanations consider the following references

Learn more about square roots in the English Wikipedia

How do we know that the root of 2 is not a fraction?

Länktips

How to find the root of a number, without the help of calculators?

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