Radicals and complex numbers
nth roots
Square roots
If you multiply a number twice, you get another number that is called square. But if you are given a number, and you find a number that you multiplied twice gives the given number, then that number is called square root of the given number. So, square root is a reverse operation of squaring.For example, $5^2=25$, here $25$ is the square of the number $5$.
And for the number $25$, $5$ and $-5$ are the square roots as we get $25$ by multiplying $5$ or $-5$, twice.
Example:Find the square root of $81$.
Solution:
We can write,$81=9^2$
and$81=(-9)^2$
So, the square root of $81$ is $9$ and $-9$
Example:Find the square root of $-36$.
Solution:
We cannot find a real number that we multiply twice gives the number $-36$. Therefore, the square root of $-36$ is not real. So, a negative number has no real square root.The radical sign
We saw that a positive number has a positive and a negative square root. For the positive square root we use a sign, $\sqrt{\hphantom{00}}\:$, called radical sign.i.e., for a positive number $a$,$\:\sqrt{a}$ is the square root of $a$, that is positive.
Examples:
$\sqrt{49}=7$
$\sqrt{36}=6$
And$\sqrt{0}=0$
Since the square root of a negative number is not a real number,
$\sqrt{-9}=$ not real.
You can have a minus sign outside the radical sign, but not inside of it.
i.e., $-\sqrt{25}=-5$
Simplifying square roots.
Since the square root of a number is just another number, you need to treat square root as a number when simplifying. Following relations hold for square roots.$\sqrt{ab}=\sqrt{a}\sqrt{b}$
$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$
Example: Simplify $\sqrt{\dfrac{64}{25}}$.
Solution:
$\begin{align*} \sqrt{\dfrac{64}{25}}&=\dfrac{\sqrt{64}}{\sqrt{25}}\\ &=\boxed{\dfrac{8}{5}}\\ \end{align*}$nth root
Like square root, we can have third root, fourth root, fifth root and so on. Third root of a number is a number that when multiplied three times is equal to the original number. Third root is also called cube root. Fourth root of a number is a number when multiplied 4 times gives the original number.If $a$ is a square root of a number, $b$, then $a^2=b$.
If $a$ is a cube root of $b$, then $a^3=b$.
If $a$ is a fourth root of $b$, then $a^4=b$, and so on.
In general, we call these roots, nth roots as if we take $n=2$, it is a square root, $n=3$ is cube root, $n=4$, is fourth root and so on.
So, if $a$ is the $n$th root of a number $b$, then$a^n=b$.
For a square root we use the radical sign, $\sqrt{\hphantom{0}}\:$, for an nth root we use the radical sign, $\sqrt[n]{\hphantom{0}}\:$, that is for the cube root, we use $\sqrt[3]{\hphantom{0}}\:$, for the fourth root, we use $\sqrt[4]{\hphantom{0}}\:$, and so on. The number $n$ is the index of the root.
Examples:
$\sqrt[3]{27}=3$
$\sqrt[3]{-64}=-4$
$\sqrt[5]{32}=2$
Not that the $n$th root with $n$ even, of a positive number is always positive, and that of a negative number is not real. But the $n$th root for $n$ odd is positive for a positive number and negative for a negative number.For $n$ odd, $\sqrt[n]{a^n}=a$
For $n$ even, $\sqrt[n]{a^n}=|a|$.
Example 1: Evaluate, $\sqrt{(-11)^2}$
Solution:
$\begin{align*} \sqrt{(-11)^2}&=|-11|\\ &=11\\ \end{align*}$Example 2: Evaluate, $\sqrt{(-3)^4}$
Solution:
$\begin{align*} \sqrt{(-3)^4}&=\sqrt{((-3)^2)^2}\\ &=|(-3)^2|\\ &=9\\ \end{align*}$Example 3: Evaluate, $\sqrt{u^6}$, where $u$ represents a positive real number.
Solution:
$\begin{align*} \sqrt{u^6}&=\sqrt{(u^3)^2}\\ &=|u^3|\\ \end{align*}$Since $u$ is positive, we can ignore the absolute value symbol, and we get
$\sqrt{u^6}=u^3$.
Example 4: Evaluate, $\sqrt{y^{30}}$, where $y$ represents any real number.
Solution:
$\begin{align*} \sqrt{y^{30}}&=\sqrt{(y^{15})2}\\ &=|y^{15}|\\ \end{align*}$It is given, $y$ is any real number, so, $y$ raised to an odd integer is positive if $y$ is positive and is negative if $y$ is negative. Since the square root is always positive, we need to put the absolute value symbol to make sure the answer is positive always. Therefore,
$\sqrt{y^{30}}=|y^{15}|$.
Simplifying nth roots
Example 1: Simplify, $\sqrt[3]{64}$
Solution:
$\sqrt[3]{64}=\sqrt[3]{4^3}$
$\hphantom{0000}=4$
Example 2: Simplify, $\sqrt[4]{8}$
Solution:
$\sqrt[3]{64}=\sqrt[4]{2^4}$
$\hphantom{0000}=2$
Example 3: Simplify, $\sqrt[3]{\dfrac{64}{27}}$
Solution:
$\begin{align*} \sqrt[3]{\dfrac{64}{27}} &= \dfrac{\sqrt[3]{64}}{\sqrt[3]{27}}\\ &= \dfrac{\sqrt[3]{4^3}}{\sqrt[3]{3^3}}\\ &=\dfrac{4}{3}\\ \end{align*}$Example 4: Simplify, $\sqrt[5]{243x^{10}}$
Solution:
$\begin{align*} \sqrt[5]{243x^{10}} &= \sqrt[5]{243}\,\sqrt[5]{x^{10}}\\ &= \sqrt[5]{3^5}\,\sqrt[5]{(x^2)^5}\\ &= 3x^2\\ \end{align*}$Example 5: Simplify, $\sqrt[8]{(13-5u)^8}$, where $u$ represents any real number.
Solution:
Since the index of the radical sign is even, the root should be positive. Therefore,
$\sqrt[8]{(13-5u)^8}=|13-5u|$.Radical function
An $n$th of an expression with a radical sign is called radical. A function that is equal to a radical is called a radical function.Examples:
$f(x)=\sqrt{x}$
$g(t)=\sqrt{3t-1}$
$h(x)=\sqrt[3]{2x+5}$
Evaluating a radical function.
Evaluating a radical function is same as evaluating any other function.Example 1: Given, $f(x)=\sqrt[3]{2x}-5$. Find $f(-4)$ and $f(32)$.
Solution:
To find $f(-4)$, substitute $x=4$ in the function, $f$:$f(-4)=\sqrt[3]{2\cdot (-4)}-5$
$\phantom{00000}=\sqrt[3]{-8}-5$
$\phantom{00000}=-2-5$
$\phantom{00000}=-7$
To find $f(32)$, substitute $x=32$ in $f(x)$:
$f(32)=\sqrt[3]{2\cdot 32}-5$
$\phantom{00000}=\sqrt[3]{64}-5$
$\phantom{00000}=4-5$
$\phantom{00000}=-1$
Domain of radical functions
Domain of a radical function, $f(x)$ is all the possible $x$ values that do not produce a not real number.If the index $n$ of a radical function is odd, then the domain is all the real numbers. But if $n$ is even, then the domain is all the real numbers that produce positive values under the radical sign.
To find the domain of an even index radical function, set the expression inside the radical sign greater than or equal to zero, and solve the resulting inequality for the variable. The solution is the domain of the radical function.
Example:1 Find the domain of the function, $f(x)=\sqrt[3]{3x-11}$
Solution:
The index in the radical sign of the function is $3$, that is an odd number, therefore, the domain of the function $f(x)$ is all the real numbers.In interval notation, the domain of $f$ is therefore, $(-\infty,\infty)$.
Example:2 Find the domain of the function, $h(x)=\sqrt{8x+4}$
Solution:
The index of the radical sign is 2 (i.e., square root), it is even, so we cannot have a negative number under the radical sign (under the square root). Therefore, to find the domain, set the expression under the radical sign greater than or equal to zero and solve for $x$. The solution will be the domain of the function $h$.Setting the expression under the radical sign greater than or equal to zero:
$8x+4\ge 0$
Solving for $x$,
$x\gt -\dfrac{1}{2}$
So, the domain is all the $x$ values that are greter than or equal to $-\dfrac{1}{2}$.
In interval notation, the domain of $h$ is $[-\dfrac{1}{2},\infty)$.
Graphing a radical function root function.
To graph a radical function, say $f(x)$, you need to exclude the $x$ values that are not in the domain. Pick the values of $x$ in the domain and find the corresponding function values, i.e., the $y$ values. Then plot the ordered pairs.Rational exponents
If $b$ is an $n$th root of $a$, then,$b=\sqrt[n]{a}$
and$a=b^n$
Now if you take the power $1/n$ on both sides, we get
$\begin{align*} a^{1/n}&=(b^n)^{1/n}\\ &=b\\ &=\sqrt[n]{a}\\ \end{align*}$i.e., $\boxed{\sqrt[n]{a}=a^{1/n}}$
The left hand side is in the radical form and the right side is in the rational exponential form. We can switch between exponential and radical form by using this relation.Further, we can write
$\begin{align*} a^{m/n}&=(a^m)^{1/n}\\ &=\sqrt[n]{a^m}\\ \end{align*}$and
$\begin{align*} a^{m/n}&=(a^{1/n})^m\\ &=(\sqrt[n]{a})^m\\ \end{align*}$The above two relations can be combined and we can write,
$a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$
Example 1: Write the following radical expressions as exponential expressions,(a) $\sqrt{13}$; (b) $\sqrt[5]{7}$
Solution:
(a). $\sqrt{13}=13^{1/2}$
(b). $\sqrt[5]{7}=7^{1/5}$
Example 2: Write the radical expression as exponential expression, $\sqrt[7]{3^4}$
Solution:
$\begin{align*} \sqrt[7]{3^4}&=(3^4)^{1/7}\\ &=3^{4\cdot 1/7}\\ &=3^{4/7}\\ \end{align*}$Simplifying rational exponents
Example 1: Simplify (write your answer without exponents), $\left(\dfrac{1}{9}\right)^{3/2}$.
Solution:
$\begin{align*} \left(\dfrac{1}{9}\right)^{3/2}&=\left(\dfrac{1}{9}\right)^{3/2}\\ &=\dfrac{1^{3/2}}{9^{3/2}}\\ &=\dfrac{1}{(3^2)^{3/2}}\\ &=\dfrac{1}{27}\\ \end{align*}$Example 2: Simplify (write your answer without exponents), $32^{-\tfrac{4}{5}}$.
Solution:
First make the exponent positive by taking the reciprocal of the base.
$\begin{align*} 32^{-\tfrac{4}{5}} &= \left(\dfrac{1}{32}\right)^{\tfrac{4}{5}}\\ &=\dfrac{1}{32^{\tfrac{4}{5}}}\\ &=\dfrac{1}{(2^5)^{\tfrac{4}{5}}}\\ &=\dfrac{1}{2^4}\\ &=\dfrac{1}{16} \end{align*}$Rational exponents: product rule
You already learned that if $m$ and $n$ are integers, then$a^m\cdot a^n=a^{m+n}$
The same applies if $m$ and $n$ are not integers but fractions.Rational exponents: quotient rule
You already learned that if $m$ and $n$ are integers, then$\dfrac{a^m}{a^n}=a^{m-n}$
The same applies if $m$ and $n$ are not integers but fractions.Example: 1 Simplify $x^{7/3}\cdot x^{2/5}$. Write your answer with positive exponents.
Solution:
$\begin{align*} x^{7/3}\cdot x^{2/5}&=x^{7/3+2/5}\\ &=x^{41/15}\\ \end{align*}$Example 2: Simplify $\dfrac{u^{11/4}}{u^{3/5}}.$ Write your answer with positive exponents.
Solution:
$\begin{align*} \dfrac{u^{11/4}}{u^{3/5}}&=u^{11/4}-u^{3/5}\\ &=u^{11/4-3/5}\\ &=u^{43/20}\\ \end{align*}$Example 3: Simplify $\dfrac{x^{-4/3}\:x^{2/5}}{x^{7/6}}$. Write your answer with positive exponents.
Solution:
$ \begin{align*} \dfrac{x^{-4/3}\:x^{2/5}}{x^{7/6}} &=\dfrac{x^{-4/3+2/5}}{x^{7/6}}\\ &=\dfrac{x^{-14/15}}{x^{7/6}}\\ &=\dfrac{1}{x^{7/6+14/15}}\\ &=\dfrac{1}{x^{7/6+14/15}}\\ &=\dfrac{1}{x^{21/10}}\\ \end{align*}$Rational exponents: power of a power rule
If $m$ and $n$ are positive integers, then you know that$(a^m)^n=a^{mn}$
This formula is valid not only for $m$ and $n$ as integers, but is valid for fractions too.Example: 1 Simplify $\bigg(x^\tfrac{5}{8}\bigg)^2$. Assume $x$ represents a positive real number.
Solution:
$\begin{align*} \left (x^\tfrac{5}{8}\right )^2 &= \left(x^\tfrac{5}{8}\right)^2\\ &= x^{\tfrac{5}{8}\cdot 2}\\ &= x^\tfrac{5}{4}\\ \end{align*}$Example: 2 Simplify $\bigg(y^\tfrac{4}{3}\bigg)^{\tfrac{3}{2}}$. Assume $y$ represents a positive real number.
Solution:
$\begin{align*} \bigg(y^\tfrac{4}{3}\bigg)^{\tfrac{3}{2}} &= y^{\tfrac{4}{3}\cdot \tfrac{3}{2}}\\ &= y^2\\ \end{align*}$Rational exponents: power of product and power of quotient rules
$(ab)^m=a^m\:b^m$
$\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$
Example:Simplify $\left(x^{\tfrac{4}{7}}y^{\tfrac{2}{5}}\right)^{-\tfrac{2}{3}}$. Assume the variables represent positive real numbers.
Solution:
$\begin{align*} \left(x^{\tfrac{4}{7}}y^{\tfrac{2}{5}}\right)^{-\tfrac{2}{3}} &= \dfrac{1}{\left(x^{\tfrac{4}{7}}y^{\tfrac{2}{5}}\right)^{\tfrac{2}{3}}} \\ &=\dfrac{1}{x^{\tfrac{4}{7}\cdot \tfrac{2}{3}}y^{\tfrac{2}{5}\cdot \tfrac{2}{3}}} \\ &=\dfrac{1}{x^{\tfrac{8}{21}}y^{\tfrac{4}{15}}} \\ \end{align*}$Multiplication property of radicals.
$\sqrt[n]{ab}=\sqrt[n]{a}\:\sqrt[n]{b}$
In the following examples, you will see how to simplify radicals using the multiplication property.Example:1 Simplify $\sqrt{50y^7z^3}$
Solution:
$\begin{align*} \sqrt{50y^7z^3}&= \sqrt{2\cdot 25y^6yz^2z}\\ &=\sqrt{25}\:\sqrt{y^6}\:\sqrt{z^2}\sqrt{2yz}\\ &=5y^3z\sqrt{2yz}\\ \end{align*}$Example:2 Simplify $\sqrt[3]{192x^{10}y^3}$
Solution:
$\begin{align*} \sqrt[3]{192x^10y^3}&=\sqrt[3]{192x^{10}y^3}\\ &=\sqrt[3]{3\cdot 64 x^9 x y^3}\\ &=\sqrt[3]{64}\:\sqrt[3]{x^9}\:\sqrt[3]{y^3}\:\sqrt[3]{3x}\\ &=4x^3y\:\sqrt[3]{3x}\\ \end{align*}$Example:3 Simplify $\sqrt[4]{112u^{15}w^9}$
Solution:
$\begin{align*} \sqrt[4]{112u^{17}w^9} &= \sqrt[4]{7\cdot 16 u^{12}u^3w^8w}\\ &= \sqrt[4]{16}\:\sqrt[4]{u^{12}}\:\sqrt[4]{w^8}\sqrt[4]{7u^3w}\\ &= 2u^3w^2\:\sqrt[4]{7u^3w}\\ \end{align*}$Addition and subtraction of radicals
We can combine like radicals as same as combining like terms. Like radicals are radicals with same index and same radicand. Radicand is the expression under the radical sign.Examples:
$4\sqrt{11}+3\sqrt{11}=7\sqrt{11}$
$2x\sqrt[5]{yz}+3x\sqrt[5]{yz}=5x\sqrt[5]{yz}$
Example 1: Simplify,$3\sqrt{20}+6\sqrt{45}-8\sqrt{5}$
Solution:
$\begin{align*} 3\sqrt{20}+6\sqrt{45}-8\sqrt{5}&=3\sqrt{4\cdot 5}+6\sqrt{9\cdot 5}-8\sqrt{5}\\ &=3\cdot 2\sqrt{5}+6\cdot 3 \sqrt{5}-8\sqrt{5}\\ &=6\sqrt{5}+18 \sqrt{5}-8\sqrt{5}\\ &=16\sqrt{5}\\ \end{align*}$Example 2:Simplify $x^3\sqrt{75y^3}-2y\sqrt{108x^6y}$
Solution:
$\begin{align*} x^3\sqrt{75y^3}-2y\sqrt{108x^6y}&=x^3\sqrt{3\cdot 25 y^2y}-2y\sqrt{3\cdot 36 x^6y}\\ &=x^3\cdot 5y\sqrt{3y}-2y\cdot 6x^3\sqrt{3y}\\ &=5x^3y\sqrt{3y}-12x^3y\sqrt{3y}\\ &=-7x^3y\sqrt{3y}\\ \end{align*}$Example 3:Simplify $2y\sqrt[3]{54u^3y^8}+8u\sqrt[3]{250y^{11}}$
Solution:
$\begin{align*} 2y\sqrt[3]{54u^3y^8}+8u\sqrt[3]{250y^{11}}&=2y\sqrt[3]{2\cdot 27u^3y^6y^2}+8u\sqrt[3]{2\cdot 125y^{9}y^2}\\ &=2y\cdot 3uy^2\sqrt[3]{2y^2}+8u\cdot 5y^3\sqrt[3]{2y^2}\\ &=6uy^3\sqrt[3]{2y^2}+40uy^3\sqrt[3]{2y^2}\\ &=46uy^3\sqrt[3]{2y^2}\\ \end{align*}$Multiplication of radicals
We can write,$\sqrt[n]{ab}=\sqrt[n]{a}\:\sqrt[n]{b}$
Proof:
$\begin{align*} \sqrt[n]{ab}&= (ab)^{1/n}\\ &=a^{1/n}b^{1/n}\\ &=\sqrt[n]{a}\:\sqrt[n]{b}\\ \end{align*}$Example: 1 Multiply and simplify $\sqrt{15}\:\sqrt{3}$
Solution:
$\begin{align*} \sqrt{15}\:\sqrt{3}&=\sqrt{15\cdot3}\\ &=\sqrt{15\cdot 3}\\ &=\sqrt{3\cdot 5\cdot 3}\\ &=3\sqrt{5}\\ \end{align*}$Example: 2 Multiply and simplify $\sqrt{24u^3y^6}\:\sqrt{32u^5y^7}$
Solution:
$\begin{align*} \sqrt{24u^3y^6}\:\sqrt{32u^5y^7}&=\sqrt{2^3\cdot 3u^3y^6}\:\sqrt{2^5u^5y^7}\\ &=\sqrt{2^3\cdot 3\cdot 2^5u^8y^{13}}\\ &=\sqrt{2^8\cdot 3u^8y^{12}y}\\ &=2^4u^4y^6\sqrt{3y}\\ &=16u^4y^6\sqrt{3y}\\ \end{align*}$Example: 3 Multiply and simplify $\sqrt[4]{18xy^3}\:\sqrt[4]{27x^5y}$
Solution:
$\begin{align*} \sqrt[4]{18xy^3}\:\sqrt[4]{27x^5y}&=\sqrt[4]{18\cdot 27 x^6y^4}\\ &=\sqrt[4]{2\cdot 3^2\cdot 3^3 x^6y^4}\\ &=\sqrt[4]{2\cdot 3^4\cdot 3 x^4x^2y^4}\\ &=3xy\sqrt[4]{2\cdot 3 x^2}\\ &=3xy\sqrt[4]{6 x^2}\\ \end{align*}$Example: 4 Multiply, $11\sqrt{7}(\sqrt{7}-2\sqrt{6})$
Solution:
$\begin{align*} 11\sqrt{7}(\sqrt{7}-2\sqrt{6})&=11\sqrt{7}\cdot \sqrt{7}-11\sqrt{7}\cdot 2\sqrt{6}\\ &=11\cdot 7-22\sqrt{7\cdot 6}\\ &=77-22\sqrt{42}\\ \end{align*}$Example: 5 Multiply, $(5\sqrt{3}+4\sqrt{11})(6\sqrt{15}-3\sqrt{6})$
Solution:
$\begin{align*} (5\sqrt{3}+4\sqrt{11})(6\sqrt{15}-3\sqrt{6})&=5\sqrt{3}\cdot 6\sqrt{15}-5\sqrt{3}\cdot 3\sqrt{6}+4\sqrt{11}\cdot 6\sqrt{15}- 4\sqrt{11}\cdot 3\sqrt{6}\\ &=30\sqrt{3\cdot 15}-15\sqrt{3\cdot 6}+24\sqrt{11\cdot 15}- 12\sqrt{11\cdot 6}\\ &=90\sqrt{5}-15\sqrt{3\cdot 6}+24\sqrt{11\cdot 15}- 12\sqrt{11\cdot 6}\\ &=90\sqrt{5}-45\sqrt{2}+24\sqrt{165}- 12\sqrt{66}\\ \end{align*}$Example: 6 Multiply, $(3\,\sqrt{a}+\sqrt{15})(3\, \sqrt{a}-\sqrt{15})$
Solution:
The expression is in the form $(a+b)(a-b)$, so we use the formula, $(a+b)(a-b)=a^2-b^2$
$\begin{align*} (3\sqrt{a}+\sqrt{15})(3\sqrt{a}-\sqrt{15})&=(3\sqrt{a})^2-(\sqrt{15})^2\\ &=9a-15\\ \end{align*}$Example: 7 Multiply, $(5\sqrt{y}+\sqrt{6})^2$
Solution:
Use the formula, $(a+b)^2=a^2+2ab+b^2$, or use $(a+b)^2=(a+b)(a-b)$
$\begin{align*} (5\sqrt{y}+\sqrt{6})^2&=(5\sqrt{y}+\sqrt{6})\cdot (5\sqrt{y}+\sqrt{6})\\ &=5\sqrt{y} \cdot 5\sqrt{y}+5\sqrt{y} \cdot \sqrt{6}+\sqrt{6}\cdot 5\sqrt{y}+\sqrt{6}\cdot \sqrt{6}\\ &=25y+5\sqrt{6y}+5\sqrt{6y}+6\\ &=25y+10\sqrt{6y}+6\\ \end{align*}$Example: 9 Simplify, $\sqrt[8]{x^5}\:\sqrt[3]{x}$
Solution:
$\begin{align*} \sqrt[8]{x^5}\:\sqrt[3]{x}&=x^{5/8}\cdot x^{1/3}\\ &=x^{5/8+1/3}\\ &=x^{23/24}\\ &=\sqrt[24]{x^{23}}\\ \end{align*}$Example: 10 Simplify, $\dfrac{\sqrt[7]{u^6}}{\sqrt[5]{u^2}}$
Solution:
$\begin{align*} \dfrac{\sqrt[7]{u^6}}{\sqrt[5]{u^2}} &= \dfrac{u^{6/7}}{u^{2/5}}\\ &= u^{6/7-2/5}\\ &=u^{16/35}\\ &=\sqrt[35]{u^{16}}\\ \end{align*}$Division of radicals
Rationalizing the denominator
Example: 1 Rationalize the denominator $\dfrac{5x}{\sqrt{7}}$
Solution:
$\dfrac{5x}{\sqrt{7}}$$=\dfrac{5x}{\sqrt{7}}$
multiplying $\sqrt{7}$ in the numerator and the denominator,
$=\dfrac{5x\cdot \sqrt{7}}{\sqrt{7}\cdot \sqrt{7}}$
$=\dfrac{5x\sqrt{7}}{7}$
Example: 2 Rationalize the denominator $\sqrt{\dfrac{3y}{11}}$
Solution:
$\sqrt{\dfrac{3y}{11}}$ $=\dfrac{\sqrt{3y}}{\sqrt{11}}$
Multiplying $\sqrt{11}$ in the numerator and the denominator,
$=\dfrac{\sqrt{3y}\cdot \sqrt{11}}{\sqrt{11}\cdot \sqrt{11}}$
$=\dfrac{\sqrt{33y}}{11}$
Example: 3 Rationalize the denominator $\dfrac{4\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
Solution:
$\dfrac{4\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
Multiplying the conjugate of $\sqrt{7}+\sqrt{5}$ in the numerator and the denominator,
$=(\dfrac{4\sqrt{7}-\sqrt{5})\cdot (\sqrt{7}-\sqrt{5}) }{(\sqrt{7}+\sqrt{5})\cdot (\sqrt{7}-\sqrt{5})}$
$=\dfrac{4\sqrt{7}\cdot \sqrt{7}-4\sqrt{7}\cdot \sqrt{5}-\sqrt{5}\cdot \sqrt{7}+\sqrt{5}\cdot\sqrt{5}}{7-5}$
$=\dfrac{4\cdot 7-4\sqrt{35}-\sqrt{35}+5}{7-5}$
$=\dfrac{33-5\sqrt{35}}{2}$
Example: 4 Rationalize the denominator $\dfrac{8\sqrt{6}+4}{\sqrt{6}-2}$
Solution:
$\dfrac{8\sqrt{6}+4}{\sqrt{6}-2}$
Multiplying the conjugate of $\sqrt{6}-2$ in the numerator and the denominator,
$=\dfrac{(8\sqrt{6}+4)\cdot (\sqrt{6}+2)}{(\sqrt{6}-2)\cdot (\sqrt{6}+2)}$
$=\dfrac{8\sqrt{6}\cdot \sqrt{6}+8\sqrt{6}\cdot 2+4 \cdot \sqrt{6}+ 4 \cdot 2}{6-2^2}$
$=\dfrac{8\cdot 6+16\sqrt{6}+4\sqrt{6}+ 8}{6-4}$
$=\dfrac{48+20\sqrt{6}+ 8}{6-4}$
$=\dfrac{56+20\sqrt{6}}{2}$
$=28+10\sqrt{6}$
Example: 5 Rationalize the denominator $\dfrac{17}{2\sqrt{y}-5}$
Solution:
$\dfrac{17}{2\sqrt{y}-5}$
Multiplying the conjugate of $2\sqrt{y}-5$ in the numerator and the denominator,
$=\dfrac{17\cdot (2\sqrt{y}+5) }{(2\sqrt{y}-5)(2\sqrt{y}+5)}$
$=\dfrac{17\cdot 2\sqrt{y}+17\cdot 5}{(2\sqrt{y})^2-5^2}$
$=\dfrac{34\sqrt{y}+85 }{4y-25}$
Example: 6 Rationalize the denominator $2\:\sqrt[3]{\dfrac{13}{7}}$
Solution:
$2\:\sqrt[3]{\dfrac{13}{7}}$
Multiplying two $7$'s in the numerator and the denominator,
$=2\:\sqrt[3]{\dfrac{13\cdot 7 \cdot 7}{7\cdot 7\cdot 7}}$
$=\dfrac{2\:\sqrt[3]{13\cdot 7 \cdot 7}}{\sqrt[3]{7\cdot 7\cdot 7}}$
$=\dfrac{2\:\sqrt[3]{13\cdot 7 \cdot 7}}{7}$
$=\dfrac{2\:\sqrt[3]{637}}{7}$
Radical equations
Example 1:Solve $\sqrt{3x-8}=\sqrt{7x-11}$
Solution:
There is a square root on the left hand side and on the right hand side, so we can remove the square root by squaring on both sides,
$\begin{align*} (\sqrt{3x-8})^2&=(\sqrt{7x-11})^2\\ 3x-8 &=7x-11\\ -8+11 &=7x-3x\\ 3 &=4x\\ x &=\dfrac{3}{4}\\ \end{align*}$Example 2:Solve $\sqrt{4u-3}+2=u$
Solution:
Keep the radical on one side and move everything else to the other side,
$\sqrt{4u-3}=u-2$
Square on both sides, that will remove the radical sign,
$4u-3=(u-2)^2$
$4u-3=u^2-4u+4$
$0=u^2-8u+7$
or$u^2-8u+7=0$
Factoring,
$(u-7)(u-1)=0$
or$u=7$ or $u=1$
There are two roots, you need to put these values in the original equation and check.
Substituting $u=7$ in the original equation,
$\sqrt{4\cdot 7-3}+2=7$
$\sqrt{25}+2=7$
$5+2=7$
$7=7$
This statement is true, therefore, $7$ is a solution of the given equation.Next, substitute the other value, $x=1$.
$\sqrt{4\cdot 1-3}+2=1$
$\sqrt{4-3}+2=1$
$\sqrt{1}+2=1$
$3=1$
This statement is not true. Therefore, $u=1$ is not a solution of the given equation.
So, the solution of the equation is $u=7$.
Example 3:Solve $\sqrt[3]{5y-9}\:-5=-3$
Solution:
$\sqrt[3]{5y-9}\:-5=-3$
Keep the radical on one side and move everything to the other side,
$\sqrt[3]{5y-9}=5-3$
$\sqrt[3]{5y-9}=2$
Take the exponent $3$ on both sides,
$(\sqrt[3]{5y-9})^3=2^3$
$5y-9=8$
Solving for $y$,
$y=\dfrac{17}{5}$
Example 4:Solve $(13y-10)^{1/4}+7=2$
Solution:
$(13y-10)^{1/4}+7=2$
Keep the rational exponent on one side and move everything to the other side,
$(13y-10)^{1/4}=2-7$
$(13y-10)^{1/4}=-7$
A fourth root is on the left hand side and a negative number on the right hand side. Since a fourth root cannot be negative, the given equation has no solution.
Example 5:Solve $(3w-10)^3=125$
Solution:
$(3w-10)^3=125$
Taking cube root on both sides,
$3w-10=5$
Solving for $w$,
$\boxed{w=5}$
Complex numbers
You saw that square root of a negative number is not a real number. But we call that number, imaginary number.An imaginary number is written in the form, $b\,i$, where $b$ is a real number and $i=\sqrt{-1}$, is an imaginary number.