Practice questions

• ${\sqrt{16}=4}$. It is a natural number, a whole number, an integer, rational, and real.
• ${-0.5=-\dfrac{1}{2}}$. It is rational and real.
• ${\sqrt{7}=2.6457\ldots}$ never ends or repeats, because ${7}$ is not a perfect square. It is irrational and real.
• ${0.\overline{6}}$ is a repeating decimal. It is rational and real. (In fact, ${0.\overline{6}=\dfrac{2}{3}}$.)

Let ${x=0.\overline{45}}$. Two digits repeat, so multiply by ${100}$:

$\begin{align*}x&=0.454545\ldots\\100x&=45.454545\ldots\end{align*}$

Subtract the first equation from the second. The repeating parts cancel:

$\begin{align*}99x&=45\\x&=\dfrac{45}{99}=\dfrac{5}{11}\end{align*}$

We divided the numerator and the denominator by their common factor ${9}$.

Since ${\sqrt{5}\approx 2.24}$, the number ${2-\sqrt{5}}$ is negative. So ${|a|=-a}$:

${|2-\sqrt{5}|=-(2-\sqrt{5})=\sqrt{5}-2}$

Use ${d(a,b)=|b-a|}$:

${d(-7,-2)=|-2-(-7)|=|5|=5}$

The endpoint ${-1}$ is not included (${<}$), so it gets a parenthesis. The endpoint ${4}$ is included (${\le}$), so it gets a bracket:

${(-1,4]}$

First square each factor of ${2x^3y^{-2}}$. Multiply the exponents:

$(2x^3y^{-2})^2=2^2x^{3\cdot 2}y^{-2\cdot 2}=4x^6y^{-4}$

Now multiply by ${x^{-1}y^5}$. Add the exponents of each variable:

$\begin{align*}&4x^6y^{-4}\cdot x^{-1}y^5\\&=4x^{6+(-1)}y^{-4+5}\\&=4x^5y\end{align*}$

Divide the numbers. Subtract the exponents of each variable, top minus bottom:

$\begin{align*}&\dfrac{12a^5b^2}{4a^{-1}b^6}\\&=3a^{5-(-1)}b^{2-6}\\&=3a^6b^{-4}\\&=\dfrac{3a^6}{b^4}\end{align*}$

Multiply the numbers, and add the exponents of ${10}$:

$\begin{align*}&(4\times 10^{-3})(6\times 10^5)\\&=24\times 10^{-3+5}\\&=24\times 10^2\end{align*}$

But ${24}$ is not between ${1}$ and ${10}$. Write ${24=2.4\times 10^1}$:

$24\times 10^2=2.4\times 10^{1+2}=2.4\times 10^3$

Take out the perfect squares: ${48=16\cdot 3}$ and ${x^3=x^2\cdot x}$.

$\begin{align*}&\sqrt{48x^3}\\&=\sqrt{16x^2}\,\sqrt{3x}\\&=4x\sqrt{3x}\end{align*}$

Multiply the numerator and the denominator by the conjugate, ${\sqrt{7}+2}$. In the denominator, use ${(a-b)(a+b)=a^2-b^2}$:

$\begin{align*}&\dfrac{5}{\sqrt{7}-2}\\&=\dfrac{5(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}\\&=\dfrac{5(\sqrt{7}+2)}{7-4}\\&=\dfrac{5(\sqrt{7}+2)}{3}\end{align*}$

The negative exponent means "one over". The denominator ${3}$ of the exponent is a cube root, and the numerator ${2}$ is a square:

$\begin{align*}&27^{-2/3}\\&=\dfrac{1}{27^{2/3}}\\&=\dfrac{1}{\left(\sqrt[3]{27}\right)^2}\\&=\dfrac{1}{3^2}=\dfrac{1}{9}\end{align*}$

Multiply every term of ${2x^2+x-4}$ by ${x}$, and then by ${-3}$:

$\begin{align*}&(x-3)(2x^2+x-4)\\&=2x^3+x^2-4x-6x^2-3x+12\end{align*}$

Combine like terms: ${x^2-6x^2=-5x^2}$ and ${-4x-3x=-7x}$. So

$\begin{align*}&(x-3)(2x^2+x-4)\\&=2x^3-5x^2-7x+12\end{align*}$

Use ${(a+b)^2=a^2+2ab+b^2}$ with ${a=2x}$ and ${b=5}$:

$\begin{align*}&(2x+5)^2\\&=(2x)^2+2(2x)(5)+5^2\\&=4x^2+20x+25\end{align*}$

Don't forget the middle term ${20x}$.

Take out the GCF, ${3x}$. What is left is a difference of squares:

$\begin{align*}&12x^3-3x\\&=3x(4x^2-1)\\&=3x(2x+1)(2x-1)\end{align*}$

We need two numbers that multiply to ${-28}$ and add to ${3}$. They are ${7}$ and ${-4}$:

${x^2+3x-28=(x+7)(x-4)}$

Use the ${ac}$ method: ${ac=3(-8)=-24}$. We need two numbers that multiply to ${-24}$ and add to ${-10}$. They are ${-12}$ and ${2}$.

Split the middle term into ${-12x+2x}$, and group:

$\begin{align*}&3x^2-10x-8\\&=3x^2-12x+2x-8\\&=3x(x-4)+2(x-4)\\&=(x-4)(3x+2)\end{align*}$

Four terms, so group them in pairs:

$\begin{align*}&x^3+5x^2-4x-20\\&=x^2(x+5)-4(x+5)\\&=(x+5)(x^2-4)\end{align*}$

The factor ${x^2-4}$ is a difference of squares. Factor it too:

$\begin{align*}&=(x+5)(x+2)(x-2)\end{align*}$

This is a sum of cubes, with ${a=2x}$ and ${b=3}$. Use ${a^3+b^3=(a+b)(a^2-ab+b^2)}$:

${8x^3+27=(2x+3)(4x^2-6x+9)}$

Factor the numerator and the denominator:

$\dfrac{x^2-x-6}{x^2-9}=\dfrac{(x-3)(x+2)}{(x-3)(x+3)}$

The denominator is ${0}$ when ${x=3}$ or ${x=-3}$. So the domain is all real numbers except ${3}$ and ${-3}$.

Cancel the common factor ${x-3}$:

${\dfrac{x^2-x-6}{x^2-9}=\dfrac{x+2}{x+3}}$,   for ${x\ne 3}$.

Multiply by the reciprocal of the second fraction, and factor everything:

$\begin{align*}&\dfrac{x^2-4}{x+1}\div\dfrac{x-2}{x^2+x}\\&=\dfrac{x^2-4}{x+1}\cdot\dfrac{x^2+x}{x-2}\\&=\dfrac{(x+2)(x-2)}{x+1}\cdot\dfrac{x(x+1)}{x-2}\end{align*}$

Cancel the common factors ${x+1}$ and ${x-2}$:

$\begin{align*}&=x(x+2)\end{align*}$

The LCD is ${(x+3)(x-1)}$. Give each fraction this denominator:

$\dfrac{2}{x+3}=\dfrac{2(x-1)}{(x+3)(x-1)}$

${\dfrac{1}{x-1}=\dfrac{x+3}{(x+3)(x-1)}}$

Subtract the numerators. Put the second one in parentheses, so both of its terms change sign:

$\begin{align*}&\dfrac{2}{x+3}-\dfrac{1}{x-1}\\&=\dfrac{2(x-1)-(x+3)}{(x+3)(x-1)}\\&=\dfrac{2x-2-x-3}{(x+3)(x-1)}\\&=\dfrac{x-5}{(x+3)(x-1)}\end{align*}$

The small fractions have denominators ${x}$ and ${x^2}$. Their LCD is ${x^2}$. Multiply the numerator and the denominator by ${x^2}$:

$\begin{align*}&\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x^2}}\cdot\dfrac{x^2}{x^2}\\&=\dfrac{x^2+x}{x^2-1}\end{align*}$

Factor, and cancel the common factor ${x+1}$:

$\begin{align*}&=\dfrac{x(x+1)}{(x+1)(x-1)}\\&=\dfrac{x}{x-1}\end{align*}$

Multiply the numerator and the denominator by the conjugate of the numerator, ${\sqrt{x}+3}$. In the numerator, ${(\sqrt{x}-3)(\sqrt{x}+3)=x-9}$:

$\begin{align*}&\dfrac{\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+3}\\&=\dfrac{x-9}{(x-9)(\sqrt{x}+3)}\end{align*}$

Cancel the common factor ${x-9}$:

$\begin{align*}&=\dfrac{1}{\sqrt{x}+3}\end{align*}$