High School Algebra I Study Guide for the STAAR test

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Algebraic Expressions

You will also need to be able to simplify expressions in radical and exponential form and be able to identify equivalent expressions in those forms.

Simplifying Numerical Expressions

The radical form you will need to know here is the square root. In exponential form, we write it as:

\(x^{\frac{1}{2}}\), where the indexes of \(1\) and \(2\) in the radical are typically understood. So you will see \(\sqrt[2]{x^1}\) written as \(\sqrt{x}\).

When simplifying the square root of a number, try to factor it so one of the factors is a perfect square, then take the square root of it. The other factor stays under the square root.

For example: \(\sqrt{252}=\sqrt{36} \cdot \sqrt{7}=6\sqrt{7}\).
Note that \(7\) stays under the square root since it isn’t a perfect square.

Also, \(162^{\frac{1}{2}}=\sqrt{162}=\sqrt{81} \cdot \sqrt{2}=9\sqrt{2}\)

Leave no square root in the denominator of a fraction.
Rationalize to accomplish this:

\[\frac{3}{\sqrt{33}}\] \[\frac{3}{\sqrt{33}} \cdot \frac{\sqrt{33}}{\sqrt{33}}\] \[\frac{3 \cdot \sqrt{33}}{\sqrt{33} \cdot \sqrt{33}}\] \[\frac{3 \cdot \sqrt{33}}{33}\] \[\frac{\sqrt{33}}{11}\]

You can also combine radicals. Remember, when adding or subtracting, the square roots must be the same.

\[3\sqrt{5}+\sqrt{80}\] \[3\sqrt{5}+4\sqrt{5}\] \[7\sqrt{5}\]

Finally, when multiplying square roots, you might simplify the roots first, then multiply the numbers outside of the square roots and the numbers inside separately, and finally simplify the square root.

For example:

\[2\sqrt{32} \cdot 3\sqrt{2}\] \[2\sqrt{16\cdot 2} \cdot3\sqrt{2}\] \[2 \cdot 4\sqrt{2} \cdot 3\sqrt{2}\] \[8\sqrt{2} \cdot 3\sqrt{2}\] \[16\sqrt{4}\] \[16 \cdot 2\] \[32\]

Using the Laws of Exponents

The rules of exponents are:

\[a^m \cdot a^n = a^{m+n}\] \[\frac{a^m}{a^n}= a^{m-n}\] \[(ab)^n=a^n \cdot b^n\] \[a^{-m} = \frac{1}{a^m}\] \[a^0 = 1\]

Examples of simplifying:

Example 1:

\[(3x^2y^4) \cdot (2xy^3) \cdot (x^{10})^\frac{1}{2}\]
\[(3x^2y^4) \cdot (2x^1y^3) \cdot x^5\] \[6x^8y^7\]

Example 2:

\[\frac{a^{3-5}b^{2-\frac{2}{3}}}{3}\] \[\frac{a^{-2}b^\frac{4}{3}}{3}\] \[\frac{b^\frac{4}{3}}{3a^2}\]

Example 3:

Write an expression for the area, \(A\), of a square, if one of the sides is \(3x^4y^3\) units long.

\[A = \text{(side)}^2\] \[A=(3x^4y^3)^2\] \[A=9x^8y^6 \text{ units}^2\]

Analyzing and Evaluating

You will need to know how to write, solve, analyze, and evaluate equations and functions, including arithmetic and geometric sequences.

Identifying a Function

Functions can be represented in many ways, including verbally, graphically, in table format, or symbolically. Additionally, you will need to know how to identify and evaluate special types of functions, including sequences, literal equations, and other mathematical equations.

Evaluating a Function

Suppose you are asked to find \(f(-2)\) if \(f(x) = 3x^2-4x+1\).

Substituting \(-2\) for \(x\) into the function, we get:

\[f(-2)=3(-2)^2-4(-2)+1\] \[f(-2)=3 \cdot 4 + 8+1\] \[f(-2)=12+8+1=21\]

Find a suitable range of values for the distance, \(d\), in miles you travel in a car in time, \(t\), in hours, if \(d(t)=30t+12\), where \(12\) is the number of miles you have traveled when you began to time your distance.

\[d(0)=30 \cdot 0+12=12\] \[d(1)=30 \cdot 1 + 12 = 42\] \[d(2) =30 \cdot 2+12=72\] \[d(3) = 30 \cdot 3+12 = 102\]

So a reasonable range of values is \(\{12, 42, 72, 102\}\) miles.

The domain is \(\{0, 1, 2, 3\}\).

Identifying Terms

An arithmetic sequence is a function whose domain is the natural (counting) numbers and the range of values shares a common difference from one term to the next. The range \(\{1,4,7,10, …\}\) is arithmetic since the common difference among the terms is \(3\).

A geometric sequence is a function whose domain is the natural (counting) numbers and the range of values shares a common ratio from one term to the next. The range \(\{1,2,4,8,16, …\}\) forms a geometric sequence since there is a common ratio of \(2\) between the terms.

Notation for a sequence:

\((a_n)=\{a_1, a_2, a_3, ... , a_n\}\) where \(a_n\) represents a term in the sequence and \(n= 1,\,2,\,3, ...\)

The common difference, \(d\), for an arithmetic sequence is \(d=a_{n+1}-a_n\) and for a geometric sequence, the common ratio, \(r\) is \(r=\dfrac{a_{n+1}}{a_n}\).

What type of sequence is formed by \(a_1=3 \;\text{and} \;a_n=4a_{n-1}\) for \(n \ge 2\)?

\[a_1=3,\; a_2=4 \cdot 3=12, \;a_3=4 \cdot 12=48, \;a_4=4 \cdot48=192\]

So the range of the terms \(\{3,\;12,\;48,\;192...\}\) form a geometric sequence since the common ratio is \(4\).

The sequence defined by \(a_1=5\) and \(a_n=3+a_{n-1}\) for \(n\ge2\) is an arithmetic sequence since the terms are \(a_1=5, \;a_2=3+5=8, \;a_3=3+8=11, \;a_4=3+11=14\) or the set \(\{5, 8, 11, 14...\}\) and have a common difference of \(3\).

Writing a Formula

We saw earlier that \(d\) and \(r\) represent the common difference and common ratio for arithmetic and geometric sequences. The formulas for the two sequences are:

  • arithmetic sequence: \(a_n=a_1+d(n-1)\), where \(a_1\) is the first term in the sequence, \(d\) = common difference among the terms, and \(n = 1, 2, 3,\) …

  • geometric sequence: \(a_n=a_1\cdot r^{n-1}\), where \(a_1\) is the first term in the sequence, \(r\) = common ratio among the terms, and \(n = 1, 2, 3,\) …

Example 1:

Write an explicit formula for the sequence \({4, 10, 16, 22, …}\).

The sequence is arithmetic; there is a common difference of \(d=6\) since \(10-4,\; 16-10, \;\) and \(\;22-16\;\) all equal \(6\). Now with \(a_1=4\), we have \(a_n=4+6(n-1)\) or \(6n-2\).

Example 2:

The half-life of an isotope is such that, each year, only one-half of the isotope’s original value remains. The sequence \(\{64, 32, 16, 4, …\}\) is the amount of the isotope left after the first four years. Write an explicit formula for the sequence.

The sequence is geometric; there is a common ratio of \(r=\frac{1}{2}\) since \(\frac{32}{64}=\frac{1}{2}\). Now with \(a_1=64\), we have \(a_n=64 \cdot \left(\frac{1}{2}\right)^{n-1}\) or \(128\cdot \left(\frac{1}{2}\right)^n\).


Literal equations involve two or more variables. You will have to know how to manipulate a literal equation that is solved for one variable for another one in the equation. There will be equations that may not represent any mathematical significance and equations such as the volume of a right cylinder, \(V= \pi r^2h\), that must be solved for \(r\) or \(h\). Use PEMDAS to help you do so.

For example:

Write a formula the perimeter of a rectangle, \(P=2l+2w\), in terms of its length, \(l\).

\[P=2l+2w\] \[P-2w=2l+2w-2w\] \[\frac{P-2w}{2} = \frac{2l}{2}\] \[\frac{P}{2} - \frac{2w}{2} = l\] \[\frac{P}{2} -w=l\]

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