• Special Offer- SAT Math Paperback Bundle
    August 15, 2017

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    Special Offer – SAT Math Paperback Bundle

    Many of our customers have been asking if it is possible to purchase several of our prep books in paperback for a discounted price. To meet this demand, we are pleased to introduce a special promotional package that includes seven of our SAT math prep books for the price of just $147. Shipping and all other fees are included in this price. By purchasing all seven of Dr. Steve Warner’s SAT math books as a bundle, you will save 30%, as compared to buying the books individually. The books included in this special promotion are:

    • 28 New SAT Math Lessons – Beginner Course
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    • 28 New SAT Math Lessons – Advanced Course
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    • New SAT Math Problems
    • 320 SAT Math Subject Test Problems – Level 1
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    Note that if you were to buy each of these books separately the total price would be $209.93, plus taxes and possibly shipping. By purchasing the books here you will get all seven books in paperback for just $147. This is a 30% savings.

    Due to the bulk discount on this special offer, all sales are final (no refunds).
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    Get 800 SAT Math Paperback Bundle

    This special offer provides the highest quality test prep materials at an affordable price. For one low price of $147 you will receive seven books containing a full arsenal of strategies, problems, and explanations. The author, Dr. Steve Warner, typically charges $500 per hour for private tutoring. For less than a third of the price of a single session, you will receive all 7 of his SAT math prep books for the current SAT and SAT math subject tests, which include all of the same strategies he teaches in private tutoring sessions. Invest in your future today and choose your college with Get 800’s prep book bundle.

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  • The Discriminant of a Quadratic Equation – Part 1
    June 28, 2017

    discriminant graphs
    The Discriminant of a Quadratic Equation
    Part 1

    Recall that the quadratic equation ax2bx + c = 0 can be solved using the quadratic formula:

    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    For more information on the quadratic formula, see the following posts:

    Quadratic Formula – Part 1   Quadratic Formula – Part 2

    Before we go on, you may also want to review the information on square roots that can be found here:

    Square Root Basics

    The discriminant of the quadratic equation ax2bx + c = 0 is the quantity Δ defined by

    Δ = b2 – 4ac

    In other words, the discriminant is simply the expression that appears under the square root in the quadratic formula.

    Although computing the discriminant of a quadratic equation does not give the roots (solutions) of the equation, it does give us a lot of information about the nature of the roots and the graph of the equation.

    For example, if the discriminant of the quadratic equation ax2bx + c = 0 is 0, then the quadratic formula simplifies to

    x = –b/2a,

    and we see that there is just one solution.

    If the coefficients a and b are integers, then the unique solution will be a rational number.

    Graphically, this means that the graph of the function y = ax2 + bx + c is a parabola that intersects the x-axis at one point.

    An example is given by the red parabola in the image above.

    Note: The discriminant does not tell us if the parabola opens upwards or downwards. However, this is easy to determine simply by looking at the value of a.

    If a > 0 (i.e. a is a positive number), then the parabola opens upwards.

    If a < 0 (i.e. a is a negative number), then the parabola opens downwards.

    So, the red parabola in the image above has an equation of the form y = ax2 + bx + c where < 0 and Δ = 0

    Example: Find the discriminant of x2 + 6+ 9 = 0. Then describe the nature of the roots of the equation, and describe the graph of the function y = x2 + 6x + 9.

    I will post a solution to this problem tomorrow, and then discuss other possibilities for the determinant. Feel free to post your own solutions in the comments.

     

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  • Setting Up a Ratio
    June 25, 2017
  • Turn Wrong Answers Into Right Answers On Standardized Tests
    June 22, 2017
  • Square Root Basics
    June 20, 2017

    Square root
    Square Root Basics

    A number a is a square root of a number x if a2 = (or equivalently, ⋅ ax).

    Example: 3 is a square root of 9 because 32 = 3 ⋅ 3 = 9.

    –3 is also a square root of 9 because (–3)2 = (–3)(–3) = 9.

    We see that 9 has two square roots: 3 and –3. We sometimes combine these two and say that the two square roots of 9 are ±3.

    Square Root Symbol

    Students sometimes get confused when using the square root symbol:

    square root symbol

    The symbol above represents the positive square root. So, for example, we have

    \sqrt{9}=3

    So even though 9 has the two square roots ±3, when we put 9 under the symbol, the result is just 3 (and not –3). You may want to compare this with the Square Root Property (students often get the square root property confused with taking the positive square root of a number).

    If we want the negative square root of a number, we need to place a minus sign before the square root symbol.

    -\sqrt{9}=-3

    And if we want both square roots of a number, we should place the symbol ± before the square root symbol.

    \pm \sqrt{9}=\pm 3

    “Types” of Square Roots

    All numbers, with the exception of 0, have two square roots. The number 0 has just one square root because both the positive and negative square root of 0 are both 0.

    When taking positive and negative square roots of numbers, it’s worth trying to determine the “type” of number that you get. For example, whenever we take the square root of a nonzero integer (the set of integers is {…–3, –2, –1, 0, 1, 2, 3,…}), the result can be an integer, an irrational number, or a pure imaginary number (see this: Complex Numbers). When we identify the type of roots of a number we usually refer to this as determining the nature of the roots.

    Examples: Determine the nature of the roots of 16, 11, –2, and 0.

    Solution: Since 16 is a positive perfect square, the two square roots of 16 are integers (in fact, they are 4 and –4).

    Since 11 is positive, but not a perfect square, the two square roots of 11 are irrational numbers.

    Since –2 is negative, the two square roots of –2 are pure imaginary numbers.

    Finally, 0 has just one square root…itself. In particular, 0 has one square root, which is an integer.

    Note: If a positive integer is not a perfect square, then it’s square root will always be irrational. The proof of this result is beyond the scope of this article. I will discuss this further in a future post. 

     

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  • Derivation of the Quadratic Formula
    June 18, 2017

    quadratic formula
    Derivation of the Quadratic Formula

    Last week we went over how to solve quadratic equations using the quadratic formula. You can see those posts here: Quadratic Formula 1   2

    Recall that the solutions to the quadratic equation ax2bx + c = 0 are

    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    At the end of the second post, I posed the following problem:

    Challenge Problem: Solve the general quadratic equation ax2bx + c = 0 by completing the square, and note that this gives a derivation of the quadratic formula.

    I will now provide a detailed solution. You may want to review the following posts before reading the following solution:

    Square Root Property     Completing the Square     Solving Quadratic Equations

    Solution:

    ax^2+bx+c=0

    ax^2+bx=-c

    x^2+\frac{b}{a} x=-\frac{c}{a}

    x^2+\frac{b}{a} x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2

    (x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2}

    (x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a} (\frac{4a}{4a})

    (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

    x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}

    x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}

    x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

     

    The derivation above was challenging. Try to understand each step. If you have any question on a particular step, feel free to post your question in the comments.

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  • The Quadratic Formula – Part 2
    June 16, 2017

    quadratic formula
    The Quadratic Formula
    Part 2

    Yesterday we began discussing the quadratic formula, and I asked you to solve a specific quadratic equation using the formula. You can see that post here: Solving Quadratic Equations with the Quadratic Formula

    Today I will provide the solution to that problem. For your reference, let me give you the quadratic formula one more time:

    The solutions to the quadratic equation ax2bx + c = 0 are

    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    Example: Solve the quadratic equation x2 – 2x – 150 by using the quadratic formula.

    Solution:

    \frac{-b\pm \sqrt{b^2-4ac}}{2a}

    =\frac{2\pm \sqrt{(-2)^2-4(1)(-15)}}{2\cdot 1}

    =\frac{2\pm \sqrt{4+60}}{2}

    =\frac{2\pm \sqrt{64}}{2}

    =\frac{2\pm 8}{2}

    =\frac{2}{2}\pm \frac{8}{2}

    =1\pm 4

    So, the two solutions are 1 + 4 = 5, and 1 – 4 = –3

    Note: This particular problem could be solved more easily by factoring. 

    Why does the quadratic formula look so messy? It’s not a very pleasant looking formula. It would have been nicer if we had a simpler formula for solving a quadratic equation. But unfortunately, this is simply what it turned out to be. As a challenging exercise, I would like you to see this for yourself firsthand. I will give a full solution tomorrow:

    Challenge Problem: Solve the general quadratic equation ax2bx + c = 0 by completing the square, and note that this gives a derivation of the quadratic formula.

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  • Solving Quadratic Equations with the Quadratic Formula
    June 15, 2017

    quadratic formula
    Solving Quadratic Equations
    with the Quadratic Formula

    A quadratic equation has the form ax2bx + c = 0.

    The quadratic formula is a formula that gives us all solutions to a quadratic equation right away. The formula is as follows:

    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    Example: Solve the quadratic equation x2 – 2x – 150 by using the quadratic formula.

    We have previously solved this problem by completing the square. You can see that solution here: Solution by Completing the Square

    Try to solve this problem by using the quadratic formula. I will give the full solution tomorrow.

     

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  • New SHSAT Verbal Book – Half Price on Amazon
    June 14, 2017
  • Complex Numbers – Solution to Division Problem
    June 10, 2017

    Complex Numbers

    Complex Numbers
    Solution to Division Problem

    Several days ago I began writing a series of posts about complex numbers. You can see that post here: Complex Numbers  Addition and Subtraction  Multiplication  Division

    Today I would like to provide a solution to Thursday’s division problem. Here is the problem one more time, followed by a solution:

    Example:

    \frac{1+5i}{2-3i}=

    Solution: 

    \frac{1+5i}{2-3i}= \frac{1+5i}{2-3i}\cdot \frac{2+3i}{2+3i}=\frac{(2-15)+(3+10)i}{4+9}=\frac{-13+13i}{13}=-\frac{13}{13}+\frac{13}{13}i=-1+i

     

     

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  • Complex Numbers – Division
    June 8, 2017

    Complex Numbers

    Complex Numbers
    Division

    A few days ago I began talking about complex numbers, and we learned how to raise the complex number i to any power. You can see that post here: Complex Numbers – Examples and Powers of i

    We then reviewed how to add, subtract, and multiply complex numbers, and I gave you a problem to try involving multiplication. You can see those posts here: Addition and Subtraction  Multiplication

    Today I will go over how to divide complex numbers. But first I will provide a solution to yesterday’s multiplication problem. Here is the problem one more time, followed by a solution:

    Example: Compute (2 – 3i)(–5 + 6i)

    Solution: (2 – 3i)(–5 + 6i) = (–10 + 18) + (12 + 15)i = 8 + 27i

     

    Division

    The conjugate of the complex number a + bi is the complex number abi.

    Examples:

    The conjugate of –5 + 6i is –5 – 6i.

    The conjugate of 3 – 2i is 3 + 2i.

    Note that when we multiply conjugates together we always get a real number. In fact, we have

    (a + bi)(abi) = a2 + b2

    We can put the quotient of two complex numbers into standard form by multiplying both the numerator and denominator by the conjugate of the denominator. This is best understood with an example.

    Example:

    \frac{1+5i}{2-3i}=

    Hint: Multiply both the numerator and denominator of the fraction by the conjugate of the denominator. The denominator will them become a real number, and the resulting fraction can then be written in the form abi

    I will provide a solution to this question tomorrow. Meanwhile, please post your attempted solutions in the comments below.

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