• Setting Up a Ratio
June 25, 2017
• Setting Up a Ratio
June 22, 2017
• Square Root Basics
June 20, 2017

### 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:

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.

• Derivation of the Quadratic Formula
June 18, 2017

### 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:

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.

• The Quadratic Formula – Part 2
June 16, 2017

### 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.

June 15, 2017

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.

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

### 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:

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

For those of you that prefer videos…

Check out the Get 800 collection of test prep books to learn how to apply this information to standardized test questions.

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• Complex Numbers – Division
June 8, 2017

### 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.

For those of you that prefer videos…

Check out the Get 800 collection of test prep books to learn how to apply this information to standardized test questions.

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• Complex Numbers – Multiplication
June 7, 2017

### Complex Numbers Multiplication

A couple of days ago I introduced 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

Yesterday we reviewed how to add and subtract complex numbers, and I gave you a problem to try involving subtraction. You can see that post here: Complex Numbers – Addition and Subtraction

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

Example: When we subtract 2 – 3i from –5 + 6i we get what complex number?

Solution: (–5 + 6i) – (2 – 3i) = –5 + 6i – 2 + 3i = –7 + 9i.

Multiplication

We can multiply two complex numbers by formally taking the product of two binomials and then replacing i2  by –1. That process leads to the following formula:

(a + bi)(c + di) = (acbd) + (ad + bc)i

Example: Compute (2 – 3i)(–5 + 6i)

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

For those of you that prefer videos…

Check out the Get 800 collection of test prep books to learn how to apply this information to standardized test questions.

Speak to you soon!