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

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 • 28 New SAT Math Lessons – Intermediate Course • 28 New SAT Math Lessons – Advanced Course • 320 SAT Math Problems • New SAT Math Problems • 320 SAT Math Subject Test Problems – Level 1 • 320 SAT Math Subject Test Problems – Level 2 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). Please allow up to 10 days for processing and shipping. To place your order simply click the following button. For more information about these books click the following link: 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. If you have any questions or concerns please feel free to contact Dr. Steve Warner directly at steve@SATPrepGet800.com Thank you all for your continued support! • Special Offer- SAT Math Paperback Bundle ### 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
• 28 New SAT Math Lessons – Intermediate Course
• 28 New SAT Math Lessons – Advanced Course
• 320 SAT Math Problems
• New SAT Math Problems
• 320 SAT Math Subject Test Problems – Level 1
• 320 SAT Math Subject Test Problems – Level 2

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).
Please allow up to 10 days for processing and shipping.

To place your order simply click the following button.

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.

If you have any questions or concerns please feel free to contact Dr. Steve Warner directly at

steve@SATPrepGet800.com

Thank you all for your continued support!

• Complex Numbers
August 10, 2017
• Special Offer- SAT Math Paperback Bundle
August 7, 2017

### 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 • 28 New SAT Math Lessons – Intermediate Course • 28 New SAT Math Lessons – Advanced Course • 320 SAT Math Problems • New SAT Math Problems • 320 SAT Math Subject Test Problems – Level 1 • 320 SAT Math Subject Test Problems – Level 2 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). Please allow up to 10 days for processing and shipping. To place your order simply click the following button. For more information about these books click the following link: 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.

If you have any questions or concerns please feel free to contact Dr. Steve Warner directly at

steve@SATPrepGet800.com

Thank you all for your continued support!

• Inverse Variation
August 1, 2017
• Direct Variation
July 24, 2017
• Math Strategy: Draw your Own Figure
July 18, 2017
• The Discriminant of a Quadratic Equation – Part 5
July 12, 2017

### The Discriminant of a Quadratic Equation Part 5

Today I’d like to solve the problem about discriminants from the last post. You can review parts 1, 2, 3, and 4 of our discussion on the discriminant here:

Last time, I asked you to solve the following problem:

Example: You are given the following system of equations.

dx + ey = f
y = x2

where ab, and c are integers. For which of the following will there be more than one (xy) solution, with real-number coordinates for the system?

Ae2 + 4df < 0
B. e2 – 4df < 0
C. d2 – 4ef < 0
D. e2 – 4df > 0
E. d2 + 4ef > 0

Make sure to try this problem yourself before reading the following solution.

Solution: By the second equation, we have y = x2, so we can replace y by xin the first equation to get

dx + ex2 = f

Writing this quadratic equation in general form gives the following.

ex2 + dx – f = 0

So, we have aebd, and c f . Thus, the discriminant is

Δ = b2 – 4ac = d2 – 4e( f) = d2 + 4ef

We want the quadratic equation to have two real solutions. Therefore, the discriminant must be positive. So, we must have

d2 + 4ef > 0

This is choice E

• The Discriminant of a Quadratic Equation – Part 4
July 10, 2017

### The Discriminant of a Quadratic Equation Part 4

Let’s continue talking about the discriminant of a quadratic equation. You can review parts 1, 2 and 3 of this discussion here:

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

Δ = b2 – 4ac

That is, the discriminant is simply the expression that appears under the square root in the quadratic formula.

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

Last week, I asked you to solve the following problem:

Example: Find the discriminant of x2 + 4+ 8 = 0. Then describe the nature of the roots of the equation, and describe the graph of the function y = x2 + 4x + 8.

Make sure to try this problem yourself before reading the following solution.

Solution: In this question, we have a = 1, b = 4, and c = 8. So the discriminant is

Δ = b2 – 4ac = 42 – 4(1)(8) = 16 – 32 = –16.

Since the discriminant is negative, it follows that the two roots of the quadratic equation are distinct complex numbers.

The graph of the function y = x2 + 4+ 8 is an upward facing parabola that does not intersect the x-axis.

Notes: (1) In this example, we can find the two complex solutions of the quadratic equation by using the quadratic formula or by completing the square. I leave it as an exercise to show that the two solutions are –2 + 2i and –2 – 2i.

(2) Notice that the two solutions are complex conjugates of each other. This will always happen when the discriminant is negative. For more information on complex conjugates, see the following post: Complex Numbers – Division

(3) We know that the parabola opens upwards because a = 1 > 0.

(4) It’s very easy to also find the y-intercept of the parabola. We simply substitute 0 in for x into the equation. So we get y = 8. It follows that the y-intercept of the parabola is the point (0,8).

Now try the following problem which is similar to a problem found on a recent ACT:

Example: You are given the following system of equations.

dx + ey = f
y = x2

where ab, and c are integers. For which of the following will there be more than one (xy) solution, with real-number coordinates for the system?

A. e2 + 4df < 0
B. e2 – 4df < 0
C. d2 – 4ef < 0
D. e2 – 4df > 0
E. d2 + 4ef > 0

I will post a solution to this problem tomorrow. Feel free to post your own solutions in the comments.

• The Discriminant of a Quadratic Equation – Part 3
July 7, 2017

### The Discriminant of a Quadratic Equation Part 3

Today I would like to continue our discussion of the discriminant of a quadratic equation. You can review parts 1 and 2 of this discussion here:

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

Δ = b2 – 4ac

That is, the discriminant is simply the expression that appears under the square root in the quadratic formula.

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

Last week, I asked you to solve the following problem:

Example: Find the discriminant of x2 + 8+ 7 = 0. Then describe the nature of the roots of the equation, and describe the graph of the function y = x2 + 8x + 7.

Make sure to try this problem yourself before reading the following solution.

Solution: In this question, we have a = 1, b = 8, and c = 7. So the discriminant is

Δ = b2 – 4ac = 82 – 4(1)(7) = 64 – 28 = 36.

Since the discriminant is positive, it follows that the two roots of the quadratic equation are distinct real numbers. Furthermore, since 36 is a perfect square (62 = 36), the roots are actually rational.

The graph of the function y = x2 + 8x + 7 is an upward facing parabola that intersects the x-axis at two points.

Notes: (1) In this example, we can easily find the two roots of the equation by factoring:

x2 + 8+ 7 = 0
(x + 1)(x + 7) =0
x + 1 = 0        or        x + 7 =0
x = –1       or       x = –7

So the two roots are –1 and –7.

(2) We know that the parabola opens upwards because a = 1 > 0.

(3) It’s very easy to also find the y-intercept of the parabola. We simply substitute 0 in for x into the equation. So we get y = 7. It follows that the y-intercept of the parabola is the point (0,7).

### Negative Discriminants

Up until now, we’ve looked at the cases where the discriminant is 0 and positive. Today let’s talk about what happens if the discriminant is negative.

If the discriminant of the quadratic equation ax2bx + c = 0 is negative, then we wind up with a negative number under the square root in the quadratic formula. The square root of a negative number is an imaginary number. We therefore wind up with two complex solutions.

Graphically, if the discriminant is negative, the graph of the function y = ax2bx +is a parabola that does not intersect the x-axis.

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

Example: Find the discriminant of x2 + 4+ 8 = 0. Then describe the nature of the roots of the equation, and describe the graph of the function y = x2 + 4x + 8.

I will post a solution to this problem tomorrow, and then discuss what happens if the determinant is negative. Feel free to post your own solutions in the comments.

• The Discriminant of a Quadratic Equation – Part 2
July 3, 2017

### The Discriminant of a Quadratic Equation Part 2

Last week, I began discussing the discriminant of a quadratic equation. You can review that post here: The Discriminant of a Quadratic Equation – Part 1

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

Δ = b2 – 4ac

That is, the discriminant is simply the expression that appears under the square root in the quadratic formula.

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

Last week, I asked you to solve the following problem:

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.

Make sure to try this problem yourself before reading the following solution.

Solution: In this question, we have a = 1, b = 6, and c = 9. So the discriminant is

Δ = b2 – 4ac = 62 – 4(1)(9) = 36 – 36 = 0.

It follows that the roots of the quadratic equation are equal (in other words, there is really just one root) and rational (a fraction, where the numerator and denominator are both integers).

The graph of the function y = x2 + 6x + 9 is an upward facing parabola that intersects the x-axis at one point.

Notes: (1) The unique rational root of this quadratic equation is

x = –b/2a = –6/2 = –3.

This means that the only x-intercept of the parabola is the point (–3,0). In this case, this point also happens to be the vertex of the parabola.

(2) We know that the parabola opens upwards because a =1 > 0.

(3) It’s very easy to also find the y-intercept of the parabola. We simply substitute 0 in for x into the equation. So we get y = 9. It follows that the y-intercept of the parabola is the point (0,9).

(4) Now that we know the x-intercepts, y-intercept, and vertex of the parabola, and we know that the parabola opens upwards, it’s very easy to sketch the graph. I leave it to the reader to draw a nice sketch.

### Positive Discriminants

So far, we’ve just looked at the case where the discriminant is 0. Today let’s talk about what happens if the discriminant is positive.

If the discriminant of the quadratic equation ax2bx + c = 0 is positive, then we wind up with a positive number under the square root in the quadratic formula. The square root of a positive number is a real number. We therefore wind up with the two real solutions

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

If the discriminant also happens to be a perfect square (such as 1, 4, 9, 16, etc.), then both solutions will be rational numbers.

Graphically, if the discriminant is positive, the graph of the function y = ax2 + bx + c is a parabola that intersects the x-axis at two points.

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

Example: Find the discriminant of x2 + 8+ 7 = 0. Then describe the nature of the roots of the equation, and describe the graph of the function y = x2 + 8x + 7.

I will post a solution to this problem tomorrow, and then discuss what happens if the determinant is negative. Feel free to post your own solutions in the comments.