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Int Algebra: Using the Quadratic Formula


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About this Lesson

  • Type: Video Tutorial
  • Length: 9:24
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 101 MB
  • Posted: 12/02/2008

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Equations & Inequalities (50 lessons, $65.34)
Intermediate Algebra Review (25 lessons, $49.50)
College Algebra: Using the Quadratic Formula (4 lessons, $5.94)

The quadratic formula is used to solve for x in quadratic equations, which come in the form ax^2+bx+c=0. This formula is most commonly used when the expression can't be easily factored for evaluation. Oftentimes, this is because the two solutions to the equation are not real numbers. In this lesson, Professor Burger will walk you through when to use the formula, what the alternatives to the formula are and how to apply the formula. He will also explain how and why the formula can give imaginary numbers as solutions and what that means.

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Intermediate Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers real numbers, equations and inequalities, exponents and polynomials, rational expressions, roots and radicals, relations and functions, the straight line, systems of equations, quadratic equations and quadratic inequalities, conic sections, inverse and exponential and logarithmic functions, and a variety of other AP algebra and advanced algebra.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Quadratic Equations and Inequalities
Solving with The Quadratic Formula
Using the Quadratic Formula Page [1 of 2]
So now we see the quadratic formula as if you're given just a general quadratic equation, ax2 + bx + c = 0. There it is.
The solution is always the following. x = -b plus or minus the square root of b2 minus 4ac, all over 2a. So that
means, remember the number a is the coefficient on the x2 term, the b there is the coefficient on the x term and that
+c is the constant coefficient. So you put all that together and you’ve got to the thing. Let’s look at some actual
examples using the quadratic equation. By the way, here’s what I do when someone gives me a quadratic equation
like this to solve; the very first thing I do is I try to factor it. And you might want to try that, too. I suggest you do it
because just so you use the quadratic equation because you know it, because actually it’s sort of a pain to use. So I
quickly try to factor it so, the same sign, their both negative. I’ve got to have an easy number here whose product is 3
and yet combined to give 5. So that’s going to be a 3 or 1 or 1 and 3. I’ll put a 3 in here. That gives me a 9 and
another 1, which would be -10. That’s no good. Put the 3 in here, that would give me what? Put a 3 in here, that
would give me a 3 and a 3, which would be a minus 6 and that’s shy, well, actually overshoots the minus 5. So this
can’t be factored in an easy way.
So I’m going to go right to the quadratic formula. The quadratic formula now, you have to understand who’s playing
what role. Think of that as the play and now we’re going to put in the actors. The role of a is going to be played 3
here because that’s the coefficient on the x2. So this is going to be a. The role of b tonight is going to be played, not
by 5, but by -5. And finally, in the role of c we have the number 3. So now we’re going to just plug in to the quadratic
formula, with these values. So what does x equal? You really want to memorize this and it’s almost like a little
mantra. x equals -b, plus or minus the square root of b2 minus
. So x = -b . So it’s negative -5, which would be
five plus or minus, they are the two solutions, the square root of what? b2. So I take minus 5 and I square it. That’s
25 - 4 x ac. Now what’s ac? Well, a is 3 and c is 3 so this times that is just 9. So 4 x 9, the whole thing over 2a. So 2
x 3 is 6. And there’s the answer. Let’s see if we can actually work that out a little bit. x = 5 plus or minus the square
root of 25 minus and what is 4 times 9? Well, that is 36 divided by 6 and what does this equal? 5 plus or minus, and
now what’s 25 minus 36? Well, that’s going to be the square root of, it’s going to be negative, you’ll notice. And it
could be negative what? I think it’s going to be -11 all over 6. So what does that mean?
Well, this means that this equation has no real solutions at all. This can’t be solved with a real number x. However,
the quadratic formula actually gives us the power to produce what the true answers really are. The true answers in
this case are going to be imaginary numbers. And we can write them like this. 5 plus or minus—remember that the
square root of -11 is just the square root of 11i, because i is the square root of -1 all over 6. And so those are our two
answers. These are imaginary answers.
So there are no real answers to this one, but in fact we found the two complex solutions. One is 5 plus the square
root of
, the other one 5 minus the square root of
Let me try one more. This one is going to be a little more exotic. The coefficients are going to be the square root of
2x2 plus 3x plus the square root of 2 = 0. And now I want to solve this. Well, I’m not even going to attempt to try to
factor that because the square root thing will just throw everything off. But we know who’s playing what role here.
The square root of 2 is in the role of a, b is being played by 3, and the square root of 2 is also going to play c. Let’s
see what happens this time. So I have x = -b. So that’s -3 plus or minus the square root of b2. So I square 3 and I get
9 minus 4 times ac. That’s going to be times the square root of 2 times the square root of 2. The square root of 2
times the square root of 2 is just 2. So times 2, all divided by 2a or 2 square root of 2. And what does this equal?
Minus 3 plus or minus the square root of, well, this is 9 - 8. That’s just 1. All divided by 2, the square root of 2. The
square root of 1 is just 1. So, in fact, this answer comes up really nicely. This is going to be -3 plus or minus 1
divided by 2, the square root of 2. So there are 2 answers there and let’s actually explore both of them. So we have x
=, what does it say over there? Minus 3 plus or minus 1—I’m checking out the actual white box, 2 square root of 2.
So let’s look at these solutions and write down what they are. So the first one, x equals minus 3 plus 1 over 2 square
root of 2. What does that equal? Well, minus 3 plus 1 is minus 2, minus 2 over 2, the square root of 2. Those 2’s
cancel, the factor on top and the factor on the bottom. I’m left on top was an invisible minus 1. Always a minus 1
factor. So minus 1 over the square root of two. Some people might actually rationalize the denominators. Some
multiply the top and bottom by the square root of 2. And then I’ll see minus the square root of 2 divided by the square
Quadratic Equations and Inequalities
Solving with The Quadratic Formula
Using the Quadratic Formula Page [2 of 2]
root, 2 times the square root of 2 is just 2. So there’s one solution. And what’s the other solution? The other solution
would be x equals, so I did the +1, now I’m doing the -1, minus 3 minus 1 over 2, the square root of 2. So -3 -1 is -4
over 2, square root of 2. I can cancel again. This factor of q with one of the factors of 2, which at least there was an
extra factor of 2 on top. And so I see -2 over the square root of 2. That’s a fine answer, but if you want to rationalize,
multiply by 1 by multiplying top and bottom by the square root of 2 and look what you get. On the bottom I get a 2.
On the top I have a factor of 2 so they cancel. I’m just left with minus the square root of 2. So the 2 answers to this
quadratic equation is x = minus the square root of 2 or x = minus the square root of 2 all divided by 2. And actually
you can check by plugging these values back in for x and seeing that each of the values when plugged in produce 0.
So the interesting thing about the quadratic formula is not only can you use it solve really sort of complicated awful
looking things like this where you have square roots as coefficients and you're still able to get a nice tidy answer like
this, but it even allows you to figure out the solutions to quadratics that before we couldn’t even find the solutions of
because they aren’t any real solutions, in fact, the solutions are imaginary. The quadratic formula will always produce
those two solutions. So now with the quadratic formula, we can see quadratics always have 2 solutions. They might
both be real numbers. They both may be the same in which case it would look like we only have 1, but it’s really 2
solutions. It just happens twice. Or in the case when there’s not 2 real solutions, it turns out it will still be there, but
now they will be imaginary. So we can see quadratics always have 2 solutions, either 2 real, and those 2 real may be
the same or they will be 2 imaginary. And the quadratic equation, quadratic formula, empowers us to actually find
those solutions always. Just be careful with who’s playing what role, a, b, c, plug in, it’s as easy as 1, 2, 3. Do, re, mi.

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