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Quadratic Equations

  1. Sep 27, 2008 #1
    I got into college as a mature student and I'm doing well in all subjects except for math. I'm struggling a good bit because I never finished high school math and theres a lot of stuff they assume everyone knows but I don't. I have to study it at home so I can keep up. Its mainly quadratic equations that are giving me trouble right now. I just wanna ask a few questions that I haven't had a chance to ask in class.

    First of all I can't understand what the teacher means when he says "make an equation = 0". I read somewhere that quadratic equations by definition will equate to zero but I must have misunderstood what I was reading. How do I make a quad equation = 0? For example say I have this equation
    "2xquared - 6x - 15" how can I make that equal zero?

    2.) I can't figure out how to plug equations into the quadratic formula
    http://home.alltel.net/okrebs/Ch1-20.gif [Broken]
    I know that its only the equation in the square root sign that I'm using but I still don't know which numbers to assign to b, a and c. Using this equation for example
    "Xsquared + 5x - 10". I was told that b is however many x's there are so b should be 5 in this case but then what about the Xsquared at the beginning? Xsquared is still x so why doesn't b represent it? I do not know which number to assign "a" in the formula and I heard that any numbers left over are assigned to c.

    Would I be right in assuming that all non squared x's are represented by "b", all squared x's fit into "a" and all left over non variable numbers fit into "c"? For example if I put this in the square root sign in the quadratic formula
    "(5)squared - 4(1)(10)"
    would I be on the right track?

    the 5 since there was 5x, the 1 since there was 1Xsquared and the 10 since that was the only non coefficient number left over.

    Any help at all would be greatly appreciated. This is my main obstacle at the moment.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 27, 2008 #2
    What your teacher wants you to do is find the "zeroes" of a function.

    Let's suppose you have a function called f. If you have the equation for f, finding different values of f is pretty easy. You just plug in the values and calculate.

    If f(x) = x^2 - 1, for example, you can immediately figure out what f(c) is for any number c. If you want, f(1), you plug in "1" for the value of "x" in the equation: f(1) = 1^2 - 1 = 0. If you want f(-3), plug in "-3" for "x": f(-3) = (-3)^2 - 1 = 9 - 1 = 8. If you want f(0), you plug in "0" for "x": f(0) = 0^2 - 1 = -1.

    So given a number c, finding the value for f(c) is easy.

    Math is more fun when it's a challenge, though!

    So instead of plugging a number into f and asking what the result is... start of with the result and try to figure out what numbers you had to plug in to get it.

    We might ask ourselves this: for what value c does f(c) = 0? Before, we were given an input (the "c") and we wanted to find the output "f(c)". Now, we've turned it around, and we are given the output "f(c)", and we want to figure out what inputs "c" we could have possibly started with.

    This is what your teacher meant by "make an equation zero". When you have a problem where you want to find values of c such that f(c) = 0, we call those the "zeroes of function f".

    If you know how to graph functions, the problem is very easy to state. Draw the graph of your function. Now, look for points where the function's curve crosses or touches the x-axis. Whatever x-value you see this happen, that x-value is a zero of the function.

    Let's do three examples of finding the zeroes by hand.

    Problem 1:
    Let f(x) = 2x - 1. The zeroes of f are all values c such that f(c) = 0.

    f(c) = 0
    f(c) = 2c - 1 (just plug in "c" for "x")
    Since those two lines are equal, we have:
    0 = 2c - 1

    Now, we try to isolate c using algebra:
    1 = 2c (add 1 to both sides)
    1/2 = c (divide both sides by 2)

    So we have our solution: the zero of f(x) = 2x - 1 is 1/2. We can double-check this by plugging 1/2 into f: f(1/2) = 2 (1/2) - 1 = 1 - 1 = 0. So that is the correct answer. This one was easy.

    Problem 2:
    Let f(x) = 6x^2. Find the zeroes of f.

    f(c) = 0
    f(c) = 6c^2.
    0 = 6c^2

    Divide both sides by 6.
    0 = c^2.
    Take the square root of both sides to remove the square.
    sqrt(0) = sqrt(c^2)
    0 = c

    So our solution is c = 0. We can double check by calculating f(0). f(0) = 6(0)^2 = 6 * 0 = 0. Good.

    Problem 3: The reason we like the quadratic formula....
    Let f(x) = x^2 - 1. Find the zeroes of f.

    0 = c^2 - 1.
    Try to solve for c..
    1 = c^2.
    Take the square root of both sides.
    sqrt(1) = c.
    1 = c.

    WARNING! This is not the correct answer! Well... it's not the full answer. It's true that f(1) = 1^1 - 1 = 0... and so 1 *is* indeed a zero of f. But functions can have more than one zero, and the problem was to find all of them, not just one of them!

    Why did this happen? It turns out, the problem is taking the square root. If c is a positive number, then it's true that sqrt(c^2) = c. But if c is a negative number, then sqrt(c^2) = -c! When we don't know if c is positive or negative, as in the problem above, we must take both into account. The end result is that -1 is also a zero of f, since f(-1) = (-1)^1 - 1 = 1 - 1 = 0.

    The moral of the story is that square root does not perfectly "undo" squaring, and whenever you take the square root of both sides, you must stop and think about the possible negative values.

    I hope you can follow most of this so far. If not, I'd be happy to clarify. But assuming you're not totally lost, let's go over the quadratic equation and what it's good for.

    Functions come in many, many flavors. The ones you are most accustomed to in algebra are the polynomials. The polynomials are ones that look like f(x) = x^3 + 2x^2 - 5x + 1.

    A quadratic function is a special kind of polynomial. (One where the highest power of x is 2). We say that for any numbers a, b, and c, the function f(x) = ax^2 + bx + c is a quadratic function. In fact, any quadratic function can be described by those three numbers: a, b, and c.

    Here's some examples:
    f(x) = x^2 + x + 1
    a = 1, b = 1, c = 1

    f(x) = 2x^2 - 5x + 7
    a = 2, b = -5, c = 7 (note that b is negative)

    f(x) = x^2 - 1
    a = 1, b = 0, c = -1 (note that b = 0 because there is no "x" term at all)

    f(x) = 2x^2
    a = 2, b = 0, c = 0

    f(x) = 0
    a = 0, b = 0, c = 0

    These a, b, and c are the same a, b, and c in the quadratic equation. The quadratic equation lets you simply plug in a, b, and c, and out comes the values of your function's zero(es).

    The last thing I'll explain about the formula is the stuff under the square root. This term "b^2 - 4ac" is called the discriminant. It's value alone tells you how many zeroes your function has. Think about how the whole equation behaves in these circumstances:

    * The discriminant is greater than 0. Because there is a plus-or-minus in front of the square root, we end up with two zeroes.

    * The discriminant is equal to 0. The square root term drops out, and we have only one solution.

    * The discriminant is less than 0. Go directly to jail for taking the square root of a negative number before Algebra II. Negative numbers do not have square roots, and so the function has no zeroes.

    I hope this helped, even if just a little!
  4. Sep 27, 2008 #3


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    Welcome to PF!

    Hi MadmanMurray! Welcome to PF! :smile:

    Yes, you're right …

    it may help to think of it in the same way as units tens and 100s when you learnt about numbers in primary school!

    the x2 position is like the 100s, the x position is like the 10s, and the constant :wink: position is like the units.

    so (a,b,c) is a shorthand way of writing the whole expression (with x2 and x), just as 312 is a shorthand way of writing three hundred and twelve. :smile:
  5. Sep 27, 2008 #4
    Tac-tics that helped a whole load actually. I learned 5 times more than I asked by reading that. Thanks alot. I'm gonna start graphing some of these equations to get a more visual idea of whats going on.

    tiny-tim: yea its been a few years since I've done this stuff but its coming back to me real quick. I left school to work in 8th grade but usually people are required to have finished highschool to get into the course I'm doing. I showed exceptional knowledge in the other subjects that they let me in but I knew I was gonna have some trouble with math. Alot of the stuff I do grasp just from seeing an example on the board but some things require knowing certain things they teach in high school so I have to go home and learn it. The teacher usually gives a little refresher for these things but he rips through them and usually moves on before I fully grasp the concept. If I keep doing this extra curricular study though I'll have no trouble keeping up.
  6. Sep 27, 2008 #5
    I'm in Year 10 and do Extended Maths at my school, so I'll add some to this question.

    Why does a quadratic equation = 0?

    The reason is that when "y=0" as it does when a quadratic equation is equal to zero allows us to find the roots, or x-intercepts of the equation.

    Take y = x^2-4x-32 = 0 for example (can also be written as f(x) = x^2-4x-32 = 0).
    In this case, y = 0 in the spatial domain (remember this is for the y-axis [i.e. vertical translation/height]).

    This means we can find the intercepts along the x-axis as y=0.

    What do the values a, b and c mean?

    "a" is the coefficient of the "x^2" term (in the above example, it is 1).
    "b" is the coefficient of the "x" term (in the above example, it is -4).
    "c" is the coefficient of the "x^0" term, or more simply the remainder of the equation (in the above example, it is -32).

    Remember here that "x^0" = 1, and so this can be omitted (as it generally is).

    What if we want to factorise this equation, or find the roots of the equation?

    To factorise we find, for this equation, it will be in the form (x +/- ?)(x +/- ?).
    The values must add up to the "b" term and multiply to give the "c" term.

    So, we find that for f(x) = x2-4x-32 = 0, we must find factors of 32 that add up to give us -4. As the "b" term is negative, we find that one of the terms in the brackets after the "x" must be negative.

    Trial and error occasionally works here, but with inspection, we see that 4 and 8 multiply to give us 32. We find then that as we are after -32, -4 or -8 multiplied by 8 or -4 give us -32.

    The only valid numbers to put in are those that give us -32 by multiplication and -4 by addition/subtraction. So here we find that this is the factored answer:

    (x + 4)(x - 8) = 0.

    Now, we use the null factor law to work out what the value of the zeroes, or x-intercepts are. For the first bracket, (x + 4) must equal 0, and so "x" is obviously equal to -4.

    For the second bracket, (x - 8) must equal 0, and so "x" is obviously equal to 8.

    To find the x-coordinate for the turning point, we use the formula -b/2a, and in this case this becomes 4/2, which then equals 2.

    For the y-coordinate of the turning point, this is merely the "c" value, as this represents the vertical translation (ie movement above or below the x-axis). And so, the turning point is at (2, -32).

    The zeroes of the function are at (-4, 0) and (8, 0), and so now to graph the equation, draw up a graph, and place points at (-4, 0) and (8, 0) and draw the parabola going through (2, -32) as the turning point.

    You have now succesfully factorised this function, found the roots and graphed it. It is always a good idea to state the domain and range, however if you're not sure of these at this stage, that should be ok, as you would likely cover it soon.

    The discriminant is also an important concept, but likely not one you have been introduced to yet. In the next post, I will talk about solving this equation using the quadratic formula.

    Good Luck!
  7. Sep 27, 2008 #6
    Ok, so now to solve this equation using a more effective method, the quadratic formula.

    How do I use the quadratic formula to solve a quadratic equation?

    First off, the quadratic formula is as follows:

    x = (-b +/- sqrt[b^2 - 4ac])/2a.

    Now, in the equation f(x) = x^2-4x-32 = 0, "a" equals 1, "b" equals -4, and "c" equals -32.

    So we find that this then becomes:

    x = (4 +/- sqrt[-4^2 - {4*1*-32}])/2*1.

    It then follows that we continue this process:

    x = (4 +/- sqrt[4 + 128])/2.
    x = (4 +/- sqrt[132])/2.
    x = (4 +/- 11.489)/2.

    From this, we then find that we will result with 2 equations, one with an addition and one with a subtraction.

    x = (4 + 11.489)/2.
    x = 15.489/2
    :. x = 7.744

    x = (4 - 11.489)/2
    :. x = -3.7445

    These final x-values (indicated by the "therefore" sign :.) are the roots of the equation, provided more accurately than the previous method. We could now plug these values in and expand them out.

    (x - 7.744)(x + 3.7445) = 0, then becomes:

    x^2 + 3.7445x - 7.744x - 29 = 0, thus becoming:

    x^2 - 3.9995x - 29 = 0

    If we rounded to the closest numbers earlier, we would have got precisely what we had in the last post, and so this method has more accuracy (if you want the greatest precision, go through the steps and use the quadratic formula. In this example the numbers had previously been rounded).

    For example, round up 7.744 to 8 and 3.7445 to 4 and then plug these in again, and we get the same as in the last post:

    (x - 8)(x + 4) = 0, and so this proves that this method is correct. If you follow the methods in the previous post, then you find that you'll get the correct answer. Huzzah!

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