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Homework Help: Help with Understanding Real and Non-Real Roots

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data

    For which values of c, cER, will the equation j(x) = c have real roots?

    2. Relevant equations

    j(x) = 2x^2 - 8x + 5

    3. The attempt at a solution

    I understand I need to get this into the form of b^2 - 4ac, yet I do not understand why this is important and such. From my understanding, if we make the answer greater than or equal to 0, the roots are real, and if we make it smaller than, the roots are negative. As I don't understand the concept of real roots beyond this, I don't know where exactly to start. This is a question from an exam from a few years ago and I could do everything apart from this for some reason. This was just never explained to me before. Any help would be greatly appreciated.
  2. jcsd
  3. Feb 17, 2017 #2
    Last edited: Feb 17, 2017
  4. Feb 17, 2017 #3
    Understanding how to complete the square, I come to 2(x-2)^2 - 3. I've used this for functions so this part all makes sense, in order to get the simplified form of the function, yet I am still confused as to how this is useful in finding c.

    if 2(x-2)^2 - 3 = c
    2(x-2)^2 - 3 - c = 0 or something like 2(x-2)^2 = c + 3

    Yet where does the discriminant come into play? As far as I understand, it is easier to apply the quadratic formula before completing the square as the equation is already in the form of ax^2 + bx + c.
  5. Feb 17, 2017 #4
    sorry friend, i was addressing the "why" aspect of your post as best as I know. I think you can answer the question by using what you already posted. This is the discriminant:

    [itex] {b^2-4ac} [/itex] these are the conditions

    [itex] {b^2-4ac} = 0 [/itex] (repeated root)

    [itex] {b^2-4ac} > 0 [/itex] (two distinct real roots)

    Through the derivation of the formula we can see why it must be true..
    Last edited by a moderator: Feb 17, 2017
  6. Feb 17, 2017 #5
    So relooking at the link I find the part where the part that is now the only part with x's is equal to what becomes, with simplification, the root, and because the root has to be positive/negative, we can say that:

    2(x-2)^2 - 3 = c
    2(x-2)^2 = c + 3

    The second part being what we are looking for.

    So if we want a positive root

    c + 3 > 0
    c > -3

    Or we need a negative root

    c + 3 < 0
    c < -3

    Is this all correct? And if so, thank you I understand now. If not then woops.

    Edit : I posted all of this before you revised your second post so my reply my seem a bit confusing
  7. Feb 17, 2017 #6
    I do not think it is correct, try plugging in the values for a and b in the formula and then finding c under the conditions

    [tex] {b^2-4a(5 - c)} = 0 [/tex]
    [tex] {b^2-4a(5 - c)} > 0 [/tex]
  8. Feb 17, 2017 #7
    If we look at the formula before completing the square, j(x) = 2x^2 - 8x + 5 = c, if I input the values that I previously worked out they somehow appear to be correct.

    If we input these values into the square, such that c > -3, for example, -2.

    j(x) = 2x^2 - 8x + 5 = -2
    2x^2 - 8x + 7=0

    If we sub these into b^2 - 4ac

    8^2 - 4(2)(7)
    64 - 56 = 8, therefore positive root.

    And we sub a value lower than -3, for example -4, we get:

    2x^2 - 8x + 5 = -4
    2x^2 - 8x + 9 = 0

    64 - 4(2)(9)
    64 - 72 = -8, negative, therefore negative root.

    If we use -3 itself, we end up with 0. Unless I did something wrong in these calculations I think what I previously stated was correct.

    We can also see this as:

    b^2 - 4ac
    64 - 4(2)(5 + c) = 0

    64 = 4(2)(5 - c) c is negative as it is brought from the left side of the equation before using the discriminant.

    What does c have to be to make the equation balanced? -3.

    64 = 4(2)(5 + 3)
    64 = 64

    Therefore if c is higher than -3, the solution and therefore root is positive.
    And if c is lower than -3 the solution and therefore root is negative.
  9. Feb 17, 2017 #8
    sorry, you are right, :sorry: my mistake for not paying more attention.
  10. Feb 17, 2017 #9
    Seriously thank you for all the help, I finally get this stuff :)
  11. Feb 17, 2017 #10


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    Gold Member

    You don't mean positive or negative root. If the discriminant is positive there are two real roots. If it is negative there are two complex roots. If it is zero there is one real root.
  12. Feb 18, 2017 #11
    Ahah, I see. My mistake. Thank you for the clarification.
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