Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Can the quadratic formula also be ax^2 - bx - c = 0?

There's example questions with that set up ^^^. Does that mean that's also the quadratic formula?

If so, can it also be ax^2 + bx - c = 0, or ax^2 - bx + c = 0?

Update:

So if they have a problem like ax^2 - bx - c = 0, I just solve it the same way as I would ax^2 + bx + c = 0?

9 Answers

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  • 1 month ago
    Favourite answer

    ax^2+bx+c=0 is called a quadratic equation, while x=[-b+/-sqr(b^2-4ac)]/(2a) is called the

    quadratic formula. Actually, the formula is the

    solutions for x of the said equation.

    a,b & c are called the coefficients of the equation, they are arbitrary constants; one word, they can be + or -. So, it is no need to

    consider ax^2+/-bx+/-c=0 but a must be =/=0.

  • 1 month ago

    The Quadratic Formula can be used to solve any quadratic equation of the form ax2 + bx + c = 0.

    The form ax2 + bx + c = 0 is called standard form of a quadratic equation. If you don't, you might use the wrong values for a, b, or c, and then the formula will give incorrect solutions.

  • DWRead
    Lv 7
    1 month ago

    You solve it using the same steps for solving ax²+bx+c=0, but instead of using b and c, you use (-b) and (-c).

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  • 1 month ago

    ax² - bx - c = 0 → where: a ≠ 0

    a.[x² - (b/a).x - (c/a)] = 0

    x² - (b/a).x - (c/a) = 0

    x² - (b/a).x + (b/2a)² - (b/2a)² - (c/a) = 0

    x² - (b/a).x + (b²/4a²) - (b²/4a²) - (c/a) = 0

    [x² - (b/a).x + (b²/4a²)] - [(b²/4a²) + (c/a)] = 0

    [x² - (b/a).x + (b²/4a²)] - [(b² + 4ac)/4a²] = 0

    [x - (b/2a)]² - [(b² + 4ac)/4a²] = 0 → let: b² + 4ac = Δ

    [x - (b/2a)]² - [Δ/4a²] = 0

    [x - (b/2a)]² - [√Δ/2a]² = 0 → recall: x² - y² = (x + y).(x - y)

    { [x - (b/2a)] + [√Δ/2a] }.{ [x - (b/2a)] - [√Δ/2a] } = 0

    {x - (b/2a) + (√Δ/2a)}.{x - (b/2a) - (√Δ/2a)} = 0

    [x - (b - √Δ)/2a].[x - (b + √Δ)/2a] = 0

    First case:

    [x - (b - √Δ)/2a] = 0

    x - (b - √Δ)/2a = 0

    x = (b - √Δ)/2a

    Second case:

    [x - (b + √Δ)/2a] = 0

    x - (b + √Δ)/2a = 0

    x = (b + √Δ)/2a

    Solution = { (b - √Δ)/2a ; (b + √Δ)/2a }

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  • 1 month ago

    It's easier just to remember it for the positive coefficients:

    ax² + bx + c = 0

    If you had something where you are subtracting, then you just plug in negative values for b and/or c.

    For example:

    2x² - 5x - 3 = 0

    a = 2

    b = -5

    c = -3

    You are much less likely to make a sign mistake if you only have to memorize *one* formula and just plug in negative coefficients instead.

    Here's the quadratic formula

    x = [-b ± √(b² - 4ac)] / (2a)

    Plugging in the values above (including the negative values):

    x = [-(-5) ± √((-5)² - 4(2)(-3))] / (2(2))

    x = [5 ± √(25 - (-24))] / 4

    x = (5 ± √49) / 4

    x = (5 ± 7)/4

    x = -2/4 = -1/2

    or

    x = 12/4 = 3

    Answer:

    Don't bother making it more complicated. Just use the quadratic formula as you learned it and plug in negative values of b and/or c as needed.

  • RockIt
    Lv 7
    1 month ago

    ax^2 + bx -c = ax^2 +bx + (-c).

    ax^2 -bx + c = ax^2 + (-bx) + c.

    Substitute into the standard formula carefully and you will get the right answers.  The important question now is "do you understand why?".

    Answer: Because the standard formula for determining quadratic roots is defined for all real a, b, and c, a<>0.

  • ?
    Lv 7
    1 month ago

    You can modify the standard formula by replacing b with -b or c with -c as needed. 

    x = -(-b) +/- sqrt((-b)^2 -4a(-c)) /(2a)

  • Jim
    Lv 7
    1 month ago

    Sure. Since the values can also be a reversed value inside there's no problem. But it just complicates things including the Quadratic Equation.

    The standard form is the standard for a reason.

  • 1 month ago

    You could do that, but then you'd need to derive a new quadratic equation to support the newly placed negatives.

    It's easier to use + in the general form and just substitute in the negative values vs. derive a modified version of the equation so your inputs are only positive.

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