Problem with quadratics (dividing by x)

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In summary, the conversation discusses the circumstances under which a quadratic can be divided by the variable, specifically in the context of projectile motion questions. The conversation also mentions the term that cannot be divided by x most of the time and the value(s) of the constants for which it can be divided by x. The conversation concludes by stating that if the vertical displacement is 0, then t = 0 will always be a solution, allowing for the quadratic to be divided by t.
  • #1
david18
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Hi, could someone please tell me the circumstances under which you can divide a quadratic by the variable?

e.g. x^2 - (15/2)x = 0. I know I'd usually have to go through the whole factorisation/formula process and get more than one answer for x, but if I know that x is positive am I able to simply divide it by x to get x - 15/2 = 0?

The reason I ask is because this is what I do in my projectile motion questions. Whenever I get a question in which the vertical displacement is 0 (e.g. ball being kicked up then falling back to ground) I end up with the equation 0 = ut + 1/2 at^2 (from the equation of motion s = ut + 1/2 at^2). I always divide this by t and get the right answer, but is this what I'm meant to be doing?

Any help would be greatly appreciated,
David
 
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  • #2
when you try to divide the general quadratic ax^2 + bx + c = 0 by x, what is the term that can't be divided by x most of the time? for what value(s) of the constants can this term also be divided by x?

when you divide by x, you really factor the quadratic, and one of the factors is x, so the quadratic still has 2 solutions, one of them is x = 0
 
  • #3
david18 said:
Whenever I get a question in which the vertical displacement is 0 …

Hi David! :smile:

If the vertical displacement is 0, then t = 0 will always be a solution, won't it (because after zero time it will always have moved zero distance! :biggrin: )?

So you will always be able to divide the quadratic by t! :smile:
 

1. What is the problem with dividing by x in quadratics?

The main issue with dividing by x in quadratics is that it can lead to undefined or infinite solutions. This is because division by 0 is undefined, and x can equal 0 in certain quadratic equations. Therefore, dividing by x can make the solution set of the quadratic equation invalid.

2. Can you give an example of a quadratic equation where dividing by x leads to an undefined solution?

One example is the equation x^2 + 4x + 4 = 0. When solving for x by factoring, we get (x + 2)(x + 2) = 0. Dividing both sides by x gives us (x + 2) = 0, which leads to the solution x = -2. However, if we divide by x from the beginning, we get x + 4 + 4/x = 0, and when x = 0, the equation becomes undefined.

3. How can we avoid dividing by x in quadratics?

One way to avoid dividing by x is to use the quadratic formula, which involves the square root function instead of division. Another method is to factor the quadratic equation and solve for x without dividing. Additionally, we can also rewrite the equation by multiplying both sides by x to eliminate the x in the denominator.

4. Is dividing by x always a problem in quadratics?

No, dividing by x is only an issue when x can equal 0 in the quadratic equation. If x cannot equal 0, then dividing by x will not lead to an undefined solution. For example, in the equation x^2 + 4x + 4 = 9, dividing by x will not cause any problems as x = 0 is not a valid solution.

5. What are some real-life applications where dividing by x in quadratics may pose a problem?

In physics, dividing by x in equations related to motion (such as v = d/t) can be problematic if the time (t) is 0, as it would lead to an undefined solution. In economics, dividing by x in cost functions can also be an issue if the quantity (x) is 0, as it would result in a division by 0 error. Therefore, when solving real-life problems involving quadratics, we must be careful when dividing by x and ensure that it does not lead to an invalid solution.

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