quadratics
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Otis
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Hi JH. That is one of the two solutions.
Dividing each side by x^2 was actually your second step.
Dividing by x^2 is okay, as long as you know in advance that x cannot be zero.However, this later step is not allowed:
The proper way to make that cancellation is to first decompose the ratio into a difference of two ratios.
\(\displaystyle \frac{3 - x}{x^2}\)
\(\displaystyle \frac{3}{x^2} - \frac{x}{x^2}\)
\(\displaystyle \frac{3}{x^2} - \frac{1}{x}\)
But, that approach doesn't really help, does it?
Your quadratic polynomial factors nicely:
2x^2 + x – 3 = (2x + 3)(x – 1)
We could now apply the Zero-Product Property.
Alternatively, we could start over using the Quadratic Formula or Completing the Square.
PS: Did you notice that I did not type 1x? It's common form to skip writing coefficients when they are 1 or -1:
2x^2 + x – 3
x^2 – x – 6
[imath]\;[/imath]
Steven G
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I need to send the same message to you that Otis stated about your arithmetic error.
\(\displaystyle \dfrac{5+4}{9+2*4} = \dfrac{9}{17}\)
\(\displaystyle \dfrac{5+1*4}{9+2*4} = \dfrac{5+1}{2+2} = \dfrac{6}{4}\) I got this by cancelling 4's like you did with the x's
Which one is correct?
\(\displaystyle \dfrac{5+4}{9+2*4} = \dfrac{9}{17}\)
\(\displaystyle \dfrac{5+1*4}{9+2*4} = \dfrac{5+1}{2+2} = \dfrac{6}{4}\) I got this by cancelling 4's like you did with the x's
Which one is correct?
Sorry for the late response.
Thank you very much for your reply, it all makes sense.
There are a lot of rules, I've discovered (or rediscovered) and study math on my own because anything
school related is mostly gone so there's a few things I'm not totally certain about when it comes to
rearranging formulas.
Through my studies I've lost touch with the basics while focusing on different concepts, which ironically
almost always involves ratios.
I haven't done any research on this question so here's an interpretation....for better or worse....
Otis, thank you, I did notice the x without the coefficient. I do know about that rule but wanted
to play it safe.
I was familiar with the Zero-Product Property and the Quadratic Formula but not Completing the Square
so thank you for that.
I'm trying to keep things a visual and geometric as possible, and breaking the solution down
into it's original form as opposed to using a formula.
Maybe I'm wrong but formulas to me are a final statement not an explanation. I like the idea
of tearing apart formulas and discovering the 'hidden message' within.
Steven G, the first is correct?
Order of operations requires to deal with the multiplication, if I do that for both I still get 9/17. The denominator product
is 8 because 4 is factoring in with the 2.
I'm not sure where the 2 + 2 comes from in the second fraction equating to (5 + 1) / (2 + 2). I wouldn't seperate
this fraction because it's not expressed as (5 + 1) / 2 which would produce 5/2 + 1/2. As expressed it would be 6/4,
but since 4 is present in the first expression they would resolve to 1, I could cancel out the 4's and end up with (5+1) / (9+ 2).
If 2 + 2 is meant to be 9 + 2, and the result is 6/11 then that would be acceptable.
Either way, if I'm correct or incorrect, the hardest part about learning is the leaning part.
Aside from that, knowledge is power.
I'm one step closer.
Thank you for your time.
Thank you very much for your reply, it all makes sense.
There are a lot of rules, I've discovered (or rediscovered) and study math on my own because anything
school related is mostly gone so there's a few things I'm not totally certain about when it comes to
rearranging formulas.
Through my studies I've lost touch with the basics while focusing on different concepts, which ironically
almost always involves ratios.
I haven't done any research on this question so here's an interpretation....for better or worse....
Otis, thank you, I did notice the x without the coefficient. I do know about that rule but wanted
to play it safe.
I was familiar with the Zero-Product Property and the Quadratic Formula but not Completing the Square
so thank you for that.
I'm trying to keep things a visual and geometric as possible, and breaking the solution down
into it's original form as opposed to using a formula.
Maybe I'm wrong but formulas to me are a final statement not an explanation. I like the idea
of tearing apart formulas and discovering the 'hidden message' within.
Steven G, the first is correct?
Order of operations requires to deal with the multiplication, if I do that for both I still get 9/17. The denominator product
is 8 because 4 is factoring in with the 2.
I'm not sure where the 2 + 2 comes from in the second fraction equating to (5 + 1) / (2 + 2). I wouldn't seperate
this fraction because it's not expressed as (5 + 1) / 2 which would produce 5/2 + 1/2. As expressed it would be 6/4,
but since 4 is present in the first expression they would resolve to 1, I could cancel out the 4's and end up with (5+1) / (9+ 2).
If 2 + 2 is meant to be 9 + 2, and the result is 6/11 then that would be acceptable.
Either way, if I'm correct or incorrect, the hardest part about learning is the leaning part.
Aside from that, knowledge is power.
I'm one step closer.
Thank you for your time.
Steven G
Elite Member
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- Dec 30, 2014
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- 14,561
I meant (5+1)/(9+2)
Steven G
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You say that the 1st is correct but then you go on to say that in \(\displaystyle \dfrac{5+1*4}{9+2*4}\) that the 4s cancel out and end up with (5+1) / (9+ 2)= 6/11. The 1st way is 9/17. Which is correct?
I didn't realize I did that...sorry.
I tried to draw this out on a line
_________________5__________4
____________________________9__________4__________4
The ratio should stay the same if I cancel the 4's but
9/17 = 0.529.... and 5/13 = 0.384....
6/11 is close to 9/17 but not close enough.
I'll say 9/17, the first one is correct.
I tried to draw this out on a line
_________________5__________4
____________________________9__________4__________4
The ratio should stay the same if I cancel the 4's but
9/17 = 0.529.... and 5/13 = 0.384....
6/11 is close to 9/17 but not close enough.
I'll say 9/17, the first one is correct.
I apologize for the mix up, I'm not sure what's actually visible but I intended for
the posting at 10:54 to be deleted because I don't believe I should take apart the fraction in that manner.
I had the thought a moment too late. ................................(It was deleted per your suggestion)
The posting at 11:36 is what I was going for.
the posting at 10:54 to be deleted because I don't believe I should take apart the fraction in that manner.
I had the thought a moment too late. ................................(It was deleted per your suggestion)
The posting at 11:36 is what I was going for.
Last edited by a moderator:
Steven G
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Both your answers are incorrect:
\(\displaystyle \dfrac{5+1}{9+2}\) = 6/11
\(\displaystyle \dfrac{5+1*4}{9+2*4}\)=9/17
Before adding, you need to know what you're adding! In 9 + 2*4, for example, you're adding 9 and 8 which is 17.
How are you getting 19/18 and 76/72?
I want to go back to the very beginning. You went
[math]2x^2 = 3 - x \implies \dfrac{2x^2}{x^2} = \dfrac{3 - x}{x^2}[/math]
That implies that x is not zero (because you are not allowed to divide by zero). At this point, you have no idea what number x may be so you are not justified in dividing by zero. You should first test whether x may be zero.
[math]\text {If } x = 0, \text { then } x^2 = 0 \text { and } 3 - x = 3 \implies x^2 \ne 3 - x \implies x \ne 0.[/math]
Dividing by zero can lead to wildly incorrect results. You need to test whether a variable may equal zero before dividing by that variable. So, you made an error at the very start, but fortunately it did no damage in this particular case.
You then went
[math]\dfrac{2x^2}{x^2} = \dfrac{3 - x}{x^2} \implies 2 = \dfrac{3 - x}{x * x} = \dfrac{3 - \cancel x}{x * \cancel x} = \dfrac{3 - 1}{x}[/math]
This cancellation is entirely incorrect. The fact that you got one of the two answers correct is pure accident. In short, you made two errors. One turned out to have no effect in this case, and the other turned out by accident to get one out of two correct answers. You were very lucky.
I think we should talk about when cancellation in fractions is valid and why, but that goes away from just quadratics.
[math]2x^2 = 3 - x \implies \dfrac{2x^2}{x^2} = \dfrac{3 - x}{x^2}[/math]
That implies that x is not zero (because you are not allowed to divide by zero). At this point, you have no idea what number x may be so you are not justified in dividing by zero. You should first test whether x may be zero.
[math]\text {If } x = 0, \text { then } x^2 = 0 \text { and } 3 - x = 3 \implies x^2 \ne 3 - x \implies x \ne 0.[/math]
Dividing by zero can lead to wildly incorrect results. You need to test whether a variable may equal zero before dividing by that variable. So, you made an error at the very start, but fortunately it did no damage in this particular case.
You then went
[math]\dfrac{2x^2}{x^2} = \dfrac{3 - x}{x^2} \implies 2 = \dfrac{3 - x}{x * x} = \dfrac{3 - \cancel x}{x * \cancel x} = \dfrac{3 - 1}{x}[/math]
This cancellation is entirely incorrect. The fact that you got one of the two answers correct is pure accident. In short, you made two errors. One turned out to have no effect in this case, and the other turned out by accident to get one out of two correct answers. You were very lucky.
I think we should talk about when cancellation in fractions is valid and why, but that goes away from just quadratics.
Steven G
Elite Member
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(5+1)/(9+2) is NOT (5/9) + 1/2)=19/18
Think about this: Since (5+1)=(1+5) we have (5+1)/(9+2)= (1+5)/(9+2)= (using your method) = (1/9) + 5/2) = 47/18
So is (5+1)/(9+2) = 19/18 or 47/18 or something else?
What if you have (5)/(9+2)=(5+0)/(9+2) =? (5/9) + (0/2) = (5/9) + 0 = 5/9????
mmm4444bot
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Hello. Here's a tip for future use. Please reference posts by their number. Timestamps differ from zone to zone and are continuously changing.
Cheers ?
[imath]\;[/imath]
Well, thank you very much
I realize my issue is a couple things but how I approach defining unknowns, I think is my problem.
I had a feeling I was not comprehending the question because of how many different attempts I made
with incorrect results. I am thankful for your time because I do like to know what things are not, and now
I have a few examples of that. I did feel like I was wasting your time at this point.
Assuming x = 0 is something I'm aware of but didn't put that piece together because I created the
question so I must be able to undo what had been done. But I find it amazing how hard it is to undo.
That is the most confusing part to me and I don't like just using a formula like the Quadratic Formula
without knowing how it was derived.
I also had a feeling I was getting lucky I just didn't know exactly why, other than improper fraction cancelling.
I now know what direction to take in terms of solving for unknowns and interpreting the question.
Your mention of testing the variable if it may equal zero before dividing, and your final statement about how and why is the answer.
I know that I don't know, but I don't know what things I don't know....lol?
I assure you I've learned from this experience and your time isn't wasted, I hope that's ok with
you guys.
I'll never forget this experience, you're all excellent teachers, I'm glad you made me think about
things and didn't just give me the answer, it means a lot.
I realize my issue is a couple things but how I approach defining unknowns, I think is my problem.
I had a feeling I was not comprehending the question because of how many different attempts I made
with incorrect results. I am thankful for your time because I do like to know what things are not, and now
I have a few examples of that. I did feel like I was wasting your time at this point.
Assuming x = 0 is something I'm aware of but didn't put that piece together because I created the
question so I must be able to undo what had been done. But I find it amazing how hard it is to undo.
That is the most confusing part to me and I don't like just using a formula like the Quadratic Formula
without knowing how it was derived.
I also had a feeling I was getting lucky I just didn't know exactly why, other than improper fraction cancelling.
I now know what direction to take in terms of solving for unknowns and interpreting the question.
Your mention of testing the variable if it may equal zero before dividing, and your final statement about how and why is the answer.
I know that I don't know, but I don't know what things I don't know....lol?
I assure you I've learned from this experience and your time isn't wasted, I hope that's ok with
you guys.
I'll never forget this experience, you're all excellent teachers, I'm glad you made me think about
things and didn't just give me the answer, it means a lot.
oops....didn't think of that, thank you.
The quadratic formula is derived from completing the square.
The general quadratic equation is [imath]ax^2 + bx + c = 0, \text { if } a \ne 0.[/imath]
Now, using basic algebra, we go
[math]ax^2 + bx + c = 0, \text { if } a \ne 0, \implies \\ \dfrac{ax^2}{a} + \dfrac{bx}{a} + \dfrac{c}{a} = \dfrac{0}{a} = 0 \implies \\ x^2 + 2 * x * \dfrac{b}{2a} = - \dfrac{c}{a}.[/math]
The truth of that restatement is easily verified, but the purpose of it may be obscure. Consider
[math](p + q)^2 = (p + q)(p + q) = p(p + q) + q(p + q) = p^2 + pq + pq + q^2 \implies \\ (p + q)^2 = p^2 + 2 * p * q + q^2.[/math]
When we square a binomial, we get a trinomial, with a middle term of 2 times the product of the two terms in the binomial.
But that is the exact pattern we have with [imath]2 * x * \dfrac{b}{2a}.[/imath]
[math]\text {Let } p = x \text { and } q = \dfrac{b}{2a} \implies \\ \left ( x + \dfrac{b}{2a} \right )^2 = (p + q)^2 = p^2 + 2 * p * q + q^2 \implies \\ \left ( x + \dfrac{b}{2a} \right )^2 = x^2 + 2 * x * \dfrac{b}{2a} + \left ( \dfrac{b}{2a} \right )^2.[/math]
So, we add [imath]\left ( \dfrac{b}{2a} \right )^2[/imath] to both sides of our equation.
[math] x^2 + 2 * x * \dfrac{b}{2a} = - \dfrac{c}{a} \implies \\ x^2 + 2 * x * \dfrac{b}{2a} + \left ( \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies \\ \left ( x + \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} = \dfrac{b^2}{4a^2} - \dfrac{4ac}{4a^2} = \dfrac{b^2 - 4ac}{4a^2} \implies \\ x + \dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}} = \dfrac{\pm \sqrt{b^2 - 4ac}}{\sqrt{4a^2}} = \dfrac{\pm \sqrt{b^2 - 4ac}} {2a} \implies \\ x = - \dfrac{b}{2a} + \dfrac{\pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. [/math]
Although it is a long sequence of manipulations, they are all simple in themselves. The quadratic formula is not a magic black box. It is what follows logically from completing the square on the general quadratic.
The general quadratic equation is [imath]ax^2 + bx + c = 0, \text { if } a \ne 0.[/imath]
Now, using basic algebra, we go
[math]ax^2 + bx + c = 0, \text { if } a \ne 0, \implies \\ \dfrac{ax^2}{a} + \dfrac{bx}{a} + \dfrac{c}{a} = \dfrac{0}{a} = 0 \implies \\ x^2 + 2 * x * \dfrac{b}{2a} = - \dfrac{c}{a}.[/math]
The truth of that restatement is easily verified, but the purpose of it may be obscure. Consider
[math](p + q)^2 = (p + q)(p + q) = p(p + q) + q(p + q) = p^2 + pq + pq + q^2 \implies \\ (p + q)^2 = p^2 + 2 * p * q + q^2.[/math]
When we square a binomial, we get a trinomial, with a middle term of 2 times the product of the two terms in the binomial.
But that is the exact pattern we have with [imath]2 * x * \dfrac{b}{2a}.[/imath]
[math]\text {Let } p = x \text { and } q = \dfrac{b}{2a} \implies \\ \left ( x + \dfrac{b}{2a} \right )^2 = (p + q)^2 = p^2 + 2 * p * q + q^2 \implies \\ \left ( x + \dfrac{b}{2a} \right )^2 = x^2 + 2 * x * \dfrac{b}{2a} + \left ( \dfrac{b}{2a} \right )^2.[/math]
So, we add [imath]\left ( \dfrac{b}{2a} \right )^2[/imath] to both sides of our equation.
[math] x^2 + 2 * x * \dfrac{b}{2a} = - \dfrac{c}{a} \implies \\ x^2 + 2 * x * \dfrac{b}{2a} + \left ( \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} \implies \\ \left ( x + \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{c}{a} = \dfrac{b^2}{4a^2} - \dfrac{4ac}{4a^2} = \dfrac{b^2 - 4ac}{4a^2} \implies \\ x + \dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}} = \dfrac{\pm \sqrt{b^2 - 4ac}}{\sqrt{4a^2}} = \dfrac{\pm \sqrt{b^2 - 4ac}} {2a} \implies \\ x = - \dfrac{b}{2a} + \dfrac{\pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. [/math]
Although it is a long sequence of manipulations, they are all simple in themselves. The quadratic formula is not a magic black box. It is what follows logically from completing the square on the general quadratic.
Otis
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Hi JH. If you check out the 'Completing the Square' link in post #2, you'll find both Jeff's explanation and a graphical representation of the steps.
Yes, you're correct. Formulas are not intended to explain; they are a direct (mechanical) path to a solution. The explanations come from taking a math course or studying bona fide math texts. Such resources (courses and texts) are available for free online.
How familiar are you with graphs of quadratic equations? There are direct (but hidden?) correspondences to discover between specific parts of formulas and specific aspects of graphs.
[imath]\;[/imath]