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- Thread starter Flo Rida
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Please explain this step.then i multiply a term with c term and I get x2-x+12=0.

ex:

2x

now, it would be (x-1)(x-4) and then i have to divide by the a term

the answer would be (2x-1)=0 and (x-2)=0,

problem with this question above is that i cant find factors that multiply to 12 and add to -1

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Again, step by step, how did 2x^{2}alone

ex:

2x^{2}-5x+2=0 becomes x^{2}-5x+4.

now, it would be (x-1)(x-4) and then i have to divide by the a term

the answer would be (2x-1)=0 and (x-2)=0,

problem with this question above is that i cant find factors that multiply to 12 and add to -1

2 is removed from the a-term (xAgain, step by step, how did 2x^{2}-5x+2=0 become x^{2}-5x+4=0?

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It's not clear to me what you are trying to do. But this approach certainly changes the equation. Do you see it? You can manipulate equations _only_ by applying the same operations to both sides. If you want to make the first coefficient 1 you need to divide both sides by 2. Maybe this is the method your teacher used?2 is removed from the a-term (x^{2}) and distributed to the c-term (2), which becomes 4.

No, she's been using this, and it worked for all the others, but I really don't understand why it isn't working for this.It's not clear to me what you are trying to do. But this approach certainly changes the equation. Do you see it? You can manipulate equations _only_ by applying the same operations to both sides. If you want to make the first coefficient 1 you need to divide both sides by 2. Maybe this is the method your teacher used?

Ok give me a few minutesCan you post a screen cap of the problem? And examples of your teacher's method if available.

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wolframalpha.com says this equation has only complex roots, that's why I suspect you have a typo.Original question:

Solve the equations by factoring.

2x^{2}+3x+6=4x

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This is ridiculous. The answer is correct, but I have no idea what this method is and why we need to make the solution more complicated than necessary.Example w/other question (#6 only)

`2x^2-32 = 0`

Divide both sides by 2:

`x^2-16 = 0`

`(x-4)(x+4) = 0`

`x = -4, 4`

Ok, well I guess that clears it up.wolframalpha.com says this equation has only complex roots, that's why I suspect you have a typo.

thank you so much! this helps with the other ones.This is ridiculous. The answer is correct, but I have no idea what this method is and why we need to make the solution more complicated than necessary.

`2x^2-32 = 0`

Divide both sides by 2:

`x^2-16 = 0`

`(x-4)(x+4) = 0`

`x = -4, 4`

I checked my calculator and it does involve roots. i asked my math teacher and she removed the question and said it was a typoHow did you solve the original problem?

First, this thread is months old.i tried solving it but unfortunately i was unable to solve it using factoring

Second, every quadratic can be factored using complex numbers although not all quadratics can be factored using real numbers.

Third, as was discussed at length in the thread, this quadratic is one of those that cannot be factored using real numbers.

So what is your question?

Thanks for you Quick reply, it cleared my doubts.First, this thread is months old.

Second, every quadratic can be factored using complex numbers although not all quadratics can be factored using real numbers. mcdvoice

Third, as was discussed at length in the thread, this quadratic is one of those that cannot be factored using real numbers.

So what is your question?