Quadratic equation help

Flo Rida

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Jun 8, 2020
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130
I haven't had trouble until I got to this question:

2x2+3x+6=4x

i subtract 4x and i get 2x2-x+6=0
then i multiply a term with c term and I get x2-x+12=0.

i dont know any factors of 12 that add to -1. Are there any?


thanks
flo
 

Flo Rida

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Jun 8, 2020
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My teacher taught me to multiply a by c to get x2 alone
ex:
2x2-5x+2=0 becomes x2-5x+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
 

lev888

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Jan 16, 2018
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My teacher taught me to multiply a by c to get x2 alone
ex:
2x2-5x+2=0 becomes x2-5x+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
Again, step by step, how did 2x2-5x+2=0 become x2-5x+4=0?
 

lev888

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Jan 16, 2018
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2 is removed from the a-term (x2) and distributed to the c-term (2), which becomes 4.
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?
 

Flo Rida

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Jun 8, 2020
Messages
130
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?
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.
 

lev888

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Can you post a screen cap of the problem? And examples of your teacher's method if available.
 

lev888

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Example w/other question (#6 only)
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`
 

Flo Rida

Junior Member
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Jun 8, 2020
Messages
130
wolframalpha.com says this equation has only complex roots, that's why I suspect you have a typo.
Ok, well I guess that clears it up.
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`
thank you so much! this helps with the other ones.
 

JeffM

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Sep 14, 2012
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6,928
i tried solving it but unfortunately i was unable to solve it using factoring
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.

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?
 

ghanasur

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Sep 24, 2021
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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?
Thanks for you Quick reply, it cleared my doubts.
 
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