anonymous
  • anonymous
FACTOR 2x^2 + x - 28 = 0 There should be 2 x's. I am stuck at 2(x^2+x-14)=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@haleyelizabeth2017 Lists steps
haleyelizabeth2017
  • haleyelizabeth2017
Wait, it was originally \(2x^2+2x-28?\)
haleyelizabeth2017
  • haleyelizabeth2017
What you have is the simplest it can go.

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More answers

anonymous
  • anonymous
Originally 2x^2+x-28=0
haleyelizabeth2017
  • haleyelizabeth2017
Okay, then what you have so far is not correct.
anonymous
  • anonymous
oh yea i made a mistake, i thought the GCF was 2 when it can't be
anonymous
  • anonymous
what can be mult. to make -56 but added to make 1
haleyelizabeth2017
  • haleyelizabeth2017
-7 and 4 multiply to get -28, okay? So then...if one side is (2x+/-_)(x+/-_) we can plug in those two numbers to try and see which works.
anonymous
  • anonymous
im guessing that another way to do it Eli?
haleyelizabeth2017
  • haleyelizabeth2017
So...let's try (2x-7)(x+4) first. We get \(2x^2+8x-7x-28\) which simplified, would work. So, (2x-7)(x+4) is the correct.
haleyelizabeth2017
  • haleyelizabeth2017
correct answer* ^^
anonymous
  • anonymous
so 2x-7=0 is x=7/2 or 3 1/2
anonymous
  • anonymous
and x=-4
haleyelizabeth2017
  • haleyelizabeth2017
Normally, it is left as an improper fraction, so x=7/2 is better. and x=-4 is correct as well.
haleyelizabeth2017
  • haleyelizabeth2017
You could have also used the quadratic formula.
haleyelizabeth2017
  • haleyelizabeth2017
It is a lot easier. Or, even complete the square works too
haleyelizabeth2017
  • haleyelizabeth2017
\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] \[ax^2+bx+c=0\]
anonymous
  • anonymous
Ok, i got a question... the formula is ax^2+bx+c... i was told that you mult. a*c and the number that you get has to be mult. by x number and x number, but those two number must also be able to be added together to equal B... did that make sense
anonymous
  • anonymous
Is there an exception to this??
haleyelizabeth2017
  • haleyelizabeth2017
Huh?
haleyelizabeth2017
  • haleyelizabeth2017
You don't have to do anything to the a b or c. just plug the coefficients into the quadratic formula and solve. Lets use your problem for example. We were given \[2x^2+x-28\] A would be 2, b would be 1, and c would be -28.
haleyelizabeth2017
  • haleyelizabeth2017
\[\frac{ -(1) \pm \sqrt{(1)^2-4(2)(-28)} }{ 2(2) }\]\[\frac{ -1 \pm \sqrt{1+224} }{ 4 }\]and so on
haleyelizabeth2017
  • haleyelizabeth2017
Did that make sense?
Hero
  • Hero
@Beleaguer did you figure out What the two numbers are?
Hero
  • Hero
In other words, can you solve m+n =1 mn =-56

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