anonymous
  • anonymous
How do I do 7x^2 + x + 2 = 0 by the quadratic formula. Thanks for your help in advance!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DanJS
  • DanJS
do you know the quadratic formula?
anonymous
  • anonymous
I think so
DanJS
  • DanJS
ax^2 + bx + c = 0 7x^2 + 1x + 2 = 0

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DanJS
  • DanJS
use those values for a, b , and c
anonymous
  • anonymous
a = 7 b = 1 c = 2?
DanJS
  • DanJS
yep.. \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
anonymous
  • anonymous
|dw:1441589861398:dw|
DanJS
  • DanJS
the stuff under the square root turns out to be negative, so no real solutions to this one, less you want the complex roots
DanJS
  • DanJS
\[\sqrt{-1}=i\]
anonymous
  • anonymous
I do not know how to solve this from start to finish.
DanJS
  • DanJS
you have it right so far with the drawing, just have to simplify it down now if needed.
anonymous
  • anonymous
I tried it but when it got down to the imaginary number it got confusing.
DanJS
  • DanJS
\[x=\frac{ -1 \pm \sqrt{-55}}{ 14 }\] \[x=\frac{ -1 }{ 14 }\pm \frac{ \sqrt{55} *\sqrt{-1}}{ 14 }\]
anonymous
  • anonymous
|dw:1441590203726:dw|
DanJS
  • DanJS
just have to remember...\[i=\sqrt{-1}\] put in i for that then solve normally
DanJS
  • DanJS
\[x =\frac{ -1 }{ 14 }+\frac{ \sqrt{55} }{ 14 }*i ~~~~~or~~~~~x =\frac{ -1 }{ 14 }-\frac{ \sqrt{55} }{ 14 }*i\]
anonymous
  • anonymous
Oh thank you I never knew what a imaginary number was.
anonymous
  • anonymous
Does it simplified more is that the final answer?
DanJS
  • DanJS
welcome, If you have to know nothing else about imaginary numbers, then just remember \[i^2=-1~~~~or~~~~~i=\sqrt{-1}\]
DanJS
  • DanJS
that is all you can do to make it more simple looking
anonymous
  • anonymous
Thanks so much! Did the i get in the problem because square root of 55 is a decimal?
DanJS
  • DanJS
no, because the quadratic formula turned out to have a negative number in the square root, remember you cant take the square root of a number smaller than zero
DanJS
  • DanJS
b^2 - 4*a*c was -55
anonymous
  • anonymous
Oh okay I see how it got in then. Thanks for your time and help!

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