Firejay5
  • Firejay5
Solve and show and explain work x^2 + bx + c = 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Try applying the quadratic equation to this one.
Firejay5
  • Firejay5
@LogicalApple
anonymous
  • anonymous
\[x = \frac{ -b \pm \sqrt{b^{2} - 4ac} }{ 2a }\]

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More answers

Firejay5
  • Firejay5
@BubbaMurphy Do you have a different way to do this problem???
anonymous
  • anonymous
LogicApple's way is definitely the way to do it.
anonymous
  • anonymous
in this case, a=1
Firejay5
  • Firejay5
Is that really how yo would solve it
Firejay5
  • Firejay5
would A, B, and C = 1
anonymous
  • anonymous
Based on the equation you supplied, a = 1, b = b, and c = c.
anonymous
  • anonymous
They are coefficients of whatever quadratic you are talking about. Did you have a quadratic in mind?
Firejay5
  • Firejay5
what you mean
anonymous
  • anonymous
\[6x^{2} - 9x + 12 = 0\] \[-5x^{2} + x - 10 = 0\] \[x^2 + x - 1 = 0\] These are some examples of quadratics, all set equal to 0.
Firejay5
  • Firejay5
Yea I know that
anonymous
  • anonymous
Oh, ok... so what is your question?
Firejay5
  • Firejay5
We have to solve x^2 + bx + c = 0 for bonus and I have no idea what to do
anonymous
  • anonymous
I suggested using the quadratic equation. The only difference between x^2 + bx + c = 0 and the quadratics that I mentioned is that the coefficients in your equation are only known by their variable letter. But that is ok, you can still apply the quadratic formula to them. The solutions to x^2 + bx + c = 0 is: \[\frac{ -b \pm \sqrt{b^{2} - 4c} }{ 2 }\] This is the same thing I wrote before, except I let a = 1 because the coefficient of x^2 in your equation is 1.
Firejay5
  • Firejay5
We didn't learn that equation you gave me
anonymous
  • anonymous
What about completing the square ?
Firejay5
  • Firejay5
so is that the answer
anonymous
  • anonymous
Yes, but you would arrive at the same answer if you complete the square. Both techniques are valid for solving quadratic equations.
Firejay5
  • Firejay5
I have to show my work, so how would I show it
anonymous
  • anonymous
Well if you're using the quadratic equation you could say "applying the quadratic equation". But if you wanted to complete the square then: |dw:1359081924838:dw|
anonymous
  • anonymous
I think you can use sum and product of roots: \[\alpha +\beta= \frac{ -b }{ a }\] \[\alpha \beta=\frac{ c }{ a }\]
Firejay5
  • Firejay5
b^2/4
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
|dw:1359082578453:dw|
anonymous
  • anonymous
(b/2)^2 = b^2/4
Firejay5
  • Firejay5
After (x + b/2)^2 = b^2/4 - c. Find the square root of both.
anonymous
  • anonymous
Yes
Firejay5
  • Firejay5
I got x + b/2 = b/4 - c. Is that correct?
anonymous
  • anonymous
no the square root of both sides gives: x + b/2 = ± sqrt(b^2/4 - c)
Firejay5
  • Firejay5
what then
anonymous
  • anonymous
All the steps are in the picture attached.
1 Attachment
Firejay5
  • Firejay5
I am sorry the last step in attachment is the answer
anonymous
  • anonymous
Oh ok did you get it figured out ?
Firejay5
  • Firejay5
so x = +/- sqrt b^2 - 4c /2 is the answer
anonymous
  • anonymous
|dw:1359083804186:dw| yes
Firejay5
  • Firejay5
Will you forgive me @LogicalApple ?
anonymous
  • anonymous
What for? It's a learning experience
Firejay5
  • Firejay5
I probably annoyed you a lot

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