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