Solve and show and explain work
x^2 + bx + c = 0

- Firejay5

Solve and show and explain work
x^2 + bx + c = 0

- chestercat

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- anonymous

Try applying the quadratic equation to this one.

- Firejay5

@LogicalApple

- anonymous

\[x = \frac{ -b \pm \sqrt{b^{2} - 4ac} }{ 2a }\]

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## More answers

- Firejay5

@BubbaMurphy Do you have a different way to do this problem???

- anonymous

LogicApple's way is definitely the way to do it.

- anonymous

in this case, a=1

- Firejay5

Is that really how yo would solve it

- Firejay5

would A, B, and C = 1

- anonymous

Based on the equation you supplied, a = 1, b = b, and c = c.

- anonymous

They are coefficients of whatever quadratic you are talking about. Did you have a quadratic in mind?

- Firejay5

what you mean

- 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

Yea I know that

- anonymous

Oh, ok... so what is your question?

- Firejay5

We have to solve x^2 + bx + c = 0 for bonus and I have no idea what to do

- 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

We didn't learn that equation you gave me

- anonymous

What about completing the square ?

- Firejay5

so is that the answer

- anonymous

Yes, but you would arrive at the same answer if you complete the square. Both techniques are valid for solving quadratic equations.

- Firejay5

I have to show my work, so how would I show it

- 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

I think you can use sum and product of roots:
\[\alpha +\beta= \frac{ -b }{ a }\]
\[\alpha \beta=\frac{ c }{ a }\]

- Firejay5

b^2/4

- anonymous

##### 1 Attachment

- anonymous

|dw:1359082578453:dw|

- anonymous

(b/2)^2 = b^2/4

- Firejay5

After (x + b/2)^2 = b^2/4 - c. Find the square root of both.

- anonymous

Yes

- Firejay5

I got x + b/2 = b/4 - c. Is that correct?

- anonymous

no the square root of both sides gives:
x + b/2 = ± sqrt(b^2/4 - c)

- Firejay5

what then

- anonymous

All the steps are in the picture attached.

##### 1 Attachment

- Firejay5

I am sorry the last step in attachment is the answer

- anonymous

Oh ok did you get it figured out ?

- Firejay5

so x = +/- sqrt b^2 - 4c /2 is the answer

- anonymous

|dw:1359083804186:dw|
yes

- Firejay5

Will you forgive me @LogicalApple ?

- anonymous

What for? It's a learning experience

- Firejay5

I probably annoyed you a lot

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