## Firejay5 Group Title Solve and show and explain work x^2 + bx + c = 0 one year ago one year ago

1. LogicalApple Group Title

Try applying the quadratic equation to this one.

2. Firejay5 Group Title

@LogicalApple

3. LogicalApple Group Title

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

4. Firejay5 Group Title

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

5. BubbaMurphy Group Title

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

6. BubbaMurphy Group Title

in this case, a=1

7. Firejay5 Group Title

Is that really how yo would solve it

8. Firejay5 Group Title

would A, B, and C = 1

9. LogicalApple Group Title

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

10. LogicalApple Group Title

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

11. Firejay5 Group Title

what you mean

12. LogicalApple Group Title

$6x^{2} - 9x + 12 = 0$ $-5x^{2} + x - 10 = 0$ $x^2 + x - 1 = 0$ These are some examples of quadratics, all set equal to 0.

13. Firejay5 Group Title

Yea I know that

14. LogicalApple Group Title

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

15. Firejay5 Group Title

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

16. LogicalApple Group Title

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.

17. Firejay5 Group Title

We didn't learn that equation you gave me

18. LogicalApple Group Title

What about completing the square ?

19. Firejay5 Group Title

so is that the answer

20. LogicalApple Group Title

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

21. Firejay5 Group Title

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

22. LogicalApple Group Title

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|

23. Azteck Group Title

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

24. Firejay5 Group Title

b^2/4

25. LogicalApple Group Title

26. LogicalApple Group Title

|dw:1359082578453:dw|

27. LogicalApple Group Title

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

28. Firejay5 Group Title

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

29. LogicalApple Group Title

Yes

30. Firejay5 Group Title

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

31. LogicalApple Group Title

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

32. Firejay5 Group Title

what then

33. LogicalApple Group Title

All the steps are in the picture attached.

34. Firejay5 Group Title

I am sorry the last step in attachment is the answer

35. LogicalApple Group Title

Oh ok did you get it figured out ?

36. Firejay5 Group Title

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

37. LogicalApple Group Title

|dw:1359083804186:dw| yes

38. Firejay5 Group Title

Will you forgive me @LogicalApple ?

39. LogicalApple Group Title

What for? It's a learning experience

40. Firejay5 Group Title

I probably annoyed you a lot