## anonymous one year ago How would you use the quadratic formula to solve x^2+5x=-2. Using compete sentences.

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1. anonymous

@SolomonZelman

2. SolomonZelman

maybe we can work the prob using math, and you do the sentences part alone?

3. anonymous

Okay

4. SolomonZelman

when the equation is $$\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }$$. you use the following formula: $$\normalsize\color{ slate }{\Huge{\bbox[5pt, lightcyan ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{b} \pm\sqrt{ \color{magenta}{b} ^2-4 \color{blue}{a} \color{red}{c}}}{2 \color{blue}{a}} }~ }}}$$ (to transfer your equation to the needed $$\large \color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }$$ form, you have to add 2 to both sides.)

5. anonymous

So it will be x^2+5+2=0

6. SolomonZelman

what do you get after adding 2 to both sides?

7. SolomonZelman

yes

8. SolomonZelman

$$\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }$$  ... comparing ...  $$\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }$$

9. SolomonZelman

Now, please plug in the a b c for me....

10. anonymous

What do you mean by plug in?

11. anonymous

I have no clue how to do this. I am just learning it

12. UsukiDoll

she meant let a = 1, b = 5, and c = 2 we will have this equation $x^2+5x+2=0$

13. anonymous

Okay and then what do I do from there?

14. UsukiDoll

Well, to solve this faster we need the discriminant $b^2-4ac$ so let b = 5, a =1, and c =2

15. SolomonZelman

who is she?

16. UsukiDoll

umm I thought that was you?

17. SolomonZelman

well, if you are thinking of me as of a good looking girl, rather than a bad looking girl, then you in a sense did tackle my identity.

18. UsukiDoll

anyway plug in b =5, c =2, and a = 1 into the discriminant formula $b^2-4ac$ @J_slate23

19. UsukiDoll

as in replace the b with 5, replace the c with 2 and replace the a with 1

20. anonymous

So 5^2=4(1)(2)?

21. UsukiDoll

yes but replace the = with -

22. UsukiDoll

$5^2-4(1)(2)$

23. anonymous

So 5^2-4(1)(2)?

24. UsukiDoll

mhm so now what's 5^2 and what's 4(1)(2)

25. SolomonZelman

$$\normalsize\color{ slate }{\Huge{\bbox[5pt, lightyellow ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{5} \pm\sqrt{ \color{magenta}{5} ^2-4 \color{blue}{(1)} \color{red}{(2)}}}{2 \color{blue}{(1)}} }~ }}}$$ and your equation is $$\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }$$.

26. anonymous

25 and 8

27. UsukiDoll

yes so now what's 25-8

28. anonymous

17

29. UsukiDoll

yes! so for the bottom denominator what is 2(1) ?

30. anonymous

2

31. UsukiDoll

yes

32. UsukiDoll

so now you have $\frac{5 +\sqrt{17}}{2}, \frac{5-\sqrt{17}}{2}$ I split the signs up because the latex for the quadratic formula is big and nasty

33. UsukiDoll

OH MAN I forgot the -!

34. UsukiDoll

$\frac{-5 +\sqrt{17}}{2}, \frac{-5-\sqrt{17}}{2}$

35. anonymous

Is that the answer? If so that was really easy

36. UsukiDoll

yeah.. I just split the sign in the middle up. the latex for the entire quadratic equation with + and - together was a pain

37. UsukiDoll

and 17 is not a perfect square... so we can't simplify .. not to mention 17 is a prime

38. anonymous

So the final answer is -5+-sqrt17/2?

39. UsukiDoll

yeah

40. anonymous

Thank you!

41. anonymous

That is easier then I expected

42. anonymous

Can you check my math for another one? @UskiDoll

43. UsukiDoll

sure

44. anonymous

Solve x^2+4x-12=0 by completing the square. X^2+4x-12=0 Add 12 X^2+4x=12 (X+2)^2-4=12 (x+2)^2=16 (x+2)= 4 or -4 X= 6 or -6 is my answer

45. anonymous

Is that correct @UskiDoll

46. UsukiDoll

$x^2+4x-12=0$ $x^2+4x=12$ $4 \times \frac{1}{2} = \frac{4}{2} = 2$ $2^2 = 4$ $x^2+4x+4=12+4$ $x^2+4x+4=16$ $(x+2)^2 = 16$ square root on both sides $\sqrt{(x+2)^2} = \sqrt{16}$ $x+2=4, x+2=-4$ $x=2, x=-6$

47. UsukiDoll

x+2 = 4 on your part is wrong. subtract 2 from both sides, but -6 is correct

48. anonymous

Okay. Thank you

49. anonymous

plug it in -b+_ b2-4ac $\sqrt{b2-4ac}$ / 4a good luck

50. UsukiDoll

The problem is already done ^_^