## anonymous one year ago Can someone help me:(? Decide which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring. -b b2 - 4ac 2a Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation: 4x2 + 6x + 2 = 0 (2 points)

1. anonymous

this was another one i dint quite get

2. pooja195

The "discriminant" gives this information. It is the square root part of the quadratic formula. $\sqrt{b^2-4ac}$

3. pooja195

From here solve the equation

4. pooja195

Start by listing your abc values a=? b=? c=?

5. pooja195

$\huge~\rm~ax^2+bx+c=0$

6. anonymous

4x^2+6x++2=0

7. anonymous

is that how it should be set up in order

8. pooja195

yes but we need to figure out the values using $\huge~\rm~ax^2+bx+c=0$ That equation can you list just the NUMBERS not the signs of variables

9. anonymous

im confused

10. arindameducationusc

@pooja195 after this, can you solve my question? http://openstudy.com/study#/updates/55b787e4e4b0f0c1f13fd1a9

11. pooja195

Ok anything ^2 is the a value anything with an "x" is the b value anything with no varible is the c value now list the numbers :)

12. anonymous

so b^2 is A, -b is B, and C is 2a?

13. pooja195

wut? o-o wow you really are confused ok lets start from scratch $\huge~\rm~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$ This is the quadratic formula $\huge~\rm~ax^2+bx+c=0$ This is standered form The discernment is The "discriminant" gives this information. It is the square root part of the quadratic formula. $\sqrt{b^2-4ac}$ We need to solve $\huge~\rm~4x^2 + 6x + 2 = 0$ Notice how it all matches up with the standered form equation? $\huge~\rm~ax^2+bx+c=0$

14. anonymous

i know, i said that already

15. pooja195

You have the numbers wrong a=4 b=6 c=2

16. anonymous

oh

17. anonymous

then why do we need just the numbers

18. pooja195

We need just the numbers so that we can solve using the discernment

19. anonymous

ok

20. arindameducationusc

@pooja195 I request you... okay after this

21. pooja195

$\sqrt{b^2-4ac}$ So now we have our abc values $\huge~\rm~ \sqrt{6^2-4(4)(2)}$ Solve that

22. anonymous

6^2-8 under a square root idk what

23. anonymous

i started with parenthesis and then multiplied 4 by (8)

24. pooja195

ok lets do this step by step whats 6^2?

25. anonymous

36

26. pooja195

ok leave that aside whats -4 times 4 ?

27. anonymous

-16

28. pooja195

Ok good -16 x 2=?

29. anonymous

-32

30. pooja195

good whats 36-32?

31. anonymous

4

32. anonymous

The 4ac

33. anonymous

4ac decides whether you get complex number or not.

34. pooja195

good now whats the $\huge~\rm \sqrt4$

35. anonymous

2

36. pooja195

@saseal you are confusing the asker we are already done with the problem.

37. anonymous

oh ok

38. pooja195

39. anonymous

40. anonymous

2a

41. pooja195

its ok :) @kinserkara now 2a it would just be x=2

42. anonymous

so the final answer is x=2

43. pooja195

yep

44. anonymous

ok thanks

45. pooja195

No problem :)