## some.random.cool.kid one year ago Solve the equation 4x2 − 12x + 3 = 0 by completing the square. Verify your answer using the quadratic formula.

1. some.random.cool.kid

@Nnesha

2. some.random.cool.kid

@nincompoop

3. some.random.cool.kid

i have a few of these this is just the first that i need help with...

4. some.random.cool.kid

@ParthKohli

5. some.random.cool.kid

@peachpi

6. some.random.cool.kid

@whpalmer4

7. some.random.cool.kid

ik the forumla is this: x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

8. Nnesha

well we should use complete the square method you can move the constant term to the right side

9. some.random.cool.kid

x is the term correct?

10. Nnesha

$\huge\rm 4x^2 − 12x + 3 = 0$ subtract 3 both sides

11. Nnesha

not just x term are divided by plus or negative sign

12. some.random.cool.kid

uh... subtract the 4x^2 by 3 and the 12 by 3?

13. Nnesha

$\huge\rm Ax^2+Bx+C=0$quadratic e quation where A=leading coefficient B=middle term C=constant term(number with no variable )

14. Nnesha

|dw:1443453772906:dw| you can't combine -3 and 4x^2 they arn't like terms

15. some.random.cool.kid

im not sure what you do next after you get -3..

16. some.random.cool.kid

divide?

17. Nnesha

|dw:1443453908051:dw| we should complete the square of the left part

18. Nnesha

$\huge\rm \color{Red}{4x^2+12x}=-3$ what is the common factor at left side ?

19. some.random.cool.kid

the x?

20. Nnesha

no just take out the common number not the varaible

21. some.random.cool.kid

uh ... im confused. am i supposed to take out the 4 and 12 and divide them?

22. some.random.cool.kid

besides the x i see nothing in common...

23. Nnesha

well it's 4x^2+12x so we just need to take out the common number

24. Nnesha

what are the factors of 4 and 12 ?

25. some.random.cool.kid

1, 2, 4 ... 4 is best factor... then i divide both by four to get 3 right?

26. Nnesha

you need GREATEST COMMON FACTOR 1,2,4 are factors of 4 1,2,3,4,6,12 are factors of 12 so which number is common ?

27. Nnesha

greatest number **

28. some.random.cool.kid

uh i said 4...

29. some.random.cool.kid

is there one greater than thta?

30. some.random.cool.kid

that*

31. Nnesha

i don't understand what u mean "divide both to get 3"

32. Nnesha

$\huge\rm 4(x^2+3x)=-3$when you take out 4 from 4 u will get 1

33. some.random.cool.kid

you still have the 12x

34. some.random.cool.kid

i was saying didvide the 4x^2 and 12x by the common factor to simplify the solution.

35. Nnesha

now we can complete the square of (x^2+3x) we should divide b by b here is an example $\huge\rm a(x^2+bx)=-C$ divide b by 2 $\huge\rm a(x+\frac{ b }{ 2 })^2=-C + (\frac{ b }{ 2})^2$ and then add(b/2)^2 to both sides

36. Nnesha

yes right. i read it wrong. so sorry

37. Nnesha

$\huge\rm 4(x^2+3x)=-3$ take half of the x term and then square it add square to the both sides

38. some.random.cool.kid

you have to x's which one do i take half of?

39. some.random.cool.kid

two*

40. Nnesha

first one is x^2 2nd one is x

41. some.random.cool.kid

so take half of the second one?

42. Nnesha

yes right

43. some.random.cool.kid

uh... ok so now you'll have- x+ 3?

44. some.random.cool.kid

thats not a negative sign

45. Nnesha

x+3 how ? we should take half of 3 in other words just divide b by 2

46. some.random.cool.kid

how am i dividing 3x by 2?

47. some.random.cool.kid

/i thought you wanted me to take away x on both sides..

48. Nnesha

here is an example $\huge\rm a(x^2+bx)=-C$ divide b by 2 $\huge\rm a(x+\frac{ b }{ 2 })^2=-C + (\frac{ b }{ 2})^2 \times a$ and then add(b/2)^2 to both sides here is an example

49. Nnesha

well first we need to divide b by 2 take square of (b/2)^2 then add (b/2)^2 to both sides

50. Nnesha

what is b in ur equation ?

51. some.random.cool.kid

the 3x?

52. Nnesha

just b

53. Nnesha

3 =b not 3x x is variable

54. Nnesha

yes right $\huge\rm 4(x^2+\color{blue}{\frac{ 3x }{ 2 }}+\color{Red}{(\frac{ 3 }{ 2 })^2}) = -3 +\color{red}{(\frac{3}{2})^2 } \times 4$ look at the blue term i took half of 3 and then add (b/2)^2 to both sides and sicne we got the common factor we should multiply (b/2)^2 by 4 at right side

55. some.random.cool.kid

do we get rid of negative 3 with the positive 3

56. Nnesha

positive 3 ?? it's 3x you can't combine 3x with -3 they are not like terms

57. some.random.cool.kid

then your left with whats in blue right?

58. Nnesha

right we can factor this quadraticc equation (x^2 +3/2x + (3/2)^2) or look at this example $\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2$

59. Nnesha

you need to convert left side in squared form

60. some.random.cool.kid

so its a bit of a leap here but i believe the simplified form would be -3 +- 2$\sqrt{6}$ over 2?

61. some.random.cool.kid

everything over 2*

62. Nnesha

and how did you get that ?

63. some.random.cool.kid

well you told me that like terms cant combine

64. some.random.cool.kid

leaving negative 3 alone.

65. Nnesha

look at this example $\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2$ and tell me how would you write $(x^2+3x) = ????$

66. some.random.cool.kid

then you said something about b being divided by2? which i believe gave me that..

67. Nnesha

how did you write $(x^2+3x)$ in squared form ?

68. some.random.cool.kid

mmm

69. some.random.cool.kid

3x^2? idk

70. some.random.cool.kid

ik your left with one x right?

71. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nnesha look at this example $\huge\rm (x^2+bx) =(x+\frac{ b }{ 2 })^2$ and tell me how would you write $(x^2+3x) = ????$ $$\color{blue}{\text{End of Quote}}$$

72. some.random.cool.kid

thats where im lost though...

73. Nnesha

yea that's obvious we have to solve for x

74. Nnesha

then how you got the final x value ?

75. some.random.cool.kid

i got what?

76. Nnesha

here is another example $(x^2+6x)=(x+\frac{ 6 }{ 2})^2=(x+3)^2$

77. some.random.cool.kid

wait but isn't that similar to what i had?

78. Nnesha

similar to what ?

79. Nnesha

$\huge\rm 4(x^2+\color{blue}{\frac{ 3x }{ 2 }}+\color{Red}{(\frac{ 3 }{ 2 })^2}) = -3 +\color{red}{(\frac{3}{2})^2 } \times 4$ $4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4$ according to what i said (x^2+bx)=(x+b/2)^2

80. Nnesha

solve this to get the answer $4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4$

81. Nnesha

here is how we can write in squared form $(x^2+bx)=(x+\frac{ b }{ 2 })^2$ $(x^2+6x)=x^2+6x+\color{reD}{9}$ i divided 6 by 2 which is equal to 3 then squared 3 that's how i got 9 now we can factor x^2+6x+9 (x+3)(x+3) are factors of that quadratic equation which is same as (x+3)^2 so in short we can do this $(x^2+bx)=(x+\frac{ b }{ 2})^2$$(x^2+6x)=(x+\frac{ 6 }{ 2 })^2=(x+3)^2$

82. Nnesha

feel free to ask question about that^let me know if you have question on that

83. some.random.cool.kid

ok thank you

84. some.random.cool.kid

i had another one mind helping?

85. some.random.cool.kid

@Nnesha

86. Nnesha

did you get this one ?

87. some.random.cool.kid

i believe i did

88. some.random.cool.kid

Solve x2 − 5x + 1 = 0 using the quadratic formula.

89. Nnesha

alright i'll try

90. some.random.cool.kid

i know the first step to this ...

91. Nnesha

can u make a new post bec this one is laggy

92. some.random.cool.kid

i have noticed

93. Nnesha

ye it keeps moving hard to use latex on this post