## anonymous one year ago Solve the equation using square roots. x^2– 15 = 34 A) z = 1/5 or z = -4 B) z = 1 or z = -4 C) z = 1/5 or z = 4 D) z = 1 or z = 4

1. anonymous

@steve816

2. steve816

What you want to do first is to get the x alone. I'll show you!|dw:1440177092884:dw|

3. steve816

Now we take the square root on both sides of the equation.|dw:1440177139414:dw|

4. anonymous

sorry just realized I put up the wrong question, I already solved that one. @steve816

5. anonymous

5z^2 - 21z + 4 = 0

6. anonymous

Solve the equation by factoring. 5z^2 - 21z + 4 = 0

7. Rushwr

wat's the answer u got ?

8. anonymous

for the first question?

9. anonymous

$\pm7$

10. Rushwr

the 2nd one

11. anonymous

I need help with that one

12. Rushwr

$5z ^{2}-21z+4=0$

13. Rushwr

$5z ^{2}-21z+4=0$$5z ^{2}-20z-1z+4=0$

14. Rushwr

did u get tht part?

15. anonymous

why did u take out the one from 21

16. Rushwr

it's like this u see the middle number 21z right? 1st we multiply 4 from 5 which is 20 right? Now we know we need 4 numbers right? So using the factors of 20z we are gonna make 21z 20z* 1 is 20z right? but we also know -20z-1z= 21z right? So we substitute those 2 numbers instead of 21z !

17. Rushwr

I'm not sure if u got this !

18. anonymous

yeah I kinda do

19. Rushwr

sure?

20. Rushwr

$5z ^{2}-20z-1z+4=0$ $5z(z-4)-1(z-4)=0$ did u get what I did in this step ?

21. anonymous

yes, if we multiply the numbers we get $5z^2 -20z -1z + 4 = 0$

22. Rushwr

yeah since we have two (z-4) Now we gonna consider only one of it ! So now:$5z(z-4)-1z(z-4)=0$$(5z-1) (z-4)=0$

23. anonymous

I don't understand this.. I have a doctor appointment now, if u can help me later on.. sorry

24. Rushwr

oki

25. Nnesha

i like ur pfp. :)

26. anonymous

I need help with this question.. I have a lot more to do :P

27. Nnesha

first you need to get like terms on one side and variable x on the other side

28. Nnesha

so move the -15 to the right how would you do that >

29. Nnesha

?*

30. anonymous

x^2 = 34 - 15?

31. Nnesha

no it's already negative at left side to cancel out you would do opposite of subtraction

32. anonymous

to add.. x^2 = 34 + 15?

33. Nnesha

yes right

34. Nnesha

and then take square root both sides to cancel out square of x.

35. anonymous

so we don't solve for 34 + 15

36. Nnesha

we should solve for x (i guess it supposed to be z)

37. Nnesha

you answer choices are wrong....

38. Nnesha

your*

39. anonymous

wait Lol the question up there is the wrong question

40. Nnesha

ye...

41. anonymous

I put it up there that it was wrong.. sorry just realized we were doing the wrong Q

42. anonymous

Solve the equation by factoring. 5z^2 - 21z + 4 = 0

43. Nnesha

there are like 4 method to solve quadratic equations you can use the quadratic formula or you can find factors of that equation

44. Nnesha

to factor it find two number when you multiply them you should get product of AC and when you add them you should get middle term $\huge\rm Ax^2+Bx+C=0$ a=leading coefficient b=middle term c=constant term so what is A and C in that equation .

45. anonymous

21z + 4 ?

46. Nnesha

hmm nope |dw:1440193043766:dw|

47. anonymous

Ooh A

48. anonymous

5z^2 + 4

49. Nnesha

no just the coefficient so A = 5 b=21 c=4 now multiply a times c = 5 times 4 = 20 find two number when you multiply them you should get 20 but when you add or subtract them you should get middle term which is 21

50. anonymous

I don't really get it, what am I supposed to do now?

51. Nnesha

what two number would you multiply to get 20 ? but when you add them you should get 21

52. anonymous

5*4 = 20 2*10 = 20

53. anonymous

when u add them all up u get 21

54. Nnesha

yes but $4+5 \cancel{=} 20$

55. Nnesha

no just 2 number

56. Nnesha

you can't add 4 number to get 20 here is an example let's say a and b are numbers so when i multiply them a times b = 20 i should add same two numbers a+b (or a-b) to get middle term

57. Nnesha

what other two numbers you can multiply to get 20 ?

58. anonymous

2 and 10

59. Nnesha

and??

60. anonymous

1 and 20.

61. Nnesha

perfect 20 times 1 =20 20+1=21

62. anonymous

Ohh I see

63. Nnesha

$\huge\rm 5z^2+1z+20z+4$ now use group method to factor it

64. Nnesha

|dw:1440193728090:dw| make a group of two terms what is common in first two terms ?

65. anonymous

so we add those with the z first?

66. Nnesha

no if you add them you will get original equation 5z^2+20x+4 :P

67. Nnesha

find common of 5x^2+1x and common factor of 20x+4

68. anonymous

okay for 5x^2 + 1x its 1?

69. anonymous

and then 2 for 20x + 4

70. Nnesha

not just one what about the variable ?

71. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @proud_yemeniah_ and then 2 for 20x + 4 $$\color{blue}{\text{End of Quote}}$$ greatest common factor yes two is common but it's not greatest common

72. anonymous

x is the common factor?

73. Nnesha

factors of 20 are: 20,5,4,2,1 factors of 4 are : 4,2,1 which one is greatest common factor ?

74. anonymous

4

75. Nnesha

yes right

76. Nnesha

|dw:1440194265638:dw| when you take out 1z from 5z^2+1x what will have left in the parentheses

77. Nnesha

remember both parentheses should be the same!

78. Nnesha

5z^2+1z not x :P

79. anonymous

5z^2

80. Nnesha

what will get get when you take out z from 5z^2 ?

81. anonymous

5^2 and then if u solve it it will be 10

82. Nnesha

in other words divide both terms by common factor

83. Nnesha

hmm no $\huge\rm \color{Red}{\frac{ 5z^2 }{ z }+\frac{ 1z }{ z }}+(\frac{ 20z }{ 4 }+\frac{ 4 }{ 4 })$ $\huge\rm \color{Red}{z(??+??)}+4(??+??)$ divide by common and then write your answer in the parentheses

84. anonymous

(5^2 + 1) + (5z + 1)

85. Nnesha

don't forget the common factor that would stay outside the parentheses do you mean 5 to the 2 power ?

86. anonymous

yes

87. Nnesha

$\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times z }{ z}$

88. Nnesha

can you cancel out z?

89. anonymous

Umm so we remove the 2 and not the z or both?

90. Nnesha

$\frac{ 2 \times 2 }{ 2}$ would you cancel both 2s's or just one ?

91. anonymous

I'm not sure

92. Nnesha

$\frac{ \heartsuit \times \cancel{\heartsuit}}{\cancel{ \heartsuit }}$ one heart canceled with one heart.

93. Nnesha

just one $\frac{ 2 \times \cancel{2} }{ \cancel{2}}$ you can multiply 2 at the top you will get 4/2 when u divide them you will get 2 if you cancel out both 2 with one you will get 1 as an answer which is wrong

94. Nnesha

$\frac{ 5z^2 }{ z }~\textrm {is~same as }\frac{ 5z \times\cancel{ z} }{\cancel{ z}}$

95. anonymous

oh okay

96. Nnesha

now both parentheses are common so you can take it out

97. Nnesha

wait there is a typo

98. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @proud_yemeniah_ (5^2 + 1) + (5z + 1) $$\color{blue}{\text{End of Quote}}$$ so it should $\huge\rm z(5z+1)+4(5z+1)$ and the hing i gave you both parentheses should be the same

99. Nnesha

|dw:1440195435458:dw| when you take out (5z+1) from z(5z+1) what will get