## jazzyfa30 Group Title Solve the equation by using the Square Root Property. x^2 = 10x + 24 one year ago one year ago

1. jazzyfa30 Group Title

@hartnn

2. hartnn Group Title

ok, so $$x^2-10x=24$$ the co-efficient of 'x' here is 10 divide it by 2 and square the result, what u get ?

3. hartnn Group Title

***the co-efficient of 'x' here is -10

4. jazzyfa30 Group Title

so it would be x^2-5x=24

5. hartnn Group Title

no no.. the co-efficient of x is -10, right ? divide it by 2, you get -5, but then square it, $$\large (-5)^2=...?$$

6. jazzyfa30 Group Title

2.23

7. jazzyfa30 Group Title

is that right????

8. hartnn Group Title

umm...noo...how u got thta ?

9. jazzyfa30 Group Title

i squared 5 isnt that what i was supposed to do

10. hartnn Group Title

5^2 =25

11. jazzyfa30 Group Title

oh i thought i was supposed to square root it not square my b now what

12. hartnn Group Title

ohh...u took square root...we are supposed to square..

13. hartnn Group Title

ok, we add 25 on both sides $$x^2-10x+25=24+25$$ notice that left side is a perfect square! of ?

14. jazzyfa30 Group Title

49

15. hartnn Group Title

49 is right side, and is the square of 7 what about left side ? x^2-10x+25 is squaree of ?

16. jazzyfa30 Group Title

idk

17. hartnn Group Title

$$(x-5)^2 =... ?$$ can u find ?

18. jazzyfa30 Group Title

x^2- 25

19. hartnn Group Title

nopes, $$(a+b)^2=a^2+2ab+b^2$$ so, $$(x-5)^2 =x^2-10x+25$$ so here we have $$(x-5)^2=7^2$$ take square root on both sides!

20. jazzyfa30 Group Title

ok

21. jazzyfa30 Group Title

then what

22. jazzyfa30 Group Title

im confused

23. hartnn Group Title

did u take square root on both sides ? what u got ?

24. jazzyfa30 Group Title

idk

25. hartnn Group Title

$$(x-5)^2=7^2 \implies x-5 = \pm 7 \\ x-5 =7 \quad \quad x-5=-7$$ get 2 values of 'x' from here!

26. jazzyfa30 Group Title

but 7 isnt apart of any of my answer choices

27. hartnn Group Title

u need to solve x-5 =7 x-5 = -7

28. jazzyfa30 Group Title

x= 12 x=-2

29. hartnn Group Title

correct!

30. jazzyfa30 Group Title

ok Solve the equation by completing the square: x2 - 8x + 15 = 0

31. hartnn Group Title

same thing, here, the co-efficient of x is -8 divide it by 2, what u get ?

32. jazzyfa30 Group Title

-4

33. hartnn Group Title

yes, then square it...what u get ?

34. jazzyfa30 Group Title

-16

35. hartnn Group Title

-4 *-4 will be +16

36. jazzyfa30 Group Title

oh

37. hartnn Group Title

so, adding 16 on both sides... $$x^2-8x+16=-15+16$$

38. jazzyfa30 Group Title

right side:1 left side: idk

39. jazzyfa30 Group Title

i think its x^2-4x-2x=1

40. jazzyfa30 Group Title

is that right??????

41. jazzyfa30 Group Title

42. hartnn Group Title

$$x^2-8x+16 = (x-4)^2$$ so, we have $$(x-4)^2=1$$ just take square root on both sides!

43. jazzyfa30 Group Title

?????????

44. hartnn Group Title

when you take square root, the square part goes away! (x+4) =+1 or x-4 =-1 find 2 values of x from here...

45. jazzyfa30 Group Title

(x-3)(x+3) i think

46. jazzyfa30 Group Title

a. x = + 5 b. x =-3 or x = 5 c. x =3 or x = -5 d. x =3 or x =5

47. hartnn Group Title

x-4 = -1 , x = -1+4 = 3 is correct but for 2nd part, you get x-4 =1 x= 1+4 = 5

48. hartnn Group Title

so, 3,5

49. jazzyfa30 Group Title

oh yea

50. jazzyfa30 Group Title

Solve the equation by completing the square: x^2 + 2x = 120

51. terenzreignz Group Title

Do you know how to complete the square? :)

52. jazzyfa30 Group Title

??????

53. terenzreignz Group Title

I see. Here's an example:$\Large x^2 + \color{red}6x =40$

54. terenzreignz Group Title

What you do is take that coefficient (number beside) the x with NO EXPONENT. In our case, 6.

55. terenzreignz Group Title

Take half of it, 3, and then square it, getting 9.

56. terenzreignz Group Title

Now, add that to BOTH SIDES of the equation. $\Large x^2 +\color{red}6x\color{blue}{+9}=40\color{blue}{+9}$

57. terenzreignz Group Title

Simplify. $\Large x^2+\color{red}6x+9=49$

58. terenzreignz Group Title

You'll notice that the left side is now what you call a 'perfect square trinomial' It may be expressed as... $\Large (x+\color{red}3)^2=49$

59. terenzreignz Group Title

Notice of course, that 3 is half of the original coefficient of x, which was 6. Now take the square root of both sides: $\Large \sqrt{(x+3)^2}=\sqrt{49}$ Notice that the square root and the square (exponent 2) cancel out. But the square root of 49 may be either positive or negative 7. $\Large x+3=\pm7$ Bring the 3 to the other side, you now get $\Large x = -3\pm7$ which means either x = -3 + 7 = 4 or x = -3 - 7 = -10 Case Closed :)

60. terenzreignz Group Title

So... go ahead now, get started with your own problem, using the same concepts. I'll walk you through it if you like :)

61. jazzyfa30 Group Title

62. terenzreignz Group Title

63. jazzyfa30 Group Title

u on skype or call me at 803-937-6271 cus by u typing it it is confusing m please

64. terenzreignz Group Title

I don't have skype. come on, we can do this... I'll get you started... $\Large x^2 +\color{red}2x = 120$

65. jazzyfa30 Group Title

then call me i dont understand

66. jazzyfa30 Group Title

@jessika96