## anonymous one year ago 1) When the sum of 5 and twice a positive number is subtracted from the square of the number, 0 results. Find the number.

1. pooja195

Let x be the number First thing it says is subtracted from the square of A NUMER (x) sooo $\huge~\rm~x^2-$

2. pooja195

*number

3. pooja195

Then it says the number is twice postivie soo 2x and the sum (adding) of 5 so it would look like this $\huge~\rm~(2x+5)$

4. pooja195

The result is 0 so =0 $\huge~\rm~x^2-(2x+5)=0$

5. pooja195

We can remove the parentheses and write it like this: $\huge~\rm~x^2-2x-5=0$

6. pooja195

Find the abc values :) a ,b , c values are $\huge\rm Ax^2+Bx+C=0$ where a =leading coefficient b= leading coefficient c= constant term

7. anonymous

a=1 b=2 c=5

8. anonymous

-2 -5

9. pooja195

Good : )

10. pooja195

$\huge~\rm~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$

11. anonymous

$x=\frac{ 2\pm \sqrt{2^2-4(1)(5)} }{ 2(1) }$

12. pooja195

$\huge~\rm~x=\frac{ -(-2) \pm \sqrt{(-2)^2-4(1)(-5)} }{ 2(1) }$

13. anonymous

oh in the sqrt its -2

14. anonymous

and -5 okay

15. pooja195

ok can you solve whats in the sqrt?

16. anonymous

$\sqrt{80}$

17. anonymous

$2\pm4\sqrt{5}$

18. pooja195

I got $\huge~\rm~1±1\sqrt{6}$

19. anonymous

..how?

20. pooja195

eh i cheated and used a calc ;p heres the work http://prntscr.com/80bm1y

21. anonymous

what calc did you use

22. pooja195
23. pooja195

Then we make 2 seperate equations one plus and one minus since we basically have to split the plus minus sign then u simplfy from here $\frac{ \huge~\rm~1+1\sqrt{6} }{ \huge~\rm~1 }$ $\frac{ \huge~\rm~1-1\sqrt{6} }{ \huge~\rm~1 }$

24. pooja195

25. anonymous

...

26. pooja195

...?

27. anonymous

how

28. pooja195

I never learned....:/ you will need to use a calculator or ask someone else...i have always used a calc for this part... @Astrophysics

29. ali2x2

@ganeshie8

30. just_one_last_goodbye

@pooja195 what do we want to convert into a decimal?

31. pooja195

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @pooja195 Then we make 2 seperate equations one plus and one minus since we basically have to split the plus minus sign then u simplfy from here $\frac{ \huge~\rm~1+1\sqrt{6} }{ \huge~\rm~1 }$ $\frac{ \huge~\rm~1-1\sqrt{6} }{ \huge~\rm~1 }$ $$\color{#0cbb34}{\text{End of Quote}}$$ This and show your stpes

32. pooja195

*steps

33. just_one_last_goodbye

alright

34. just_one_last_goodbye

both on top of each other like fractions correct?

35. ali2x2

@just_one_last_goodbye i think so

36. pooja195

Actually a better way to solve it Apply this rule $\huge~\rm~\frac{ a }{ 1 }=a$

37. pooja195

$\huge~\rm~\frac{1+\sqrt{6}1}{1}=1+\sqrt{6}1$

38. pooja195

$\huge~\rm~1+1\cdot \sqrt{6}$

39. pooja195

$\huge~\rm~1\cdot \:a=a$

40. pooja195

$\huge~\rm~1+\sqrt{6}$

41. pooja195

Now turn it into decimal form

42. pooja195

We dont even need to solve fore the subtraction problem because we know that the number will be negative

43. anonymous

ok

44. ali2x2

ill try it myself

45. just_one_last_goodbye

I got 3.44949

46. pooja195

Thats right

47. anonymous

I was gonna ask if we round it to 3.45

48. pooja195

Yep :)

49. ali2x2

i got 3.44949

50. anonymous

okay, thank you for being so patient.

51. ali2x2

52. pooja195

lol

53. ali2x2

i cri