jenejij 3 years ago If the graph of the quadratic function f(x) = x^2 + dx +3d has its vertex on the x axis what are the possible values of d? What if f(x) = x^2 + 3dx +1?

1. mukushla

complete the square for x^2 + dx +3d to get its vertex

mukushla is right :) try doing ...

3. jenejij

I have the answers. Easily accessible in the back of the book. How to solve it is the problem. When you say complete the square what do you mean? I've been trying x^2 + dx +3d = ax^2 + (-2ah)x+ (ah^2+k) And trying to identify the vertex.. that waayy... But what do you mean by complete the square?

4. k.rajabhishek

the vertex of quadratic equation is given by (-b/2a, c-b^2/4a) since it has vertex on X axis therefore y ordinate =0 . hence c-b^2=0

5. jenejij

Thanks! I'll try it again.

6. k.rajabhishek

in 1st case the value of d=0 and 3 in 2nd case d=1/3 and -1/3

7. mukushla

@jenejij take a look http://www.mathsisfun.com/algebra/completing-square.html

8. jenejij

Oh okay. I know how to "complete the square" haha. Got it. =)

9. jenejij

Thanks everyone =)

10. jenejij

Actually - still having problems.. =/ I don't think x=3. x= 0, and x= 12.

11. k.rajabhishek

12. jenejij

thats what the answer in the back of my text book is... but I don't know how to get there.

13. k.rajabhishek

what is the final answer of first part?

14. jenejij

d = 0 and d= 12. According the the textbook...

15. k.rajabhishek

and for second part ??

16. jenejij

$d = + or - 2 / \sqrt{13}$

17. k.rajabhishek

wait

18. k.rajabhishek

I got the answer: plz check it

19. k.rajabhishek

the value of x coordinate of vertex is given by -b/2a; from the given question we have a=1 b=d c=3d substituting the values in -b/2a we have -d/2

20. k.rajabhishek

to find y ordinate substitute this value in the given eq. we have |dw:1341572810912:dw|

21. jenejij

|dw:1341572956794:dw|

22. jenejij

I circled that thing.. is it d?

23. k.rajabhishek

since it has vertex on x axis thus y=0 hence we have d^2/4 - d^2/2 +3d = 0 on solving we have d=0 and 12

24. k.rajabhishek

yes it is d

25. k.rajabhishek

have u understood ??

26. jenejij

yes Thank you!!!

27. k.rajabhishek

similarly u can do second part .. plz do it and tell me have u got the answer or not

28. jenejij

okay I will work on it

29. k.rajabhishek

give me a medal .........lol

30. jenejij

$x^2 + 3dx - d^2 +1 then 0 = (-d/2)^2 +3d(-d/2) - d^2 + 1$

31. jenejij

$(d/4)^2 + -3d^2/2 -d^2 +1 = 0$

32. k.rajabhishek

no; for second part u again need to find the x coordinate and follow the same steps

33. k.rajabhishek

since it was a different equation

34. jenejij

okay let me try again. toss that grain of salt my way (I've been working all night.. it's 4:21 am) I'm not usually this bad haha

35. k.rajabhishek

no problem: from which country u r from ??

36. jenejij

US.

37. jenejij

okay so then is 3d = -b/2a?

38. jenejij

no. wait.

39. jenejij

I can't think. I have to do this in the morning. Thanks so far.

40. k.rajabhishek

give me 1 min, let me tell you

41. k.rajabhishek

in second part : a=1 b=3d c=1 now subs. in -b/2a u have -3d/2

42. jenejij

okay. I was on the right track. but not quite there.

43. jenejij

hmm

44. k.rajabhishek

then put this value of x in original equation x^2+3dx+1=0|dw:1341574077931:dw|

45. jenejij

Okay. I see the answer. And I understand, but without sleep I won't reeeally get it. Again thank you for your help.

46. k.rajabhishek

bye i had sent u mail plz check it