## jenejij Group Title 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? 2 years ago 2 years ago

1. mukushla Group Title

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

mukushla is right :) try doing ...

3. jenejij Group Title

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 Group Title

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 Group Title

Thanks! I'll try it again.

6. k.rajabhishek Group Title

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

7. mukushla Group Title

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

8. jenejij Group Title

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

9. jenejij Group Title

Thanks everyone =)

10. jenejij Group Title

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

11. k.rajabhishek Group Title

12. jenejij Group Title

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

13. k.rajabhishek Group Title

what is the final answer of first part?

14. jenejij Group Title

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

15. k.rajabhishek Group Title

and for second part ??

16. jenejij Group Title

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

17. k.rajabhishek Group Title

wait

18. k.rajabhishek Group Title

I got the answer: plz check it

19. k.rajabhishek Group Title

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 Group Title

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

21. jenejij Group Title

|dw:1341572956794:dw|

22. jenejij Group Title

I circled that thing.. is it d?

23. k.rajabhishek Group Title

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 Group Title

yes it is d

25. k.rajabhishek Group Title

have u understood ??

26. jenejij Group Title

yes Thank you!!!

27. k.rajabhishek Group Title

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

28. jenejij Group Title

okay I will work on it

29. k.rajabhishek Group Title

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

30. jenejij Group Title

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

31. jenejij Group Title

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

32. k.rajabhishek Group Title

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

33. k.rajabhishek Group Title

since it was a different equation

34. jenejij Group Title

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 Group Title

no problem: from which country u r from ??

36. jenejij Group Title

US.

37. jenejij Group Title

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

38. jenejij Group Title

no. wait.

39. jenejij Group Title

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

40. k.rajabhishek Group Title

give me 1 min, let me tell you

41. k.rajabhishek Group Title

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

42. jenejij Group Title

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

43. jenejij Group Title

hmm

44. k.rajabhishek Group Title

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

45. jenejij Group Title

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 Group Title

bye i had sent u mail plz check it