anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
complete the square for x^2 + dx +3d to get its vertex
anonymous
  • anonymous
mukushla is right :) try doing ...
anonymous
  • anonymous
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?

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anonymous
  • anonymous
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
anonymous
  • anonymous
Thanks! I'll try it again.
anonymous
  • anonymous
in 1st case the value of d=0 and 3 in 2nd case d=1/3 and -1/3
anonymous
  • anonymous
@jenejij take a look http://www.mathsisfun.com/algebra/completing-square.html
anonymous
  • anonymous
Oh okay. I know how to "complete the square" haha. Got it. =)
anonymous
  • anonymous
Thanks everyone =)
anonymous
  • anonymous
Actually - still having problems.. =/ I don't think x=3. x= 0, and x= 12.
anonymous
  • anonymous
x=12 is not right answer
anonymous
  • anonymous
thats what the answer in the back of my text book is... but I don't know how to get there.
anonymous
  • anonymous
what is the final answer of first part?
anonymous
  • anonymous
d = 0 and d= 12. According the the textbook...
anonymous
  • anonymous
and for second part ??
anonymous
  • anonymous
\[d = + or - 2 / \sqrt{13}\]
anonymous
  • anonymous
wait
anonymous
  • anonymous
I got the answer: plz check it
anonymous
  • anonymous
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
anonymous
  • anonymous
to find y ordinate substitute this value in the given eq. we have |dw:1341572810912:dw|
anonymous
  • anonymous
|dw:1341572956794:dw|
anonymous
  • anonymous
I circled that thing.. is it d?
anonymous
  • anonymous
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
anonymous
  • anonymous
yes it is d
anonymous
  • anonymous
have u understood ??
anonymous
  • anonymous
yes Thank you!!!
anonymous
  • anonymous
similarly u can do second part .. plz do it and tell me have u got the answer or not
anonymous
  • anonymous
okay I will work on it
anonymous
  • anonymous
give me a medal .........lol
anonymous
  • anonymous
\[x^2 + 3dx - d^2 +1 then 0 = (-d/2)^2 +3d(-d/2) - d^2 + 1 \]
anonymous
  • anonymous
\[(d/4)^2 + -3d^2/2 -d^2 +1 = 0 \]
anonymous
  • anonymous
no; for second part u again need to find the x coordinate and follow the same steps
anonymous
  • anonymous
since it was a different equation
anonymous
  • anonymous
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
anonymous
  • anonymous
no problem: from which country u r from ??
anonymous
  • anonymous
US.
anonymous
  • anonymous
okay so then is 3d = -b/2a?
anonymous
  • anonymous
no. wait.
anonymous
  • anonymous
I can't think. I have to do this in the morning. Thanks so far.
anonymous
  • anonymous
give me 1 min, let me tell you
anonymous
  • anonymous
in second part : a=1 b=3d c=1 now subs. in -b/2a u have -3d/2
anonymous
  • anonymous
okay. I was on the right track. but not quite there.
anonymous
  • anonymous
hmm
anonymous
  • anonymous
then put this value of x in original equation x^2+3dx+1=0|dw:1341574077931:dw|
anonymous
  • anonymous
Okay. I see the answer. And I understand, but without sleep I won't reeeally get it. Again thank you for your help.
anonymous
  • anonymous
bye i had sent u mail plz check it

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