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

mukushla is right :)
try doing ...

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Thanks! I'll try it again.

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

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

Thanks everyone =)

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

x=12 is not right answer

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

what is the final answer of first part?

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

and for second part ??

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

wait

I got the answer: plz check it

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

|dw:1341572956794:dw|

I circled that thing.. is it d?

yes it is d

have u understood ??

yes
Thank you!!!

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

okay I will work on it

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

\[x^2 + 3dx - d^2 +1 then 0 = (-d/2)^2 +3d(-d/2) - d^2 + 1 \]

\[(d/4)^2 + -3d^2/2 -d^2 +1 = 0 \]

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

since it was a different equation

no problem:
from which country u r from ??

US.

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

no. wait.

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

give me 1 min, let me tell you

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

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

hmm

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

bye
i had sent u mail
plz check it

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