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?

- anonymous

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- anonymous

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

- anonymous

mukushla is right :)
try doing ...

- 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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## More answers

- 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

Thanks! I'll try it again.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Thanks everyone =)

- anonymous

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

- anonymous

x=12 is not right answer

- anonymous

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

- anonymous

what is the final answer of first part?

- anonymous

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

- anonymous

and for second part ??

- anonymous

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

- anonymous

wait

- anonymous

I got the answer: plz check it

- 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

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

- anonymous

|dw:1341572956794:dw|

- anonymous

I circled that thing.. is it d?

- 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

yes it is d

- anonymous

have u understood ??

- anonymous

yes
Thank you!!!

- anonymous

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

- anonymous

okay I will work on it

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

since it was a different equation

- 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

no problem:
from which country u r from ??

- anonymous

US.

- anonymous

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

- anonymous

no. wait.

- anonymous

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

- anonymous

give me 1 min, let me tell you

- anonymous

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

- anonymous

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

- anonymous

hmm

- anonymous

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

- 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

bye
i had sent u mail
plz check it

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