- #1

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y = x^2 + 3x + 5/2

Not looking for an answer, just looking for how to do this (I don't know how to do it if it has rationals in it). Thank you.

- Thread starter Caldus
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- #1

- 106

- 0

y = x^2 + 3x + 5/2

Not looking for an answer, just looking for how to do this (I don't know how to do it if it has rationals in it). Thank you.

- #2

- 362

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Complete the square.

x^{2}+3x +k=(x+3/2)^{2}

What is k?

Also, you can multiply through by two to get rid of the rationals, but don't forget to divide it out at the end.

x

What is k?

Also, you can multiply through by two to get rid of the rationals, but don't forget to divide it out at the end.

Last edited:

- #3

Hurkyl

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IOW you do it the same as if there weren't rationals.

- #4

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I am sorry, but, how are those two equal ? and how are they connected to the original question ?Originally posted by StephenPrivitera

x^{2}+3x +k=(x+3/2)^{2}

To convert a quadratic to the form (a(x-h)^2 + k) you must (as StephenPrivitera said) complete the square.

If you have a quadratic on the form of :

ax^2 + bx + c

Then, it is a complete square if c=(b/2)^2

So, to turn any quadratic to a complete square you need to make (

In your case, (b/2)^2 = (3/2)^2 = 9/4

To turn 5/2 into 9/4, you will need to add (

Here you go:

y = x^2 + 3x + 5/2

y = x^2 + 3x + 5/2 + 0

y = x^2 + 3x + 5/2 - 1/4 + 1/4

y = x^2 + 3x + (5/2 - 1/4) + 1/4

y = x^2 + 3x + (10/4 - 1/4) + 1/4

y = x^2 + 3x + 9/4 + 1/4

y = (x^2 + 3x + 9/4) + 1/4

y = ((x + 3/2)*(x + 3/2)) + 1/4

y = (x + 3/2)^2 + 1/4

Which is on the form that you asked for .

- #5

Hurkyl

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I am sorry, but, how are those two equal ? and how are they connected to the original question ?

Notice that you proved:

x^2 + 3x + 5/2 = (x + 3/2)^2 + 1/4

Do a little rearrangement and you'll see that's (essentially) of the form

x^2+3x +k=(x+3/2)^2

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