Caldus

Trying to rewrite a quadratic equation in the form a(x - h)^2 + k. The equation I'm trying to rewrite is:

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.

StephenPrivitera

Complete the square.
x2+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.

Last edited:

Hurkyl

Staff Emeritus
Gold Member
IOW you do it the same as if there weren't rationals.

STAii

Originally posted by StephenPrivitera
x2+3x +k=(x+3/2)2
I am sorry, but, how are those two equal ? and how are they connected to the original question ?

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 (c) in it equal to ((b/2)^2)
In your case, (b/2)^2 = (3/2)^2 = 9/4
To turn 5/2 into 9/4, you will need to add ((9/4)-(5/2)=(9/4)-(10/4)=-(1/4)) to it. But if you add any number to the quadratic you will actually change its value. So, to maintain the value, you will subtract the same number again, therefore leaving the qudratic unchanged (adding and subtracting the same number is like adding 0, it does nothing to the quadratic).
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 .

Hurkyl

Staff Emeritus
Gold Member
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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