Perfect square trinomial..?

Question:What number can be added to q^2-30q to make a perfect square trinomial?

Explain how you found the answer please.

Answers:
225..

First I took the half of b and squared it or the coefficient of the equation q^2-30q..
General equation: ax^2+bx+c=0
Since b is 30, then its half is 15..Then I squared it and got 225..
That's how I got it..
Hope this helps..
q^2 - 30q + 225 = (q - 15)^2

Perfect square trinomial


Think about the theory using only variables..

(a - b)^2 = a^2 - 2ab + b^2

So...with your we know a is q. and we know -2ab = -30q. Well, we can solve that equation by substiting q for a (which cancels the q on both sides of the equation leaving us with
-2b = -30. Therefore b must be 15. SO plug that in

(q - 15)^2 and then multiply to get the last number...
b^2 = 225.
q^2 - 30q + 225

Take a look at -30q, then divide in half
You get -15. You know this because -15 + -15 = -30

Take -15 and multiply it by -15, itself.

Now you can factor this

q(-30 + 255)
q(-30 + 255 - 15 + 15)

Take the 15 out, multiplying it by 1. (Because of the value of a)

q(-30 + 255 - 15) + 15

Then you want to show that in vector form/ complete the square

y= q(-30 -15)^2 + 15

Therefore the vertex equals v = (30, 15)

// I'm hoping I did that right, someone please tell me if I didn't.

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