Solve x^2 + 6x + 9 using difference of perfect squares

I remember the basics of solving with a perfect squares; I just can't remeber the exact detail. If someone could refresh my memory, I would be grateful. Thank you!

Eg. Solve x²+6x+9 using a difference of perfect squares.

Re: perfect squares

Hello, James!

I'm not sure what you're asking . . .

I remember the basics of solving with a perfect square.
I just can't remember the exact details.
If someone could refresh my memory, I would be greatful.

e.g. Solve \(\displaystyle x^2\,+\,6x\,+\,9\)
Use a difference of perfect squares. ?

Click to expand...

Those are the wrong directions . . .

A perfect square trinomial comes from: \(\displaystyle \,(x\,+\,a)^2\)

The result is: \(\displaystyle \:x^2\,+\,2ax\,+\,a^2\)
. . . . . . . . . . .\(\displaystyle \uparrow\) . . . . . . .\(\displaystyle \uparrow\)
. . . . . . . . . square . . . . . square

We note that there is a "square" on the ends.
. . This will suggest that a trinomial might be a perfect square.
And we need to check the middle term.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Your problem has: \(\displaystyle \:x^2\,+\,6x\,+\,9\)

There are square on the ends: \(\displaystyle x^2\) and \(\displaystyle 3^2\)

This suggests that it might come from: \(\displaystyle \,(x\,+\,3)(x\,+\,3)\)
Multiply: \(\displaystyle \,x^2\,+\,3x\,+\,3x\,+\,9 \:=\:x^2\,+\,6x\,+\,9\) . . . yes!

Therefore: \(\displaystyle \:x^2\,+\,6x\,+\,9 \;=\;(x\,+\,3)^2\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

"Squares on the ends" is not a guarentee of a perfect square.

Example: \(\displaystyle \:x^2\,-\,13x\,+\,36\)

This has "squares on the ends": \(\displaystyle \,x^2\) and \(\displaystyle 6^2\)

We suspect that it came from: \(\displaystyle \x\,-\,6)(x\,-\,6)\)

But this product is: \(\displaystyle \,x^2\,-\,12x\,+\,36\) . . . . it is not a perfect square!
. . . . . . . . . . . . . . . . . . .\(\displaystyle \uparrow?\)

[Edit: I had a terrible blunder here . . . sorry!]

Re: perfect squares

soroban said:

The correct factoring is: \(\displaystyle \x^2\,-\,4)(x^2\,-\,9) \:=\x \,-\,2)(x\,+\,2)(x\,-\,3)(x\,+\,3)\)

Click to expand...

Can you explain what you mean, Soroban?
Shouldn't that be (x - 4)(x - 9) ?