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

#### james cook

##### New member
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.

#### soroban

##### Elite Member
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. ?

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!]

#### Denis

##### Senior Member
Re: perfect squares

soroban said:
The correct factoring is: $$\displaystyle \x^2\,-\,4)(x^2\,-\,9) \:=\x \,-\,2)(x\,+\,2)(x\,-\,3)(x\,+\,3)$$
Can you explain what you mean, Soroban?
Shouldn't that be (x - 4)(x - 9) ?