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

[james cook]

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]

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]

Re: perfect squares

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) ?