I need help with the intermediates steps of a sum of squares problem. I have a trinomial and the answers it factors into but cannot work it backwards correctly

Question

anonymous · Answer

a sum of squares will have complex numbers in the factorized form

anonymous · Answer

the squar is always equal
for example: For the first one:
The goal of these problems is to get the left hand side of the equation to be a perfect square and then take the square root of both sides and solve for answers.
x^2+16x+64=81
The left side of the equation factors into (x+8)^2
so taking the square root of both sides
x+8=+ or - 9
so x= 1 or x= -17

anonymous · Answer

are you talking about factoring a sum of squares or completing the square?

anonymous · Answer

can i get a medal plzzzzzzzzzzzzzzzzzzz

anonymous · Answer

if y^2-2xy-x^2=0 and the answers are y=(-1+,- 2^.5)x what  are the intermediate steps working backward from those answers?

anonymous · Answer

you have to try to get the left side of the equation to a perfect square then you square root both sides then solve

anonymous · Answer

medal plzzz

anonymous · Answer

sum of squares!
$$f(1)=1^2+f(0)$$
$$f(2)=2^2+f(1)$$
.
.
.
$$f(n)=n^2+(f(n-1))$$
Sum all the expressions above:
$$\sum_{1}^{n}(x^n)=f(n)-f(0)$$
$$f(n)=ax^3+bx^2+cx+d$$
$$\sum_{1}^{n}(x^n)=ax^3+bx^2+cx+d$$

this is the notion you need. then you compare polynomials and get to the answer.
i think it is $$f(n)=(n^3)/6+(n^2)/2+n/3$$

anonymous · Answer

just dunno if I get the question right! hehe

anonymous · Answer

thanks both of you but I need to work backward from the answer

anonymous · Answer

to rearrive at y^2-2xy-x^2=0?

anonymous · Answer

yes

anonymous · Answer

I think I am totally off topic haha, can't wait to see the answer

anonymous · Answer

$$y^2-2xy-x^2=0$$

$$(y-x)^2 -2x^2 = 0$$

$$y - x = \pm (\sqrt{2})x$$
$$y = x \pm (\sqrt{2})x = (1 \pm \sqrt{2})x$$

anonymous · Answer

thanks but I'm timed out on my library computer

anonymous · Answer

ahh

anonymous · Answer

y^2−2xy−x^2=0


(y−x)^2 − 2x^2=0


y−x=±(2^0.5)x

y=x±(2^0.5)x=(1±(2^0.5))x

anonymous · Answer

i think some people got confused when you called it "sum of squares" instead of "completing the square" :)

anonymous · Answer

this works as y^2 - 2xy -x^2 = y^2 - 2xy + x^2  -2x^2 = (y-x)^2 - 2x^2