i have (x^2+6x)+(y^2+4y)+(z^2-2z)=1
how do i complete the square

Question

i have (x^2+6x)+(y^2+4y)+(z^2-2z)=1
how do i complete the square

campbell_st · Answer

if you have a quadratic 

$$x^2 + bx + ?    $$

to complete the square you need to find 

$$\frac{b}{2} $$

then square it.... and add it to both sides of the equation 

so 

$$x^2 + bx + (\frac{b}{2})^2 = (x + \frac{b}{2})^2$$

hope this helps

jdoe0001 · Answer

freddy_eighty7    do you know what  a $\bf \text{perfect square trinomial}$  is ?

anonymous · Answer

so i do each one at a time? (x,y,z) i don't know about the perfect square

anonymous · Answer

$$
x^2+bx = \left(x+\frac b 2\right)^2-\left(\frac b 2\right)^2
$$

campbell_st · Answer

yes... some process for each

jdoe0001 · Answer

hmmmm.... well... you'd need to, otherwise completing the square won't be of much use http://www.youtube.com/watch?v=7rffjYDxW5I

campbell_st · Answer

but make sure you add the term (b/2)^2 to both sides of the equation... to keep it in balance

anonymous · Answer

so the pattern I'm seeing is half the x, then square it, and that becomes the b to where you also add to the other side of the = to even it out, yes?

anonymous · Answer

and its always +?

campbell_st · Answer

well if you have a perfect square then milled term is always double to product of the 2 terms in the binomial... the last term is always positive... 


$$(m - n)2 = m^2 - 2mn + n^2$$

the middle term always has the same sign as the binomial...

anonymous · Answer

i think i got it, thank you ill prob bug you more later since your answering the way i want to by giving me the base formula