Question

y= 1/4x^2 + 3x + 6... how do you complete the square for that?

Answer 1

factor the 1/4, or mutliply both sides by 4

Answer 2

4y= x^2 +3x+ 6

Answer 3

?

Answer 4

close :)

4y = x^2 +12x +24

Answer 5

now complete the square as usual and then divide off the 4 if wanted

Answer 6

oh okay
how do you do that?

Answer 7

do what, complete the square?

Answer 8

yes

Answer 9

well, lets consider a generic square setup:

(x+n)^2 , can you expand that out? to a trinomial setup?

Answer 10

yes x^2 +2xn+n^2

Answer 11

perfect, so our format needs to look like this

x^2 + 2n x + n^2 ,but they we have
x^2 +12 x + 24

by equating like terms we get:

x^2 = x^2
2n x = 12 x
n^2 = 24; for get this last line at the moment

the crucial thing is to determine a suitable value for n:

so what is a solution to: 2n = 12 ?

Answer 12

6

Answer 13

good, then n^2 = 6^2 = 36; but we cant really go changing the inherent value of the setup, so we would LOVE to add zero to it. it just so happens that 36 - 36 = 0

so lets complete the square for this

x^2 +12x + 24

x^2 +12x + 24 + 36 - 36

x^2 +12x + 36 + 24 - 36

(x^2 +12x + 36) + 24 - 36

(x+6)^2 - 12

therefore

4y = (x+6)^2 - 12

Answer 14

ok

Answer 15

now, the next step depends on why you would want to complete it to begin with; if its just to place the quadratic in vertex form then divide off the 4

\[y=\frac14(x+6)^2-3\]
if its to find the roots ....
\[0=(x+6)^2-12\]
\[12=(x+6)^2\]
\[\pm\sqrt{12}=x+6\]
\[-6\pm\sqrt{12}=x\]

Answer 16

the key to completing a square is that: 2n = middle term setup

notice that if we add and subtract the square of half that term

(12/2)^2 = 6^2 = 36