Need Pre-Calc help...

[Pastore]

Someone please please explain to me how to complete the square. My teacher went through it so fast that my head was spinning, and my book doesn't explain it. Please help me!

If you could use the following quadratic as an example I would appreciate it much.

f(x)=2(x^2+4x)+7

 

[chiwawa626]

Im in calc AB and dont remmeber lol...but searching is ur firend : Text

 

[gwlam12]

i can do it, but cant explain it it just comes to me

 

[DanFungus]

f(x)=2(x^2+4x)+7

4x => 4/2
4/2 => 2^2

f(x)=2(x^2+4x+4)+7-4
f(x)=2(x+2)^2+3

 

[johnjohn320]

DanFungus beat me to it.

 

[DanFungus]

Originally posted by: johnjohn320
DanFungus beat me to it.

Damn I'm good

 

[da loser]

this is the same method, but here's a formula you could use, I just learned it
get the equation in this form, you might have to divide the coefficient in front of x^2
x^2+a*x+b = (x+a/2)^2 + (b-a^2/4)

so in your example you have x^2 + 4x -> (x+4/2)^2 + (0-4^2/4) = (x+2)^2 - 4

since that was multipled by 2 plus 7, then 2*(x+2)^2 - 2*4 + 7 = 2*(x+2)^2 -1

i think dan fangus is wrong, choose x=3, your answer is 2(9+12)+7=49, his is 2(5)^2+3= 53
mine is 2(5)^2-1=49

 

[DanFungus]

Originally posted by: da loser
this is the same method, but here's a formula you could use, I just learned it
get the equation in this form, you might have to divide the coefficient in front of x^2
x^2+a*x+b = (x+a/2)^2 + (b-a^2/4)

so in your example you have x^2 + 4x -> (x+4/2)^2 + (0-4^2/4) = (x+2)^2 - 4

since that was multipled by 2 plus 7, then 2*(x+2)^2 - 2*4 + 7 = 2*(x+2)^2 -1

i think dan fangus is wrong, choose x=3, your answer is 2(9+12)+7=49, his is 2(5)^2+3= 53
mine is 2(5)^2-1=49

lol. I quickly did the quadratic on it after making it f(x) = 2x^2 + 8x + 7

and I got the X roots as being;

-4 + (sq.root of 2)
______________
2

and

-4 - (sq.root of 2)
______________
2