# Complete the Square

• Feb 13th 2006, 02:09 AM
phgao
Complete the Square
1. x^2 + 2x + k
2. kx^2 + 2x + k
3. x^2 - 3kx + 1

For the first I get: -1+- root(1+k)
Second : [-1+- root(1-k)]/k

But I'm told im wrong. Help!
• Feb 13th 2006, 02:16 AM
jacs
are you supposed to just find k? or solve for k?

x² + 2x + k
x² + 2x + 1 + k - 1
(x + 1)² + k -1

not sure if that is what you mean

since there is no equation, there is nothing to solve

hope that helps maybe?
• Feb 13th 2006, 02:17 AM
phgao
solve for x,
• Feb 13th 2006, 06:03 AM
earboth
Quote:

Originally Posted by phgao
1. x^2 + 2x + k
2. kx^2 + 2x + k
3. x^2 - 3kx + 1

For the first I get: -1+- root(1+k)
Second : [-1+- root(1-k)]/k

But I'm told im wrong. Help!

Hello,

the title of your message suggests, that you are looking for a value of k so you got a square. Is this right? If so:

to 1) $\displaystyle k=1 \ \mbox{then you get} \ (x+1)^2$
to 2) $\displaystyle k=1 \ \mbox{then you get} \ (x+1)^2$
to 3) $\displaystyle k=\frac{2}{3} \ \mbox{then you get} \ (x-1)^2$

I hope that this here is of some help.

Bye
• Feb 13th 2006, 10:49 AM
ticbol
Quote:

Originally Posted by phgao
1. x^2 + 2x + k
2. kx^2 + 2x + k
3. x^2 - 3kx + 1

For the first I get: -1+- root(1+k)
Second : [-1+- root(1-k)]/k

But I'm told im wrong. Help!

"Complete the square" means complete the perfect square. Perfect square here is in the form (x+a)^2, where "a" is plus or minus, depending on the given data.

Before attempting completing the square, make sure first that the coeff(icient of the x^2 is 1 only. Meaning, it it were originally b(x^2), then divide it by "b"......

Completing the square of (x^2 +bx) is adding, and subtracting at the same time, the square of half of the coefficient of linear x.
x^2 +bx +(b/2)^2 -(b/2)^2 ------------***

------------------------
x^2 +2x +k = 0
x^2 +2x +(2/2)^2 -(2/2)^2 +k = 0
x^2 +2x +1 -1 +k = 0
(x+1)^2 -1 +k = 0
(x+1)^2 = 1-k
Take the square roots of both sides,
x+1 = +,-sqrt(1-k)
So,

----------------------
kx^2 +2x +k = 0

The x^2 has a coefficient k which is not 1, so divide kx^2 by k. In so doing, divide also all of the other terms of the original equation by k to retain the equality of the original equation.
x^2 +(2/k)x +1 = 0
Now that the coefficient of the x^2 is 1 only, we can start completing the square,
x^2 +(2/k)x +(1/k)^2 -(1/k)^2 +1 = 0
(x +1/k)^2 -1/(k^2) +1 = 0
(x +1/k)^2 = 1/(k^2) -1
x +1/k = +,-sqrt[1/(k^2) -1]
x = -1/k +,-sqrt[1/(k^2) -1] ------------answer.

--------------------------------
x^2 -3kx +1 = 0
x^2 -3kx +(-3k/2)^2 -(-3k/2)^2 +1 = 0
(x -3k/2)^2 -(9/4)k^2 +1 = 0
(x -3k/2)^2 = (9/4)k^2 -1
x -3k/2 = +,-sqrt[(9/4)k^2 -1]
x = 3k/2 +,-sqrt[(9/4)k^2 -1] ----------answer.