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,Originally Posted by phgao
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
"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.Originally Posted by phgao
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 ------------***
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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,
x = -1 +,-sqrt(1-k) ----------answer.
----------------------
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.