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?
"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.
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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.
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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.