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Mathematics 9 Online
OpenStudy (anonymous):

solve the equation for x by completing the square x^2 = kx + 1

OpenStudy (anonymous):

Do you know how to complete the square?

OpenStudy (anonymous):

@wombat yeee

OpenStudy (anonymous):

So what is confusing?

OpenStudy (anonymous):

i did this x^2 = kx-1 =0 x^2 - kx = 1 x^2 - kx + k^2/x = 1 and I'm stuck there

OpenStudy (anonymous):

The problem is that you have added the term k^2/x spontaneously. In mathematics, it is important that you perform the same operation to both sides.

OpenStudy (anonymous):

i meant k^2/4 not k^2/x so it should be x^2 - kx + k^2/4 = 1 - k^2/4

OpenStudy (anonymous):

? *

OpenStudy (anonymous):

That looks to be correct.

Directrix (directrix):

x^2 - kx + k^2/4 = 1 - k^2/4 [x - (k/2)] ^2 = 4/4 - k^2/4 [x - (k/2)] ^2 = (4 - k^2)/4 What comes next?

OpenStudy (anonymous):

[x- (k/2)]^2 = (4/x - k^2/4) x- k/2 = √ k^2/2 + 4 x = (k + √k^2 +4)/2, x=(k-√k^2+4)/2

Directrix (directrix):

That's what I got. You may need another grouping symbol. x =[ (k + √(k^2 +4)] / 2 x =[ (k - √(k^2 +4)] / 2 That shows clearly what the radicand is.

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