- #1

autodidude

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## Homework Statement

Find the exact (surd) answer

## Homework Equations

-2kx^2 + 4x +6k = 0

## The Attempt at a Solution

Attempt 1 (quadratic formula method)

-2kx^2 + 4x +6k = 0

-> -(-2) +- sqrt((-2)^2 - (4 x k x -3k)0 / 2k

-> 2 +- sqrt(4+12k^2) / 2k

-> 2 +- sqrt(4(1+3k^2) / 2k

-> 2 +- sqrt(4) sqrt(1+3k^2) / 2k

-> 2 +- 2sqrt(1+3k^2) / 2k

-> x = sqrt(1 + 3k^2) / k

Given solutions are:

x = sqrt(3k^2 + 1 + 1) / k

and

x = -sqrt(3k^2 + 1 - 1) / k

Attempt 2 (completing the square method)

-2kx^2 + 4x + 6k = 0

-> kx^2 - 2x - 3k = 0

-> kx^2 - 2x + 1 -1 - 3k = 0

-> kx^2 - 2x + 1 = 1 + 3k

Not sure how to factor kx^2 - 2x + 1 but if I move k to the RHS, wouldn't the k's cancel out? I'm interested in solving it both ways as I'm trying to teach myself this and I don't want any holes in my knowledge

Thanks a lot