Inactive Tutor answered 07/29/19
kx^2 - (1+k)x + (3k+2) = 0
kx^2 - x - kx + 3k + 2 = 0
kx^2 - kx + 3k - x + 2 = 0
kx^2 - kx + 3k = x -2
Factor out the k:
k(x^2 - x + 3) = x - 2
Divide both sides by (x^2 - x + 3)
k = (x - 2)/(x^2 - x + 3)
Amirou S.
asked 07/29/19for the equation kx^2 - (1+k)x + (3k+2) = 0, how do i find the value of k ?
Inactive Tutor answered 07/29/19
kx^2 - (1+k)x + (3k+2) = 0
kx^2 - x - kx + 3k + 2 = 0
kx^2 - kx + 3k - x + 2 = 0
kx^2 - kx + 3k = x -2
Factor out the k:
k(x^2 - x + 3) = x - 2
Divide both sides by (x^2 - x + 3)
k = (x - 2)/(x^2 - x + 3)
If you multiply the expression, you get
kx2 -x -kx + 3k + 2 = 0 Then k(x2 -x + 3) = x - 2
So, k = (x-2)/(x2-x+3)
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Amirou S.
I might've not been specific enough, so that's my bad on my part. But I'm using the HL Core Mathmatics Haese Pearsons text book, and I'm practicing on the Quadratic chapter under the section of "The Sum And product of The Roots". In this particular question it ask as I quote "The equation kx^2 - (1+k)x + (3k+2) = 0 is such that the sum of its root is twice their product. Find k and the two roots. " So I think that somehow the variable 'k' has an actual numerical value but I don't understand how to solve that?07/30/19