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Mathematics 8 Online
OpenStudy (anonymous):

equation has two equal roots, find the values of k... x^2-6x+3=k(2^x-1)

OpenStudy (bahrom7893):

If equation has to equal roots, the discriminant must be 0.

OpenStudy (bahrom7893):

x^2 - 6x + 3 = kx^2-k

OpenStudy (bahrom7893):

by the way is that a typo? 2^x?

OpenStudy (anonymous):

typo- 2x ..not 2^x~

OpenStudy (bahrom7893):

oh ok so: x^2-6x+3=2kx-k x^2 - 6x - 2kx + 3+ k = 0

OpenStudy (bahrom7893):

or grouping the terms or whatever u call that, x^2 + (-6-2k)x + (3+k) = 0 Now for the equation to have two equal roots, the discriminant = b^2 - 4ac = 0.

OpenStudy (bahrom7893):

b = -6-2k a=1 c = 3+k (-6-2k)^2 - 4 * 1 * (3+k) = (-6-2k)^2 - 12 - 4k = 0

OpenStudy (bahrom7893):

((-1)*(6+2k))^2 - 12 - 4k = 0 (6+2k)^2 - 12 - 4k = 0 36 + 24k + 4k^2 - 12 - 4k = 0

OpenStudy (bahrom7893):

4k^2 + 20k + 24 = 0 k^2 + 5k + 6 = 0

OpenStudy (bahrom7893):

factoring that gets you: (k+2)(k+3) = 0, so: k = -2, or k = -3 will get u two equal roots

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