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# Determine the value of k that will give the indicated solution.?

Using b^2 - 4ac (c being k) what is the value of k.

kx^2 - 2x + 1 =0 Two real distinct roots (b^2 - 4ac = a positive number)

x^2 + (k - 1)x + 1 = 0 Two real distinct roots

(k + 1)x^2 - 2x - 3 = 0 imaginary roots(or no roots;b^2 - 4ac = a negative number)

Update:

My teacher did it as k is the c.

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- railbuffLv 71 decade agoFavorite Answer
k is not c

kx^2 - 2x + 1 = 0

k = a

b^2 - 4ac

= (-2)^2 - (4)(k)(1)

= 4 - 4k

4-4k > 0

==> 4 > 4k

==> 1 > k

or k < 1

b^2 - 4ac

= (k - 1)^2 - (4)(1)(1)

= k^2 - 2k + 1 - 4

= k^2 - 2k - 3

k^2 - 2k - 3 > 0

Set k^2 - 2k - 3 = 0

(k - 3)(k + 1) = 0

==> k = 3 or k = -1

for k^2 - 2k - 3 > 0

for k = -2 true

for k = 0 false

for k = 4 true

so -1 < k < 3

b^2 - 4ac < 0

==> 2 -4(k + 1)(-3) < 0

==> 2 + 12k +1 < 0

==> 12k < -3

==> k < -1/4

Source(s): retired math teacher

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