? asked in Science & MathematicsMathematics · 1 decade ago

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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  • 1 decade ago
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    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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