Problem 1
The discriminant formula is
d = b^2 - 4ac
from the original expression given to us, it is in the form ax^2+bx+c with
a = k-1
b = -k
c = -k
So we have a discriminant of
d = b^2 - 4ac
d = (-k)^2 - 4(k-1)(-k)
d = k^2 + 4k(k-1)
d = k^2 + 4k^2 - 4k
d = 5k^2 - 4k
Set this equal to 0 and solve for k. We set d equal to zero because a discriminant of 0 means we have two repeated roots.
d = 0
5k^2 - 4k = 0
k(5k - 4) = 0
k = 0 or 5k-4 = 0
k = 0 or 5k = 4
k = 0 or k = 4/5
<h3>There are two possible answers here: k = 0 or k = 4/5</h3>
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Problem 2
For this problem, I'll replace every c with k
Also, I'll replace every y with x
The expression turns into (2k+3)x^2-6x+4-k
We'll use the same idea as problem 1. Match it with ax^2+bx+c to find
a = 2k+3
b = -6
c = 4-k
the discriminant is
d = b^2 - 4ac
d = (-6)^2 - 4(2k+3)(4-k)
d = 36 - 4(-2k^2 + 5k + 12)
d = 36 + 8k^2 - 20k - 48
d = 8k^2 - 20k - 12
Set this equal to zero and solve for k
8k^2 - 20k - 12 = 0
4(2k^2 - 5k - 3) = 0
2k^2 - 5k - 3 = 0
2k^2 - 6k + k - 3 = 0
(2k^2-6k) + (k-3) = 0
2k(k-3) + 1(k-3) = 0
(2k+1)(k-3) = 0
2k+1 = 0 or k-3 = 0
2k = -1 or k = 3
k = -1/2 or k = 3
We ignore k = -1/2 as the instructions state the value of c (which I changed to k) is positive.
<h3>Answer: 3</h3>
