Question

pls I need it like now
1. for What values of
the Constant k
does the equation
$(k - 1)x - k {x}^{2} - k$
have equal roots?

2. find the Positive value of "C" if the
expression
$(2c + 3)y}^{2} - 6y + 4 - c$
is a perfect
Square​

Answer 1

Answer:

Quadratic equation has equal roots

We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero. We know that two roots of quadratic equation are equal only if discriminant is equal to zero.

Step-by-step explanation:

144

So, c must be 144 to make the trinomial a perfect square.

Answer 2

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

There are two possible answers here: k = 0 or k = 4/5

======================================================

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.

Answer:   3