Question

I can't find the two distinct and equal roots,I ended up with answer than can't be solved...
Can someone use questions below as examples..
This is the formula:b^2-4ac=0
1.Find the values of k for which (k+1)x^2+kx-2k=0 has two equal roots.
2.........3x^2-4x+5-k=0 has two distinct roots.

Answer 1

Just sub in your values:
b^2 - 4ac = 0
k^2 - 4(k+1)(-k) = 0
k^2 + 4k^2 + 4k = 0
5k^2 + 4k = 0
k(5k+4) = 0
k = 0 or k = -4/5

2 , this time b^2 - 4ac > 0
16 - 4(3)(-k) > 0
16 + 12k > 0
k > -16/12
k > -4/3

Answer 2

ok, you knew to use the discriminant. Too bad you didn't show your work. I hope you did

b^2 - 4ac = 0
k^2 - 4(-2k)(k+1) = 0
k(9k+8) = 0
It is clear that k=0 works
But we could have known that from the start:
y = (k+1)x^2+kx-2k
It's not quite so obvious that k = -8/9 also works. But to check it,
y = 1/9 x^2 - 8/9 x + 16/9
= 1/9 (x^2 - 8x + 16) = 1/9 (x-4)^2
which has two roots of x=4.
If k=0, y = x^2 which has two roots x=0

For the second one,
3x^2-4x+5-k=0
you need a positive discriminant.
16 - 4(3)(5-k) > 0
16 - 60 + 12k > 0
12k > 44
k > 11/3
Or, complete the square. You will need something of the form
y = a(x-h)^2 - n
3x^2-4x+5-k = 1/3 (3x-2)^2 + 11/3 - k
You can see that if k = 11/3 then y is a perfect square with two roots of x = 2/3
If k > 11/3, the graph dips below the x-axis, giving two distinct roots

Answer 3

Mr/Ms Reiny, the first questions c is 2k thought isn't it supposed to be
k^2 - 4(k+1)(-2k)

Answer 4

Ohh thanks

Answer 5

go with oobleck on the both,

I read c = -k instead of c = -2k
I read c = -k instead of c = 5-k

time to get new reading glasses.

Answer 6

How do you make the -4(k+1)(-2k) into (9k+8)?