Question

Help !? Find the value of k for which the equation has a real double root:
3x2 - 6x - k = 0

Answer 1

Are you familiar with the quadratic equation?

Answer 2

yes but how do i use that if i dont know the value of k?

Answer 3

k is your c.

Answer 4

\[ax^2 + bx + c = 0 \implies x = {-(b) \pm \sqrt{(b)^2 - 4(a)(c)} \over 2(a)}\]

So what is a, b, and c?

Answer 5

From your problem I mean?

Answer 6

A is3

Answer 7

B is 6

Answer 8

is the answer 33

Answer 9

a = 3
b = -6
c = -k
\[x = {-(-6) \pm \sqrt{(-6)^2 - 4(3)(-k)}\over 2(3)}\]

So when will you get 2 values for x? When will you get 1?

Answer 10

I'll give you a hint. What happens when the part inside the square root equals 0?

Answer 11

doesnt it cancle out ?
is it -3

Answer 12

K= 2/3

Answer 13

It would be x = 6/6 = 1.

So that would have only one solution. So what does k need to be to make the part inside the square root 0?

Answer 14

thanks i got it now :)