maths question, higher maths help!?

Question

find the value of k which results in the equation

kx^2 + 2kx - 1 = 0

having equal roots, given that k does not = 0


i did b^2 - 4ac = 0

therefore 2k^2 - 4k + 4 = 0

but i cant factorise this :s is this correct? what now if so :s

do i use -b +/b^2-4ac / 2a?

Answer 1

write equation as x^2 + 2x -1/k = 0

Then roots are a and a. So (x-a)^2 = above so x^2 -2ax +a^2 =above So a = -1 which gives a value for k= -1.

x^2 +2x +1 is equation

Two factors are x +1, x+1.

x= -1

Answer 2

Div by k> x^2+2x-1/k=0
by inspection x^2+2x+1=0> (x+1)^2=0> equal roots (x=-1)
>k=-1 equating coeff

Answer 3

The roots of the 2. degree eqatation is:
x= (2k +-root of((-2k)^2-4k))/2k

X= 1+-((root of(4k^2-4k))k
=1+-root of(4k-1)

Then k=1/4 gives both roots x=1

Sorry for bad mathematical presentation. Do not know all possibilities of symbols in this dokument.

Answer 4

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Answer 5

i think k = - 1

Answer 6

love to help but sadly i can't =[


Thats to hard for me what does ^ mean anyway????