SOLUTION: For what value(s) of k will the equation kx^2+(k+1)x-1=0 have exactly one solution? My teacher has never taught us this. Please help!

Question 836283: For what value(s) of k will the equation kx^2+(k+1)x-1=0 have exactly one solution?
My teacher has never taught us this. Please help!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
ax^2 + bx + c = 0 will only have one solution if and only if the discriminant D is 0, so D = 0

D = b^2 - 4ac

0 = b^2 - 4ac

0 = (k+1)^2 - 4(k)(-1)

0 = (k+1)^2 + 4k

0 = k^2 + 2k + 1 + 4k

0 = k^2 + 6k + 1

k^2 + 6k + 1 = 0

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Now use the quadratic formula to solve for k

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for k:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=6, and c=1

Square 6 to get 36

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

So the solutions are:

or

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Final Answer:

The two solutions are or
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