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!

Algebra ->  Quadratic Equations and Parabolas -> 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!      Log On


   

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) About Me  (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


ak%5E2%2Bbk%2Bc=0


the general solution using the quadratic equation is:


k+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve k%5E2%2B6%2Ak%2B1=0 ( notice a=1, b=6, and c=1)





k+=+%28-6+%2B-+sqrt%28+%286%29%5E2-4%2A1%2A1+%29%29%2F%282%2A1%29 Plug in a=1, b=6, and c=1




k+=+%28-6+%2B-+sqrt%28+36-4%2A1%2A1+%29%29%2F%282%2A1%29 Square 6 to get 36




k+=+%28-6+%2B-+sqrt%28+36%2B-4+%29%29%2F%282%2A1%29 Multiply -4%2A1%2A1 to get -4




k+=+%28-6+%2B-+sqrt%28+32+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




k+=+%28-6+%2B-+4%2Asqrt%282%29%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




k+=+%28-6+%2B-+4%2Asqrt%282%29%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


k+=+%28-6+%2B+4%2Asqrt%282%29%29%2F2 or k+=+%28-6+-+4%2Asqrt%282%29%29%2F2



Now break up the fraction



k=-6%2F2%2B4%2Asqrt%282%29%2F2 or k=-6%2F2-4%2Asqrt%282%29%2F2



Simplify



k=-3%2B2%2Asqrt%282%29 or k=-3-2%2Asqrt%282%29



So the solutions are:

k=-3%2B2%2Asqrt%282%29 or k=-3-2%2Asqrt%282%29





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Final Answer:

The two solutions are k+=+-3%2B2%2Asqrt%282%29 or k+=+-3-2%2Asqrt%282%29