Question 405308: For which value or values of k, if any, does the quadratic equation in x,
(k-8)x^2+kx -1 = 0 have exactly one solution?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! For which value or values of k, if any, does the quadratic equation in x,
(k-8)x^2+kx -1 = 0 have exactly one solution?
:
We know that when the discriminant = 0, we have a single value for the roots
b^2 - 4*a*c = 0
In this equation a = (k-8), b = k, c = -1
:
k^2 - (4*(k-8)*-1) = 0
:
k^2 - (-4(k-8))
:
k^2 - (-4k + 32)
Remove brackets, change the signs
k^2 + 4k - 32 = 0
factors to
(k+8)(k-4) = 0
two values for k
k = -8
k = +4
:
we can prove this
k=-8
(-8-8)x^2 - 8x - 1 = 0
-16x^2 - 8x - 1 = 0
Multiply by -1
16x^2 + 8x + 1 = 0
factors to
(4x + 1)(4x + 1) = 0
4x = -1
x =  a double root with one value
:
and for k = 4
(4-8)x^2 + 4x - 1 = 0
-4x^2 + 4x - 1 = 0
Multiply by -1
4x^2 - 4x + 1 = 0
factors to
(2x - 1)(2x - 1) = 0
2x = 1
x =  a double root with one value
:
:
We can say: k=-8 or k=4 will give one solution for the equation
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